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Concept explainers
a.
Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.
a.
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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
Conditions for 95% confidence interval:
1. Values of nˆp and n(1−ˆp):
nˆp=1,003(0.68)=682.04n(1−ˆp)=1,003(1−0.68)=320.96
Since nˆp and n(1−ˆp) 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.
ˆp±1.96√ˆp(1−ˆp)n
Substitute the values of n and ˆp in confidence interval formula as shown below.
ˆp±1.96√ˆp(1−ˆp)n=0.68±1.96√0.68(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.
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.
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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=ˆp−p√p(1−p)n=ˆp−0.5√0.5(1−0.5)1,003
6. Assumptions:
- Here, nˆp=501.5 (1,003 × 0.5) and n(1−ˆp)=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, ˆp=0.52. Substitute the values in the test statistic formula as shown below.
z=0.52−0.5√0.5(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:
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.
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.
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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=ˆp1−ˆp2√ˆpc(1−ˆpc)n1+ˆpc(1−ˆpc)n2
Where, ˆpc is the combined estimate of common population proportion.
6. Assumptions:
Here, n1 is 1,003, n2 is 1,003, ˆp1 is 0.68, and ˆp2 is 0.47
n1ˆp1=1,003(0.68)=682.04n1(1−ˆp1)=1,003(1−0.68)=320.96
n2ˆp2=1,003(0.47)=471.41n2(1−ˆp2)=1,003(1−0.47)=531.59
- Since the values of n1ˆp1, n1(1−ˆp1), n2ˆp2, and n2(1−ˆp2) 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.
ˆpc=n1ˆp1+n2ˆp2n1+n2=1,003(0.68)+1,003(0.47)1,003+1,003=1,153.452,006=0.575z=0.68−0.47√0.575(1−0.575)1,003+0.575(1−0.575)1,003=0.21√0.000244+0.000244=0.21√0.000487=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:
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.
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Chapter 13 Solutions
Introduction To Statistics And Data Analysis
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