Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

Question

Want to see more full solutions like this?

Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 13, Problem 5CRE

a.

To determine

Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

a.

Expert Solution

Answer to Problem 5CRE

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089).

Explanation of Solution

It is given that out of 1,003 adults, 68% thought a landline phone was a necessity.

Here, the sample size, $n$ is 1,003 and the proportion of adults thinks a landline phone is a necessity, $p^$ is 0.68 (= 68%).

Conditions for 95% confidence interval:

1. Values of $np^$ and $n(1−p^)$ :

$np^=1,003(0.68)=682.04n(1−p^)=1,003(1−0.68)=320.96$

Since $np^$ and $n(1−p^)$ are greater than 10, the sample size is large enough to proceed.

2. The sample size $n=1,003$ is much smaller than 10% of the number of adults (population size).

3. Since it is given that the survey is nationally representative, it is reasonable to assume the sample as a random sample from the population of adults.

Calculation:

The 95% confidence interval for p is given below.

$p^±1.96p^(1−p^)n$

Substitute the values of $n$ and $p^$ in confidence interval formula as shown below.

$p^±1.96p^(1−p^)n=0.68±1.960.68(1−0.68)1,003=0.68±(1.96×0.0147)=0.68±0.0289=(0.68−0.0289,0.68+0.0289)=(0.6511,0.7089)$

Interpretation:

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089). Therefore, one can be 95% confident that the proportion of all adults who view a landline phone as a necessity is between 0.6511 and 0.7089.

b.

To determine

Check whether there is any convincing evidence that a majority of adults view a television set as a necessity at 0.05 significance level.

b.

Expert Solution

Answer to Problem 5CRE

There is no convincing evidence that a majority of adults consider a TV set as necessity.

Explanation of Solution

In order to test majority of adults view a television set as a necessity, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p$ denotes the proportion of all adults who consider a TV set as necessity.

2. Null hypothesis:

$H0:p=0.5$

That is, the proportion of adults who view a television set as a necessity is 0.5.

3. Alternative hypothesis:

$Ha:p>0.5$

That is, the proportion of adults who view a television set as a necessity is greater than 0.5.

That is, majority of adults consider a TV set as necessity.

4. Significance level:

$α=0.05$

5. Test statistic:

$z=p^−pp(1−p)n=p^−0.50.5(1−0.5)1,003$

6. Assumptions:

Here, $np^=501.5$ (1,003 × 0.5) and $n(1−p^)=501.5$ (1,003 × (1−0.5)). Since, both values are greater than 10, the sample is large enough to proceed.
The sample is nationally representative. Hence, the sample can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 52% (= 0.52) said they viewed a television set as a necessity. Therefore, $p^=0.52$ . Substitute the values in the test statistic formula as shown below.

$z=0.52−0.50.5(1−0.5)1,003=0.020.01579=1.267$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose Right tail and give X value as 1.267.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 1

From the MINITAB output, the P-value is 0.1026.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀.

Here the significance level is 0.05.

Hence, $0.103>0.05$ .

Since P-value is greater than 0.05, fail to reject the null hypothesis. Therefore, there is no convincing evidence that a majority of adults consider a TV set as necessity.

c.

To determine

Check whether there is any convincing evidence that the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009 at 0.01 significance level.

c.

Expert Solution

Answer to Problem 5CRE

There is a convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Explanation of Solution

In order to test the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p1$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2003 and $p2$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2009.

2. Null hypothesis:

$H0:p1−p2=0$

That is, the proportions of adults who regarded a microwave oven as a necessity in 2003 and 2009 are equal.

3. Alternative hypothesis:

$Ha:p1−p2>0$

That is, the proportion of adults who regarded a microwave oven as a necessity in 2003 is greater than the proportion of adults who regarded a microwave oven as a necessity in 2009.

4. Significance level:

$α=0.01$

5. Test statistic:

$z=p^1−p^2p^c(1−p^c)n1+p^c(1−p^c)n2$

Where, $p^c$ is the combined estimate of common population proportion.

6. Assumptions:

Here, $n1$ is 1,003, $n2$ is 1,003, $p^1$ is 0.68, and $p^2$ is 0.47

$n1p^1=1,003(0.68)=682.04n1(1−p^1)=1,003(1−0.68)=320.96$

$n2p^2=1,003(0.47)=471.41n2(1−p^2)=1,003(1−0.47)=531.59$

Since the values of $n1p^1$ , $n1(1−p^1)$ , $n2p^2$ , and $n2(1−p^2)$ are greater than 10, the samples are large enough to proceed.
Also, the sample is nationally representative. Hence, the samples can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 68% of the 2003 sample regarded a microwave oven as a necessity and 47% of the 2009 sample regarded a microwave oven as a necessity.

$p^c=n1p^1+n2p^2n1+n2=1,003(0.68)+1,003(0.47)1,003+1,003=1,153.452,006=0.575z=0.68−0.470.575(1−0.575)1,003+0.575(1−0.575)1,003=0.210.000244+0.000244=0.210.000487=0.210.022075=9.5132$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose left tail and give X value as 9.5132.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 2

From the MINITAB output, the P-value is 0.000.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀. Otherwise, fail to reject $H0$ .

Here, the significance level is 0.01.

Here, the P-value is less than the level of significance.

Hence, $0.000<0.01$ .

Since P-value is less than 0.01, reject the null hypothesis. Therefore, there is convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Questions An insurance company's cumulative incurred claims for the last 5 accident years are given in the following table: Development Year Accident Year 0 2018 1 2 3 4 245 267 274 289 292 2019 255 276 288 294 2020 265 283 292 2021 263 278 2022 271 It can be assumed that claims are fully run off after 4 years. The premiums received for each year are: Accident Year Premium 2018 306 2019 312 2020 318 2021 326 2022 330 You do not need to make any allowance for inflation. 1. (a) Calculate the reserve at the end of 2022 using the basic chain ladder method. (b) Calculate the reserve at the end of 2022 using the Bornhuetter-Ferguson method. 2. Comment on the differences in the reserves produced by the methods in Part 1.

From a sample of 26 graduate students, the mean number of months of work experience prior to entering an MBA program was 34.67. The national standard deviation is known to be18 months. What is a 90% confidence interval for the population mean? Question content area bottom Part 1 A 9090% confidence interval for the population mean is left bracket nothing comma nothing right bracketenter your response here,enter your response here. (Use ascending order. Round to two decimal places as needed.)

A test consists of 10 questions made of 5 answers with only one correct answer. To pass the test, a student must answer at least 8 questions correctly. (a) If a student guesses on each question, what is the probability that the student passes the test? (b) Find the mean and standard deviation of the number of correct answers. (c) Is it unusual for a student to pass the test by guessing? Explain.

Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 13, Problem 5CRE

a.

To determine

Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

a.

Expert Solution

Answer to Problem 5CRE

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089).

Explanation of Solution

It is given that out of 1,003 adults, 68% thought a landline phone was a necessity.

Here, the sample size, $n$ is 1,003 and the proportion of adults thinks a landline phone is a necessity, $p^$ is 0.68 (= 68%).

Conditions for 95% confidence interval:

1. Values of $np^$ and $n(1−p^)$ :

$np^=1,003(0.68)=682.04n(1−p^)=1,003(1−0.68)=320.96$

Since $np^$ and $n(1−p^)$ are greater than 10, the sample size is large enough to proceed.

2. The sample size $n=1,003$ is much smaller than 10% of the number of adults (population size).

3. Since it is given that the survey is nationally representative, it is reasonable to assume the sample as a random sample from the population of adults.

Calculation:

The 95% confidence interval for p is given below.

$p^±1.96p^(1−p^)n$

Substitute the values of $n$ and $p^$ in confidence interval formula as shown below.

$p^±1.96p^(1−p^)n=0.68±1.960.68(1−0.68)1,003=0.68±(1.96×0.0147)=0.68±0.0289=(0.68−0.0289,0.68+0.0289)=(0.6511,0.7089)$

Interpretation:

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089). Therefore, one can be 95% confident that the proportion of all adults who view a landline phone as a necessity is between 0.6511 and 0.7089.

b.

To determine

Check whether there is any convincing evidence that a majority of adults view a television set as a necessity at 0.05 significance level.

b.

Expert Solution

Answer to Problem 5CRE

There is no convincing evidence that a majority of adults consider a TV set as necessity.

Explanation of Solution

In order to test majority of adults view a television set as a necessity, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p$ denotes the proportion of all adults who consider a TV set as necessity.

2. Null hypothesis:

$H0:p=0.5$

That is, the proportion of adults who view a television set as a necessity is 0.5.

3. Alternative hypothesis:

$Ha:p>0.5$

That is, the proportion of adults who view a television set as a necessity is greater than 0.5.

That is, majority of adults consider a TV set as necessity.

4. Significance level:

$α=0.05$

5. Test statistic:

$z=p^−pp(1−p)n=p^−0.50.5(1−0.5)1,003$

6. Assumptions:

Here, $np^=501.5$ (1,003 × 0.5) and $n(1−p^)=501.5$ (1,003 × (1−0.5)). Since, both values are greater than 10, the sample is large enough to proceed.
The sample is nationally representative. Hence, the sample can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 52% (= 0.52) said they viewed a television set as a necessity. Therefore, $p^=0.52$ . Substitute the values in the test statistic formula as shown below.

$z=0.52−0.50.5(1−0.5)1,003=0.020.01579=1.267$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose Right tail and give X value as 1.267.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 1

From the MINITAB output, the P-value is 0.1026.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀.

Here the significance level is 0.05.

Hence, $0.103>0.05$ .

Since P-value is greater than 0.05, fail to reject the null hypothesis. Therefore, there is no convincing evidence that a majority of adults consider a TV set as necessity.

c.

To determine

Check whether there is any convincing evidence that the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009 at 0.01 significance level.

c.

Expert Solution

Answer to Problem 5CRE

There is a convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Explanation of Solution

In order to test the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p1$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2003 and $p2$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2009.

2. Null hypothesis:

$H0:p1−p2=0$

That is, the proportions of adults who regarded a microwave oven as a necessity in 2003 and 2009 are equal.

3. Alternative hypothesis:

$Ha:p1−p2>0$

That is, the proportion of adults who regarded a microwave oven as a necessity in 2003 is greater than the proportion of adults who regarded a microwave oven as a necessity in 2009.

4. Significance level:

$α=0.01$

5. Test statistic:

$z=p^1−p^2p^c(1−p^c)n1+p^c(1−p^c)n2$

Where, $p^c$ is the combined estimate of common population proportion.

6. Assumptions:

Here, $n1$ is 1,003, $n2$ is 1,003, $p^1$ is 0.68, and $p^2$ is 0.47

$n1p^1=1,003(0.68)=682.04n1(1−p^1)=1,003(1−0.68)=320.96$

$n2p^2=1,003(0.47)=471.41n2(1−p^2)=1,003(1−0.47)=531.59$

Since the values of $n1p^1$ , $n1(1−p^1)$ , $n2p^2$ , and $n2(1−p^2)$ are greater than 10, the samples are large enough to proceed.
Also, the sample is nationally representative. Hence, the samples can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 68% of the 2003 sample regarded a microwave oven as a necessity and 47% of the 2009 sample regarded a microwave oven as a necessity.

$p^c=n1p^1+n2p^2n1+n2=1,003(0.68)+1,003(0.47)1,003+1,003=1,153.452,006=0.575z=0.68−0.470.575(1−0.575)1,003+0.575(1−0.575)1,003=0.210.000244+0.000244=0.210.000487=0.210.022075=9.5132$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose left tail and give X value as 9.5132.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 2

From the MINITAB output, the P-value is 0.000.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀. Otherwise, fail to reject $H0$ .

Here, the significance level is 0.01.

Here, the P-value is less than the level of significance.

Hence, $0.000<0.01$ .

Since P-value is less than 0.01, reject the null hypothesis. Therefore, there is convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 13, Problem 5CRE

a.

To determine

Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

a.

Expert Solution

Answer to Problem 5CRE

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089).

Explanation of Solution

It is given that out of 1,003 adults, 68% thought a landline phone was a necessity.

Here, the sample size, $n$ is 1,003 and the proportion of adults thinks a landline phone is a necessity, $p^$ is 0.68 (= 68%).

Conditions for 95% confidence interval:

1. Values of $np^$ and $n(1−p^)$ :

$np^=1,003(0.68)=682.04n(1−p^)=1,003(1−0.68)=320.96$

Since $np^$ and $n(1−p^)$ are greater than 10, the sample size is large enough to proceed.

2. The sample size $n=1,003$ is much smaller than 10% of the number of adults (population size).

3. Since it is given that the survey is nationally representative, it is reasonable to assume the sample as a random sample from the population of adults.

Calculation:

The 95% confidence interval for p is given below.

$p^±1.96p^(1−p^)n$

Substitute the values of $n$ and $p^$ in confidence interval formula as shown below.

$p^±1.96p^(1−p^)n=0.68±1.960.68(1−0.68)1,003=0.68±(1.96×0.0147)=0.68±0.0289=(0.68−0.0289,0.68+0.0289)=(0.6511,0.7089)$

Interpretation:

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089). Therefore, one can be 95% confident that the proportion of all adults who view a landline phone as a necessity is between 0.6511 and 0.7089.

b.

To determine

Check whether there is any convincing evidence that a majority of adults view a television set as a necessity at 0.05 significance level.

b.

Expert Solution

Answer to Problem 5CRE

There is no convincing evidence that a majority of adults consider a TV set as necessity.

Explanation of Solution

In order to test majority of adults view a television set as a necessity, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p$ denotes the proportion of all adults who consider a TV set as necessity.

2. Null hypothesis:

$H0:p=0.5$

That is, the proportion of adults who view a television set as a necessity is 0.5.

3. Alternative hypothesis:

$Ha:p>0.5$

That is, the proportion of adults who view a television set as a necessity is greater than 0.5.

That is, majority of adults consider a TV set as necessity.

4. Significance level:

$α=0.05$

5. Test statistic:

$z=p^−pp(1−p)n=p^−0.50.5(1−0.5)1,003$

6. Assumptions:

Here, $np^=501.5$ (1,003 × 0.5) and $n(1−p^)=501.5$ (1,003 × (1−0.5)). Since, both values are greater than 10, the sample is large enough to proceed.
The sample is nationally representative. Hence, the sample can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 52% (= 0.52) said they viewed a television set as a necessity. Therefore, $p^=0.52$ . Substitute the values in the test statistic formula as shown below.

$z=0.52−0.50.5(1−0.5)1,003=0.020.01579=1.267$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose Right tail and give X value as 1.267.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 1

From the MINITAB output, the P-value is 0.1026.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀.

Here the significance level is 0.05.

Hence, $0.103>0.05$ .

Since P-value is greater than 0.05, fail to reject the null hypothesis. Therefore, there is no convincing evidence that a majority of adults consider a TV set as necessity.

c.

To determine

Check whether there is any convincing evidence that the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009 at 0.01 significance level.

c.

Expert Solution

Answer to Problem 5CRE

There is a convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Explanation of Solution

In order to test the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p1$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2003 and $p2$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2009.

2. Null hypothesis:

$H0:p1−p2=0$

That is, the proportions of adults who regarded a microwave oven as a necessity in 2003 and 2009 are equal.

3. Alternative hypothesis:

$Ha:p1−p2>0$

That is, the proportion of adults who regarded a microwave oven as a necessity in 2003 is greater than the proportion of adults who regarded a microwave oven as a necessity in 2009.

4. Significance level:

$α=0.01$

5. Test statistic:

$z=p^1−p^2p^c(1−p^c)n1+p^c(1−p^c)n2$

Where, $p^c$ is the combined estimate of common population proportion.

6. Assumptions:

Here, $n1$ is 1,003, $n2$ is 1,003, $p^1$ is 0.68, and $p^2$ is 0.47

$n1p^1=1,003(0.68)=682.04n1(1−p^1)=1,003(1−0.68)=320.96$

$n2p^2=1,003(0.47)=471.41n2(1−p^2)=1,003(1−0.47)=531.59$

Since the values of $n1p^1$ , $n1(1−p^1)$ , $n2p^2$ , and $n2(1−p^2)$ are greater than 10, the samples are large enough to proceed.
Also, the sample is nationally representative. Hence, the samples can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 68% of the 2003 sample regarded a microwave oven as a necessity and 47% of the 2009 sample regarded a microwave oven as a necessity.

$p^c=n1p^1+n2p^2n1+n2=1,003(0.68)+1,003(0.47)1,003+1,003=1,153.452,006=0.575z=0.68−0.470.575(1−0.575)1,003+0.575(1−0.575)1,003=0.210.000244+0.000244=0.210.000487=0.210.022075=9.5132$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose left tail and give X value as 9.5132.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 2

From the MINITAB output, the P-value is 0.000.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀. Otherwise, fail to reject $H0$ .

Here, the significance level is 0.01.

Here, the P-value is less than the level of significance.

Hence, $0.000<0.01$ .

Since P-value is less than 0.01, reject the null hypothesis. Therefore, there is convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 13, Problem 5CRE

a.

To determine

Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

a.

Expert Solution

Answer to Problem 5CRE

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089).

Explanation of Solution

It is given that out of 1,003 adults, 68% thought a landline phone was a necessity.

Here, the sample size, $n$ is 1,003 and the proportion of adults thinks a landline phone is a necessity, $p^$ is 0.68 (= 68%).

Conditions for 95% confidence interval:

1. Values of $np^$ and $n(1−p^)$ :

$np^=1,003(0.68)=682.04n(1−p^)=1,003(1−0.68)=320.96$

Since $np^$ and $n(1−p^)$ are greater than 10, the sample size is large enough to proceed.

2. The sample size $n=1,003$ is much smaller than 10% of the number of adults (population size).

3. Since it is given that the survey is nationally representative, it is reasonable to assume the sample as a random sample from the population of adults.

Calculation:

The 95% confidence interval for p is given below.

$p^±1.96p^(1−p^)n$

Substitute the values of $n$ and $p^$ in confidence interval formula as shown below.

$p^±1.96p^(1−p^)n=0.68±1.960.68(1−0.68)1,003=0.68±(1.96×0.0147)=0.68±0.0289=(0.68−0.0289,0.68+0.0289)=(0.6511,0.7089)$

Interpretation:

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089). Therefore, one can be 95% confident that the proportion of all adults who view a landline phone as a necessity is between 0.6511 and 0.7089.

b.

To determine

Check whether there is any convincing evidence that a majority of adults view a television set as a necessity at 0.05 significance level.

b.

Expert Solution

Answer to Problem 5CRE

There is no convincing evidence that a majority of adults consider a TV set as necessity.

Explanation of Solution

In order to test majority of adults view a television set as a necessity, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p$ denotes the proportion of all adults who consider a TV set as necessity.

2. Null hypothesis:

$H0:p=0.5$

That is, the proportion of adults who view a television set as a necessity is 0.5.

3. Alternative hypothesis:

$Ha:p>0.5$

That is, the proportion of adults who view a television set as a necessity is greater than 0.5.

That is, majority of adults consider a TV set as necessity.

4. Significance level:

$α=0.05$

5. Test statistic:

$z=p^−pp(1−p)n=p^−0.50.5(1−0.5)1,003$

6. Assumptions:

Here, $np^=501.5$ (1,003 × 0.5) and $n(1−p^)=501.5$ (1,003 × (1−0.5)). Since, both values are greater than 10, the sample is large enough to proceed.
The sample is nationally representative. Hence, the sample can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 52% (= 0.52) said they viewed a television set as a necessity. Therefore, $p^=0.52$ . Substitute the values in the test statistic formula as shown below.

$z=0.52−0.50.5(1−0.5)1,003=0.020.01579=1.267$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose Right tail and give X value as 1.267.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 1

From the MINITAB output, the P-value is 0.1026.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀.

Here the significance level is 0.05.

Hence, $0.103>0.05$ .

Since P-value is greater than 0.05, fail to reject the null hypothesis. Therefore, there is no convincing evidence that a majority of adults consider a TV set as necessity.

c.

To determine

Check whether there is any convincing evidence that the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009 at 0.01 significance level.

c.

Expert Solution

Answer to Problem 5CRE

There is a convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Explanation of Solution

In order to test the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p1$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2003 and $p2$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2009.

2. Null hypothesis:

$H0:p1−p2=0$

That is, the proportions of adults who regarded a microwave oven as a necessity in 2003 and 2009 are equal.

3. Alternative hypothesis:

$Ha:p1−p2>0$

That is, the proportion of adults who regarded a microwave oven as a necessity in 2003 is greater than the proportion of adults who regarded a microwave oven as a necessity in 2009.

4. Significance level:

$α=0.01$

5. Test statistic:

$z=p^1−p^2p^c(1−p^c)n1+p^c(1−p^c)n2$

Where, $p^c$ is the combined estimate of common population proportion.

6. Assumptions:

Here, $n1$ is 1,003, $n2$ is 1,003, $p^1$ is 0.68, and $p^2$ is 0.47

$n1p^1=1,003(0.68)=682.04n1(1−p^1)=1,003(1−0.68)=320.96$

$n2p^2=1,003(0.47)=471.41n2(1−p^2)=1,003(1−0.47)=531.59$

Since the values of $n1p^1$ , $n1(1−p^1)$ , $n2p^2$ , and $n2(1−p^2)$ are greater than 10, the samples are large enough to proceed.
Also, the sample is nationally representative. Hence, the samples can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 68% of the 2003 sample regarded a microwave oven as a necessity and 47% of the 2009 sample regarded a microwave oven as a necessity.

$p^c=n1p^1+n2p^2n1+n2=1,003(0.68)+1,003(0.47)1,003+1,003=1,153.452,006=0.575z=0.68−0.470.575(1−0.575)1,003+0.575(1−0.575)1,003=0.210.000244+0.000244=0.210.000487=0.210.022075=9.5132$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose left tail and give X value as 9.5132.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 2

From the MINITAB output, the P-value is 0.000.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀. Otherwise, fail to reject $H0$ .

Here, the significance level is 0.01.

Here, the P-value is less than the level of significance.

Hence, $0.000<0.01$ .

Since P-value is less than 0.01, reject the null hypothesis. Therefore, there is convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 13, Problem 5CRE

a.

To determine

Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

a.

Expert Solution

Answer to Problem 5CRE

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089).

Explanation of Solution

It is given that out of 1,003 adults, 68% thought a landline phone was a necessity.

Here, the sample size, $n$ is 1,003 and the proportion of adults thinks a landline phone is a necessity, $p^$ is 0.68 (= 68%).

Conditions for 95% confidence interval:

1. Values of $np^$ and $n(1−p^)$ :

$np^=1,003(0.68)=682.04n(1−p^)=1,003(1−0.68)=320.96$

Since $np^$ and $n(1−p^)$ are greater than 10, the sample size is large enough to proceed.

2. The sample size $n=1,003$ is much smaller than 10% of the number of adults (population size).

3. Since it is given that the survey is nationally representative, it is reasonable to assume the sample as a random sample from the population of adults.

Calculation:

The 95% confidence interval for p is given below.

$p^±1.96p^(1−p^)n$

Substitute the values of $n$ and $p^$ in confidence interval formula as shown below.

$p^±1.96p^(1−p^)n=0.68±1.960.68(1−0.68)1,003=0.68±(1.96×0.0147)=0.68±0.0289=(0.68−0.0289,0.68+0.0289)=(0.6511,0.7089)$

Interpretation:

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089). Therefore, one can be 95% confident that the proportion of all adults who view a landline phone as a necessity is between 0.6511 and 0.7089.

b.

To determine

Check whether there is any convincing evidence that a majority of adults view a television set as a necessity at 0.05 significance level.

b.

Expert Solution

Answer to Problem 5CRE

There is no convincing evidence that a majority of adults consider a TV set as necessity.

Explanation of Solution

In order to test majority of adults view a television set as a necessity, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p$ denotes the proportion of all adults who consider a TV set as necessity.

2. Null hypothesis:

$H0:p=0.5$

That is, the proportion of adults who view a television set as a necessity is 0.5.

3. Alternative hypothesis:

$Ha:p>0.5$

That is, the proportion of adults who view a television set as a necessity is greater than 0.5.

That is, majority of adults consider a TV set as necessity.

4. Significance level:

$α=0.05$

5. Test statistic:

$z=p^−pp(1−p)n=p^−0.50.5(1−0.5)1,003$

6. Assumptions:

Here, $np^=501.5$ (1,003 × 0.5) and $n(1−p^)=501.5$ (1,003 × (1−0.5)). Since, both values are greater than 10, the sample is large enough to proceed.
The sample is nationally representative. Hence, the sample can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 52% (= 0.52) said they viewed a television set as a necessity. Therefore, $p^=0.52$ . Substitute the values in the test statistic formula as shown below.

$z=0.52−0.50.5(1−0.5)1,003=0.020.01579=1.267$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose Right tail and give X value as 1.267.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 1

From the MINITAB output, the P-value is 0.1026.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀.

Here the significance level is 0.05.

Hence, $0.103>0.05$ .

Since P-value is greater than 0.05, fail to reject the null hypothesis. Therefore, there is no convincing evidence that a majority of adults consider a TV set as necessity.

c.

To determine

Check whether there is any convincing evidence that the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009 at 0.01 significance level.

c.

Expert Solution

Answer to Problem 5CRE

There is a convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Explanation of Solution

In order to test the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p1$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2003 and $p2$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2009.

2. Null hypothesis:

$H0:p1−p2=0$

That is, the proportions of adults who regarded a microwave oven as a necessity in 2003 and 2009 are equal.

3. Alternative hypothesis:

$Ha:p1−p2>0$

That is, the proportion of adults who regarded a microwave oven as a necessity in 2003 is greater than the proportion of adults who regarded a microwave oven as a necessity in 2009.

4. Significance level:

$α=0.01$

5. Test statistic:

$z=p^1−p^2p^c(1−p^c)n1+p^c(1−p^c)n2$

Where, $p^c$ is the combined estimate of common population proportion.

6. Assumptions:

Here, $n1$ is 1,003, $n2$ is 1,003, $p^1$ is 0.68, and $p^2$ is 0.47

$n1p^1=1,003(0.68)=682.04n1(1−p^1)=1,003(1−0.68)=320.96$

$n2p^2=1,003(0.47)=471.41n2(1−p^2)=1,003(1−0.47)=531.59$

Since the values of $n1p^1$ , $n1(1−p^1)$ , $n2p^2$ , and $n2(1−p^2)$ are greater than 10, the samples are large enough to proceed.
Also, the sample is nationally representative. Hence, the samples can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 68% of the 2003 sample regarded a microwave oven as a necessity and 47% of the 2009 sample regarded a microwave oven as a necessity.

$p^c=n1p^1+n2p^2n1+n2=1,003(0.68)+1,003(0.47)1,003+1,003=1,153.452,006=0.575z=0.68−0.470.575(1−0.575)1,003+0.575(1−0.575)1,003=0.210.000244+0.000244=0.210.000487=0.210.022075=9.5132$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose left tail and give X value as 9.5132.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 2

From the MINITAB output, the P-value is 0.000.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀. Otherwise, fail to reject $H0$ .

Here, the significance level is 0.01.

Here, the P-value is less than the level of significance.

Hence, $0.000<0.01$ .

Since P-value is less than 0.01, reject the null hypothesis. Therefore, there is convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Answer 6

Chapter 13, Problem 5CRE

a.

To determine

Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

a.

Expert Solution

Answer to Problem 5CRE

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089).

Explanation of Solution

It is given that out of 1,003 adults, 68% thought a landline phone was a necessity.

Here, the sample size, $n$ is 1,003 and the proportion of adults thinks a landline phone is a necessity, $p^$ is 0.68 (= 68%).

Conditions for 95% confidence interval:

1. Values of $np^$ and $n(1−p^)$ :

$np^=1,003(0.68)=682.04n(1−p^)=1,003(1−0.68)=320.96$

Since $np^$ and $n(1−p^)$ are greater than 10, the sample size is large enough to proceed.

2. The sample size $n=1,003$ is much smaller than 10% of the number of adults (population size).

3. Since it is given that the survey is nationally representative, it is reasonable to assume the sample as a random sample from the population of adults.

Calculation:

The 95% confidence interval for p is given below.

$p^±1.96p^(1−p^)n$

Substitute the values of $n$ and $p^$ in confidence interval formula as shown below.

$p^±1.96p^(1−p^)n=0.68±1.960.68(1−0.68)1,003=0.68±(1.96×0.0147)=0.68±0.0289=(0.68−0.0289,0.68+0.0289)=(0.6511,0.7089)$

Interpretation:

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089). Therefore, one can be 95% confident that the proportion of all adults who view a landline phone as a necessity is between 0.6511 and 0.7089.

Answer 7

Chapter 13, Problem 5CRE

a.

To determine

Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

Answer 8

a.

Expert Solution

Answer to Problem 5CRE

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089).

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

It is given that out of 1,003 adults, 68% thought a landline phone was a necessity.

Here, the sample size, $n$ is 1,003 and the proportion of adults thinks a landline phone is a necessity, $p^$ is 0.68 (= 68%).

Conditions for 95% confidence interval:

1. Values of $np^$ and $n(1−p^)$ :

$np^=1,003(0.68)=682.04n(1−p^)=1,003(1−0.68)=320.96$

Since $np^$ and $n(1−p^)$ are greater than 10, the sample size is large enough to proceed.

2. The sample size $n=1,003$ is much smaller than 10% of the number of adults (population size).

3. Since it is given that the survey is nationally representative, it is reasonable to assume the sample as a random sample from the population of adults.

Calculation:

The 95% confidence interval for p is given below.

$p^±1.96p^(1−p^)n$

Substitute the values of $n$ and $p^$ in confidence interval formula as shown below.

$p^±1.96p^(1−p^)n=0.68±1.960.68(1−0.68)1,003=0.68±(1.96×0.0147)=0.68±0.0289=(0.68−0.0289,0.68+0.0289)=(0.6511,0.7089)$

Interpretation:

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089). Therefore, one can be 95% confident that the proportion of all adults who view a landline phone as a necessity is between 0.6511 and 0.7089.

Answer 13

b.

To determine

Check whether there is any convincing evidence that a majority of adults view a television set as a necessity at 0.05 significance level.

b.

Expert Solution

Answer to Problem 5CRE

There is no convincing evidence that a majority of adults consider a TV set as necessity.

Explanation of Solution

In order to test majority of adults view a television set as a necessity, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p$ denotes the proportion of all adults who consider a TV set as necessity.

2. Null hypothesis:

$H0:p=0.5$

That is, the proportion of adults who view a television set as a necessity is 0.5.

3. Alternative hypothesis:

$Ha:p>0.5$

That is, the proportion of adults who view a television set as a necessity is greater than 0.5.

That is, majority of adults consider a TV set as necessity.

4. Significance level:

$α=0.05$

5. Test statistic:

$z=p^−pp(1−p)n=p^−0.50.5(1−0.5)1,003$

6. Assumptions:

Here, $np^=501.5$ (1,003 × 0.5) and $n(1−p^)=501.5$ (1,003 × (1−0.5)). Since, both values are greater than 10, the sample is large enough to proceed.
The sample is nationally representative. Hence, the sample can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 52% (= 0.52) said they viewed a television set as a necessity. Therefore, $p^=0.52$ . Substitute the values in the test statistic formula as shown below.

$z=0.52−0.50.5(1−0.5)1,003=0.020.01579=1.267$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose Right tail and give X value as 1.267.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 1

From the MINITAB output, the P-value is 0.1026.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀.

Here the significance level is 0.05.

Hence, $0.103>0.05$ .

Since P-value is greater than 0.05, fail to reject the null hypothesis. Therefore, there is no convincing evidence that a majority of adults consider a TV set as necessity.

Answer 14

b.

To determine

Check whether there is any convincing evidence that a majority of adults view a television set as a necessity at 0.05 significance level.

Answer 15

b.

Expert Solution

Answer to Problem 5CRE

There is no convincing evidence that a majority of adults consider a TV set as necessity.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

In order to test majority of adults view a television set as a necessity, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p$ denotes the proportion of all adults who consider a TV set as necessity.

2. Null hypothesis:

$H0:p=0.5$

That is, the proportion of adults who view a television set as a necessity is 0.5.

3. Alternative hypothesis:

$Ha:p>0.5$

That is, the proportion of adults who view a television set as a necessity is greater than 0.5.

That is, majority of adults consider a TV set as necessity.

4. Significance level:

$α=0.05$

5. Test statistic:

$z=p^−pp(1−p)n=p^−0.50.5(1−0.5)1,003$

6. Assumptions:

Here, $np^=501.5$ (1,003 × 0.5) and $n(1−p^)=501.5$ (1,003 × (1−0.5)). Since, both values are greater than 10, the sample is large enough to proceed.
The sample is nationally representative. Hence, the sample can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 52% (= 0.52) said they viewed a television set as a necessity. Therefore, $p^=0.52$ . Substitute the values in the test statistic formula as shown below.

$z=0.52−0.50.5(1−0.5)1,003=0.020.01579=1.267$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose Right tail and give X value as 1.267.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 1

From the MINITAB output, the P-value is 0.1026.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀.

Here the significance level is 0.05.

Hence, $0.103>0.05$ .

Since P-value is greater than 0.05, fail to reject the null hypothesis. Therefore, there is no convincing evidence that a majority of adults consider a TV set as necessity.

Answer 20

c.

To determine

Check whether there is any convincing evidence that the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009 at 0.01 significance level.

c.

Expert Solution

Answer to Problem 5CRE

There is a convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Explanation of Solution

In order to test the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p1$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2003 and $p2$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2009.

2. Null hypothesis:

$H0:p1−p2=0$

That is, the proportions of adults who regarded a microwave oven as a necessity in 2003 and 2009 are equal.

3. Alternative hypothesis:

$Ha:p1−p2>0$

That is, the proportion of adults who regarded a microwave oven as a necessity in 2003 is greater than the proportion of adults who regarded a microwave oven as a necessity in 2009.

4. Significance level:

$α=0.01$

5. Test statistic:

$z=p^1−p^2p^c(1−p^c)n1+p^c(1−p^c)n2$

Where, $p^c$ is the combined estimate of common population proportion.

6. Assumptions:

Here, $n1$ is 1,003, $n2$ is 1,003, $p^1$ is 0.68, and $p^2$ is 0.47

$n1p^1=1,003(0.68)=682.04n1(1−p^1)=1,003(1−0.68)=320.96$

$n2p^2=1,003(0.47)=471.41n2(1−p^2)=1,003(1−0.47)=531.59$

Since the values of $n1p^1$ , $n1(1−p^1)$ , $n2p^2$ , and $n2(1−p^2)$ are greater than 10, the samples are large enough to proceed.
Also, the sample is nationally representative. Hence, the samples can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 68% of the 2003 sample regarded a microwave oven as a necessity and 47% of the 2009 sample regarded a microwave oven as a necessity.

$p^c=n1p^1+n2p^2n1+n2=1,003(0.68)+1,003(0.47)1,003+1,003=1,153.452,006=0.575z=0.68−0.470.575(1−0.575)1,003+0.575(1−0.575)1,003=0.210.000244+0.000244=0.210.000487=0.210.022075=9.5132$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose left tail and give X value as 9.5132.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 2

From the MINITAB output, the P-value is 0.000.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀. Otherwise, fail to reject $H0$ .

Here, the significance level is 0.01.

Here, the P-value is less than the level of significance.

Hence, $0.000<0.01$ .

Since P-value is less than 0.01, reject the null hypothesis. Therefore, there is convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Answer 21

c.

To determine

Check whether there is any convincing evidence that the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009 at 0.01 significance level.

Answer 22

c.

Expert Solution

Answer to Problem 5CRE

There is a convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

In order to test the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p1$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2003 and $p2$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2009.

2. Null hypothesis:

$H0:p1−p2=0$

That is, the proportions of adults who regarded a microwave oven as a necessity in 2003 and 2009 are equal.

3. Alternative hypothesis:

$Ha:p1−p2>0$

That is, the proportion of adults who regarded a microwave oven as a necessity in 2003 is greater than the proportion of adults who regarded a microwave oven as a necessity in 2009.

4. Significance level:

$α=0.01$

5. Test statistic:

$z=p^1−p^2p^c(1−p^c)n1+p^c(1−p^c)n2$

Where, $p^c$ is the combined estimate of common population proportion.

6. Assumptions:

Here, $n1$ is 1,003, $n2$ is 1,003, $p^1$ is 0.68, and $p^2$ is 0.47

$n1p^1=1,003(0.68)=682.04n1(1−p^1)=1,003(1−0.68)=320.96$

$n2p^2=1,003(0.47)=471.41n2(1−p^2)=1,003(1−0.47)=531.59$

Since the values of $n1p^1$ , $n1(1−p^1)$ , $n2p^2$ , and $n2(1−p^2)$ are greater than 10, the samples are large enough to proceed.
Also, the sample is nationally representative. Hence, the samples can be treated as a random sample from the population.
The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 68% of the 2003 sample regarded a microwave oven as a necessity and 47% of the 2009 sample regarded a microwave oven as a necessity.

$p^c=n1p^1+n2p^2n1+n2=1,003(0.68)+1,003(0.47)1,003+1,003=1,153.452,006=0.575z=0.68−0.470.575(1−0.575)1,003+0.575(1−0.575)1,003=0.210.000244+0.000244=0.210.000487=0.210.022075=9.5132$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

Select Graph > Probability Distribution Plots.
Choose View Probability.
In Distribution, select Normal under Distribution.
In Shaded Area, choose left tail and give X value as 9.5132.
Click OK.

The MINITAB output is as follows:

Introduction To Statistics And Data Analysis, Chapter 13, Problem 5CRE , additional homework tip 2

From the MINITAB output, the P-value is 0.000.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H₀. Otherwise, fail to reject $H0$ .

Here, the significance level is 0.01.

Here, the P-value is less than the level of significance.

Hence, $0.000<0.01$ .

Since P-value is less than 0.01, reject the null hypothesis. Therefore, there is convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Answer 27

Ch. 13.1 - Let x be the size of a house (in square feet) and...Ch. 13.1 - Consider the variables and population regression...Ch. 13.1 - The flow rate in a device used for air quality...Ch. 13.1 - The paper Predicting Yolk Height, Yolk Width,...Ch. 13.1 - A sample of small cars was selected, and the...Ch. 13.1 - Prob. 6E Ch. 13.1 - Suppose that a simple linear regression model is...Ch. 13.1 - a. Explain the difference between the line y x...Ch. 13.1 - Prob. 9E Ch. 13.1 - Hormone replacement therapy (HRT) is thought to...

Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

Concept explainers

Videos

Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

Answer to Problem 5CRE

Explanation of Solution

Check whether there is any convincing evidence that a majority of adults view a television set as a necessity at 0.05 significance level.

Answer to Problem 5CRE

Explanation of Solution

Check whether there is any convincing evidence that the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009 at 0.01 significance level.

Answer to Problem 5CRE

Explanation of Solution

Want to see more full solutions like this?

Chapter 13 Solutions