The sample proportion on the usage of cell phone while in a store.

Question

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Answer 1

Question

Chapter 8.1, Problem 15E

(a)

Section 1:

To determine

To find: The sample proportion on the usage of cell phone while in a store.

(a)

Section 1:

Expert Solution

Answer to Problem 15E

Solution: The sample proportion of the usage of cell phone while in a store is $p^=0.461_$ .

Explanation of Solution

Given: A Pew Internet poll surveyed 1003 adults. 462 responded that they used their cell phone while in the store within the last 30 days to make call.

Explanation:

Calculation: The formula for sample proportion is defined as:

$p^=Xn$

Here,

$X=number of succeses in the samplen=sample size$

Substitute $X=462 and n=1003$ in the above formula, to get the required sample proportion. So,

$p^=Xn=4621003=0.461$

Therefore, the sample proportion $p^$ is obtained as 0.461.

Section 2:

To determine

To find: The standard error $SEp^$ of sample proportion $p^$ .

Section 2:

Expert Solution

Answer to Problem 15E

Solution: The standard error $SEp^$ of sample proportion $p^$ is $SEp^=0.0157_$ .

Explanation of Solution

Calculation: The formula for standard error $SEp^$ of sample proportion $p^$ and sample size n is defined as:

$SEp^=p^(1−p^)n$

The sample proportion $p^$ is obtained as 0.461 in the section 1. Substitute the obtained sample proportion of 0.461 and sample size of 1003 in the standard error formula. So,

$SEp^=p^(1−p^)n=0.461(1−0.461)1003=0.2484791003=0.0157$

Therefore, the standard error is obtained as 0.0157.

Section 3:

To determine

To find: The margin of error for 95% confidence level.

Section 3:

Expert Solution

Answer to Problem 15E

Solution: The margin of error for 95% confidence level is $m=0.0308_$ .

Explanation of Solution

Calculation: The formula for margin of error m is defined as:

$m=z*×SEp^$

Here, $z*$ is the critical value of the standard normal density curve.

The standard error is obtained as $SEp^=0.0157$ in previous part. The value of $z*$ for 95% confidence level is $z*=1.96$ from the standard normal table provided in the book.

So, the margin of error is obtained as:

$m=z*×SEp^=1.96×0.0157=0.0308$

Therefore, the margin of error is obtained as 0.0308.

(b)

To determine

To explain: Whether the guidelines to use the large-sample confidence interval for population proportion are satisfied.

(b)

Expert Solution

Answer to Problem 15E

Solution: Yes, the guidelines are satisfied to use the large-sample confidence interval for the population proportion.

Explanation of Solution

The guidelines required to be satisfied to use large-sample confidence interval for the population proportion is that the count of successes and the count of failures both should be at least 10.

In the provided problem of cell phone survey, the number of successes is defined as the number of respondents who used their cell phone while in the store within the last 30 days to make call So, the number of successes is 462.

The number of failures is obtained as,

$Number of failures=1003−462=541$

The obtained number of successes and failures shows that they are more than 10.

Therefore, the guidelines to use the large-sample confidence interval are satisfied for a population proportion.

(c)

To determine

To find: The 95% large-sample confidence interval for the population proportion.

(c)

Expert Solution

Answer to Problem 15E

Solution: The 95% large-sample confidence interval is $(0.4302,0.4918)_$ .

Explanation of Solution

Calculation: The formula for large-sample confidence interval for population proportion p is defined as:

$p^±m$

Here, $p^$ is the sample proportion and m is is the margin of error.

The sample proportion $p^$ is obtained as 0.461 and the margin of error is obtained as 0.0308 in part (a).

Substitute the values of margin of error and sample proportion in the formula for confidence interval. Therefore, the large-sample confidence interval is obtained as:

$p^±m=0.461±0.0308=(0.461−0.0308,0.461+0.0308)=(0.4302,0.4918)$

Therefore, the large-sample confidence interval for the population proportion is obtained as $(0.4302,0.4918)$ .

(d)

To determine

To explain: A short statement on the meaning of the obtained confidence interval.

(d)

Expert Solution

Answer to Problem 15E

Solution: The obtained confidence interval shows that it is 95% confident that between 43.02% an 49.18% of cell phone owners used their cell phone while in a store within last 30 days to make call to any friend or family member for advice on their purchase.

Explanation of Solution

The 95% confidence interval is obtained as

$(0.4302,0.4918)$ in the previous part. So, the 95% confidence interval in the form of percents is calculated as:

$(0.4302,0.4918)=(43.02%,49.18%)$

This shows that there is 95% confidence that the percentage of people who used their cell phone while in a store within the last 30 days to make call to any friend or family member for advice on their purchase will lie between 43.02% and 49.18%.

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***Please do not just simply copy and paste the other solution for this problem posted on bartleby as that solution does not have all of the parts completed for this problem. Please answer this I will leave a like on the problem. The data needed to answer this question is given in the following link (file is on view only so if you would like to make a copy to make it easier for yourself feel free to do so) https://docs.google.com/spreadsheets/d/1aV5rsxdNjHnkeTkm5VqHzBXZgW-Ptbs3vqwk0SYiQPo/edit?usp=sharing

The data needed to answer this question is given in the following link (file is on view only so if you would like to make a copy to make it easier for yourself feel free to do so) https://docs.google.com/spreadsheets/d/1aV5rsxdNjHnkeTkm5VqHzBXZgW-Ptbs3vqwk0SYiQPo/edit?usp=sharing

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Answer 2

Question

Chapter 8.1, Problem 15E

(a)

Section 1:

To determine

To find: The sample proportion on the usage of cell phone while in a store.

(a)

Section 1:

Expert Solution

Answer to Problem 15E

Solution: The sample proportion of the usage of cell phone while in a store is $p^=0.461_$ .

Explanation of Solution

Given: A Pew Internet poll surveyed 1003 adults. 462 responded that they used their cell phone while in the store within the last 30 days to make call.

Explanation:

Calculation: The formula for sample proportion is defined as:

$p^=Xn$

Here,

$X=number of succeses in the samplen=sample size$

Substitute $X=462 and n=1003$ in the above formula, to get the required sample proportion. So,

$p^=Xn=4621003=0.461$

Therefore, the sample proportion $p^$ is obtained as 0.461.

Section 2:

To determine

To find: The standard error $SEp^$ of sample proportion $p^$ .

Section 2:

Expert Solution

Answer to Problem 15E

Solution: The standard error $SEp^$ of sample proportion $p^$ is $SEp^=0.0157_$ .

Explanation of Solution

Calculation: The formula for standard error $SEp^$ of sample proportion $p^$ and sample size n is defined as:

$SEp^=p^(1−p^)n$

The sample proportion $p^$ is obtained as 0.461 in the section 1. Substitute the obtained sample proportion of 0.461 and sample size of 1003 in the standard error formula. So,

$SEp^=p^(1−p^)n=0.461(1−0.461)1003=0.2484791003=0.0157$

Therefore, the standard error is obtained as 0.0157.

Section 3:

To determine

To find: The margin of error for 95% confidence level.

Section 3:

Expert Solution

Answer to Problem 15E

Solution: The margin of error for 95% confidence level is $m=0.0308_$ .

Explanation of Solution

Calculation: The formula for margin of error m is defined as:

$m=z*×SEp^$

Here, $z*$ is the critical value of the standard normal density curve.

The standard error is obtained as $SEp^=0.0157$ in previous part. The value of $z*$ for 95% confidence level is $z*=1.96$ from the standard normal table provided in the book.

So, the margin of error is obtained as:

$m=z*×SEp^=1.96×0.0157=0.0308$

Therefore, the margin of error is obtained as 0.0308.

(b)

To determine

To explain: Whether the guidelines to use the large-sample confidence interval for population proportion are satisfied.

(b)

Expert Solution

Answer to Problem 15E

Solution: Yes, the guidelines are satisfied to use the large-sample confidence interval for the population proportion.

Explanation of Solution

The guidelines required to be satisfied to use large-sample confidence interval for the population proportion is that the count of successes and the count of failures both should be at least 10.

In the provided problem of cell phone survey, the number of successes is defined as the number of respondents who used their cell phone while in the store within the last 30 days to make call So, the number of successes is 462.

The number of failures is obtained as,

$Number of failures=1003−462=541$

The obtained number of successes and failures shows that they are more than 10.

Therefore, the guidelines to use the large-sample confidence interval are satisfied for a population proportion.

(c)

To determine

To find: The 95% large-sample confidence interval for the population proportion.

(c)

Expert Solution

Answer to Problem 15E

Solution: The 95% large-sample confidence interval is $(0.4302,0.4918)_$ .

Explanation of Solution

Calculation: The formula for large-sample confidence interval for population proportion p is defined as:

$p^±m$

Here, $p^$ is the sample proportion and m is is the margin of error.

The sample proportion $p^$ is obtained as 0.461 and the margin of error is obtained as 0.0308 in part (a).

Substitute the values of margin of error and sample proportion in the formula for confidence interval. Therefore, the large-sample confidence interval is obtained as:

$p^±m=0.461±0.0308=(0.461−0.0308,0.461+0.0308)=(0.4302,0.4918)$

Therefore, the large-sample confidence interval for the population proportion is obtained as $(0.4302,0.4918)$ .

(d)

To determine

To explain: A short statement on the meaning of the obtained confidence interval.

(d)

Expert Solution

Answer to Problem 15E

Solution: The obtained confidence interval shows that it is 95% confident that between 43.02% an 49.18% of cell phone owners used their cell phone while in a store within last 30 days to make call to any friend or family member for advice on their purchase.

Explanation of Solution

The 95% confidence interval is obtained as

$(0.4302,0.4918)$ in the previous part. So, the 95% confidence interval in the form of percents is calculated as:

$(0.4302,0.4918)=(43.02%,49.18%)$

This shows that there is 95% confidence that the percentage of people who used their cell phone while in a store within the last 30 days to make call to any friend or family member for advice on their purchase will lie between 43.02% and 49.18%.

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Answer 3

Question

Chapter 8.1, Problem 15E

(a)

Section 1:

To determine

To find: The sample proportion on the usage of cell phone while in a store.

(a)

Section 1:

Expert Solution

Answer to Problem 15E

Solution: The sample proportion of the usage of cell phone while in a store is $p^=0.461_$ .

Explanation of Solution

Given: A Pew Internet poll surveyed 1003 adults. 462 responded that they used their cell phone while in the store within the last 30 days to make call.

Explanation:

Calculation: The formula for sample proportion is defined as:

$p^=Xn$

Here,

$X=number of succeses in the samplen=sample size$

Substitute $X=462 and n=1003$ in the above formula, to get the required sample proportion. So,

$p^=Xn=4621003=0.461$

Therefore, the sample proportion $p^$ is obtained as 0.461.

Section 2:

To determine

To find: The standard error $SEp^$ of sample proportion $p^$ .

Section 2:

Expert Solution

Answer to Problem 15E

Solution: The standard error $SEp^$ of sample proportion $p^$ is $SEp^=0.0157_$ .

Explanation of Solution

Calculation: The formula for standard error $SEp^$ of sample proportion $p^$ and sample size n is defined as:

$SEp^=p^(1−p^)n$

The sample proportion $p^$ is obtained as 0.461 in the section 1. Substitute the obtained sample proportion of 0.461 and sample size of 1003 in the standard error formula. So,

$SEp^=p^(1−p^)n=0.461(1−0.461)1003=0.2484791003=0.0157$

Therefore, the standard error is obtained as 0.0157.

Section 3:

To determine

To find: The margin of error for 95% confidence level.

Section 3:

Expert Solution

Answer to Problem 15E

Solution: The margin of error for 95% confidence level is $m=0.0308_$ .

Explanation of Solution

Calculation: The formula for margin of error m is defined as:

$m=z*×SEp^$

Here, $z*$ is the critical value of the standard normal density curve.

The standard error is obtained as $SEp^=0.0157$ in previous part. The value of $z*$ for 95% confidence level is $z*=1.96$ from the standard normal table provided in the book.

So, the margin of error is obtained as:

$m=z*×SEp^=1.96×0.0157=0.0308$

Therefore, the margin of error is obtained as 0.0308.

(b)

To determine

To explain: Whether the guidelines to use the large-sample confidence interval for population proportion are satisfied.

(b)

Expert Solution

Answer to Problem 15E

Solution: Yes, the guidelines are satisfied to use the large-sample confidence interval for the population proportion.

Explanation of Solution

The guidelines required to be satisfied to use large-sample confidence interval for the population proportion is that the count of successes and the count of failures both should be at least 10.

In the provided problem of cell phone survey, the number of successes is defined as the number of respondents who used their cell phone while in the store within the last 30 days to make call So, the number of successes is 462.

The number of failures is obtained as,

$Number of failures=1003−462=541$

The obtained number of successes and failures shows that they are more than 10.

Therefore, the guidelines to use the large-sample confidence interval are satisfied for a population proportion.

(c)

To determine

To find: The 95% large-sample confidence interval for the population proportion.

(c)

Expert Solution

Answer to Problem 15E

Solution: The 95% large-sample confidence interval is $(0.4302,0.4918)_$ .

Explanation of Solution

Calculation: The formula for large-sample confidence interval for population proportion p is defined as:

$p^±m$

Here, $p^$ is the sample proportion and m is is the margin of error.

The sample proportion $p^$ is obtained as 0.461 and the margin of error is obtained as 0.0308 in part (a).

Substitute the values of margin of error and sample proportion in the formula for confidence interval. Therefore, the large-sample confidence interval is obtained as:

$p^±m=0.461±0.0308=(0.461−0.0308,0.461+0.0308)=(0.4302,0.4918)$

Therefore, the large-sample confidence interval for the population proportion is obtained as $(0.4302,0.4918)$ .

(d)

To determine

To explain: A short statement on the meaning of the obtained confidence interval.

(d)

Expert Solution

Answer to Problem 15E

Solution: The obtained confidence interval shows that it is 95% confident that between 43.02% an 49.18% of cell phone owners used their cell phone while in a store within last 30 days to make call to any friend or family member for advice on their purchase.

Explanation of Solution

The 95% confidence interval is obtained as

$(0.4302,0.4918)$ in the previous part. So, the 95% confidence interval in the form of percents is calculated as:

$(0.4302,0.4918)=(43.02%,49.18%)$

This shows that there is 95% confidence that the percentage of people who used their cell phone while in a store within the last 30 days to make call to any friend or family member for advice on their purchase will lie between 43.02% and 49.18%.

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Answer 4

Question

Chapter 8.1, Problem 15E

(a)

Section 1:

To determine

To find: The sample proportion on the usage of cell phone while in a store.

(a)

Section 1:

Expert Solution

Answer to Problem 15E

Solution: The sample proportion of the usage of cell phone while in a store is $p^=0.461_$ .

Explanation of Solution

Given: A Pew Internet poll surveyed 1003 adults. 462 responded that they used their cell phone while in the store within the last 30 days to make call.

Explanation:

Calculation: The formula for sample proportion is defined as:

$p^=Xn$

Here,

$X=number of succeses in the samplen=sample size$

Substitute $X=462 and n=1003$ in the above formula, to get the required sample proportion. So,

$p^=Xn=4621003=0.461$

Therefore, the sample proportion $p^$ is obtained as 0.461.

Section 2:

To determine

To find: The standard error $SEp^$ of sample proportion $p^$ .

Section 2:

Expert Solution

Answer to Problem 15E

Solution: The standard error $SEp^$ of sample proportion $p^$ is $SEp^=0.0157_$ .

Explanation of Solution

Calculation: The formula for standard error $SEp^$ of sample proportion $p^$ and sample size n is defined as:

$SEp^=p^(1−p^)n$

The sample proportion $p^$ is obtained as 0.461 in the section 1. Substitute the obtained sample proportion of 0.461 and sample size of 1003 in the standard error formula. So,

$SEp^=p^(1−p^)n=0.461(1−0.461)1003=0.2484791003=0.0157$

Therefore, the standard error is obtained as 0.0157.

Section 3:

To determine

To find: The margin of error for 95% confidence level.

Section 3:

Expert Solution

Answer to Problem 15E

Solution: The margin of error for 95% confidence level is $m=0.0308_$ .

Explanation of Solution

Calculation: The formula for margin of error m is defined as:

$m=z*×SEp^$

Here, $z*$ is the critical value of the standard normal density curve.

The standard error is obtained as $SEp^=0.0157$ in previous part. The value of $z*$ for 95% confidence level is $z*=1.96$ from the standard normal table provided in the book.

So, the margin of error is obtained as:

$m=z*×SEp^=1.96×0.0157=0.0308$

Therefore, the margin of error is obtained as 0.0308.

(b)

To determine

To explain: Whether the guidelines to use the large-sample confidence interval for population proportion are satisfied.

(b)

Expert Solution

Answer to Problem 15E

Solution: Yes, the guidelines are satisfied to use the large-sample confidence interval for the population proportion.

Explanation of Solution

The guidelines required to be satisfied to use large-sample confidence interval for the population proportion is that the count of successes and the count of failures both should be at least 10.

In the provided problem of cell phone survey, the number of successes is defined as the number of respondents who used their cell phone while in the store within the last 30 days to make call So, the number of successes is 462.

The number of failures is obtained as,

$Number of failures=1003−462=541$

The obtained number of successes and failures shows that they are more than 10.

Therefore, the guidelines to use the large-sample confidence interval are satisfied for a population proportion.

(c)

To determine

To find: The 95% large-sample confidence interval for the population proportion.

(c)

Expert Solution

Answer to Problem 15E

Solution: The 95% large-sample confidence interval is $(0.4302,0.4918)_$ .

Explanation of Solution

Calculation: The formula for large-sample confidence interval for population proportion p is defined as:

$p^±m$

Here, $p^$ is the sample proportion and m is is the margin of error.

The sample proportion $p^$ is obtained as 0.461 and the margin of error is obtained as 0.0308 in part (a).

Substitute the values of margin of error and sample proportion in the formula for confidence interval. Therefore, the large-sample confidence interval is obtained as:

$p^±m=0.461±0.0308=(0.461−0.0308,0.461+0.0308)=(0.4302,0.4918)$

Therefore, the large-sample confidence interval for the population proportion is obtained as $(0.4302,0.4918)$ .

(d)

To determine

To explain: A short statement on the meaning of the obtained confidence interval.

(d)

Expert Solution

Answer to Problem 15E

Solution: The obtained confidence interval shows that it is 95% confident that between 43.02% an 49.18% of cell phone owners used their cell phone while in a store within last 30 days to make call to any friend or family member for advice on their purchase.

Explanation of Solution

The 95% confidence interval is obtained as

$(0.4302,0.4918)$ in the previous part. So, the 95% confidence interval in the form of percents is calculated as:

$(0.4302,0.4918)=(43.02%,49.18%)$

This shows that there is 95% confidence that the percentage of people who used their cell phone while in a store within the last 30 days to make call to any friend or family member for advice on their purchase will lie between 43.02% and 49.18%.

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Answer 5

Chapter 8.1, Problem 15E

(a)

Section 1:

To determine

To find: The sample proportion on the usage of cell phone while in a store.

(a)

Section 1:

Expert Solution

Answer to Problem 15E

Solution: The sample proportion of the usage of cell phone while in a store is $p^=0.461_$ .

Explanation of Solution

Given: A Pew Internet poll surveyed 1003 adults. 462 responded that they used their cell phone while in the store within the last 30 days to make call.

Explanation:

Calculation: The formula for sample proportion is defined as:

$p^=Xn$

Here,

$X=number of succeses in the samplen=sample size$

Substitute $X=462 and n=1003$ in the above formula, to get the required sample proportion. So,

$p^=Xn=4621003=0.461$

Therefore, the sample proportion $p^$ is obtained as 0.461.

Section 2:

To determine

To find: The standard error $SEp^$ of sample proportion $p^$ .

Section 2:

Expert Solution

Answer to Problem 15E

Solution: The standard error $SEp^$ of sample proportion $p^$ is $SEp^=0.0157_$ .

Explanation of Solution

Calculation: The formula for standard error $SEp^$ of sample proportion $p^$ and sample size n is defined as:

$SEp^=p^(1−p^)n$

The sample proportion $p^$ is obtained as 0.461 in the section 1. Substitute the obtained sample proportion of 0.461 and sample size of 1003 in the standard error formula. So,

$SEp^=p^(1−p^)n=0.461(1−0.461)1003=0.2484791003=0.0157$

Therefore, the standard error is obtained as 0.0157.

Section 3:

To determine

To find: The margin of error for 95% confidence level.

Section 3:

Expert Solution

Answer to Problem 15E

Solution: The margin of error for 95% confidence level is $m=0.0308_$ .

Explanation of Solution

Calculation: The formula for margin of error m is defined as:

$m=z*×SEp^$

Here, $z*$ is the critical value of the standard normal density curve.

The standard error is obtained as $SEp^=0.0157$ in previous part. The value of $z*$ for 95% confidence level is $z*=1.96$ from the standard normal table provided in the book.

So, the margin of error is obtained as:

$m=z*×SEp^=1.96×0.0157=0.0308$

Therefore, the margin of error is obtained as 0.0308.

(b)

To determine

To explain: Whether the guidelines to use the large-sample confidence interval for population proportion are satisfied.

(b)

Expert Solution

Answer to Problem 15E

Solution: Yes, the guidelines are satisfied to use the large-sample confidence interval for the population proportion.

Explanation of Solution

The guidelines required to be satisfied to use large-sample confidence interval for the population proportion is that the count of successes and the count of failures both should be at least 10.

In the provided problem of cell phone survey, the number of successes is defined as the number of respondents who used their cell phone while in the store within the last 30 days to make call So, the number of successes is 462.

The number of failures is obtained as,

$Number of failures=1003−462=541$

The obtained number of successes and failures shows that they are more than 10.

Therefore, the guidelines to use the large-sample confidence interval are satisfied for a population proportion.

(c)

To determine

To find: The 95% large-sample confidence interval for the population proportion.

(c)

Expert Solution

Answer to Problem 15E

Solution: The 95% large-sample confidence interval is $(0.4302,0.4918)_$ .

Explanation of Solution

Calculation: The formula for large-sample confidence interval for population proportion p is defined as:

$p^±m$

Here, $p^$ is the sample proportion and m is is the margin of error.

The sample proportion $p^$ is obtained as 0.461 and the margin of error is obtained as 0.0308 in part (a).

Substitute the values of margin of error and sample proportion in the formula for confidence interval. Therefore, the large-sample confidence interval is obtained as:

$p^±m=0.461±0.0308=(0.461−0.0308,0.461+0.0308)=(0.4302,0.4918)$

Therefore, the large-sample confidence interval for the population proportion is obtained as $(0.4302,0.4918)$ .

(d)

To determine

To explain: A short statement on the meaning of the obtained confidence interval.

(d)

Expert Solution

Answer to Problem 15E

Solution: The obtained confidence interval shows that it is 95% confident that between 43.02% an 49.18% of cell phone owners used their cell phone while in a store within last 30 days to make call to any friend or family member for advice on their purchase.

Explanation of Solution

The 95% confidence interval is obtained as

$(0.4302,0.4918)$ in the previous part. So, the 95% confidence interval in the form of percents is calculated as:

$(0.4302,0.4918)=(43.02%,49.18%)$

This shows that there is 95% confidence that the percentage of people who used their cell phone while in a store within the last 30 days to make call to any friend or family member for advice on their purchase will lie between 43.02% and 49.18%.

Answer 6

Chapter 8.1, Problem 15E

(a)

Section 1:

To determine

To find: The sample proportion on the usage of cell phone while in a store.

(a)

Section 1:

Expert Solution

Answer to Problem 15E

Solution: The sample proportion of the usage of cell phone while in a store is $p^=0.461_$ .

Explanation of Solution

Given: A Pew Internet poll surveyed 1003 adults. 462 responded that they used their cell phone while in the store within the last 30 days to make call.

Explanation:

Calculation: The formula for sample proportion is defined as:

$p^=Xn$

Here,

$X=number of succeses in the samplen=sample size$

Substitute $X=462 and n=1003$ in the above formula, to get the required sample proportion. So,

$p^=Xn=4621003=0.461$

Therefore, the sample proportion $p^$ is obtained as 0.461.

Answer 7

Chapter 8.1, Problem 15E

(a)

Section 1:

To determine

To find: The sample proportion on the usage of cell phone while in a store.

Answer 8

(a)

Section 1:

Expert Solution

Answer to Problem 15E

Solution: The sample proportion of the usage of cell phone while in a store is $p^=0.461_$ .

Answer 9

(a)

Section 1:

Expert Solution

Answer 10

(a)

Section 1:

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given: A Pew Internet poll surveyed 1003 adults. 462 responded that they used their cell phone while in the store within the last 30 days to make call.

Explanation:

Calculation: The formula for sample proportion is defined as:

$p^=Xn$

Here,

$X=number of succeses in the samplen=sample size$

Substitute $X=462 and n=1003$ in the above formula, to get the required sample proportion. So,

$p^=Xn=4621003=0.461$

Therefore, the sample proportion $p^$ is obtained as 0.461.

Answer 13

Section 2:

To determine

To find: The standard error $SEp^$ of sample proportion $p^$ .

Section 2:

Expert Solution

Answer to Problem 15E

Solution: The standard error $SEp^$ of sample proportion $p^$ is $SEp^=0.0157_$ .

Explanation of Solution

Calculation: The formula for standard error $SEp^$ of sample proportion $p^$ and sample size n is defined as:

$SEp^=p^(1−p^)n$

The sample proportion $p^$ is obtained as 0.461 in the section 1. Substitute the obtained sample proportion of 0.461 and sample size of 1003 in the standard error formula. So,

$SEp^=p^(1−p^)n=0.461(1−0.461)1003=0.2484791003=0.0157$

Therefore, the standard error is obtained as 0.0157.

Answer 14

Section 2:

To determine

To find: The standard error $SEp^$ of sample proportion $p^$ .

Answer 15

Section 2:

Expert Solution

Answer to Problem 15E

Solution: The standard error $SEp^$ of sample proportion $p^$ is $SEp^=0.0157_$ .

Answer 16

Section 2:

Expert Solution

Answer 17

Section 2:

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation: The formula for standard error $SEp^$ of sample proportion $p^$ and sample size n is defined as:

$SEp^=p^(1−p^)n$

The sample proportion $p^$ is obtained as 0.461 in the section 1. Substitute the obtained sample proportion of 0.461 and sample size of 1003 in the standard error formula. So,

$SEp^=p^(1−p^)n=0.461(1−0.461)1003=0.2484791003=0.0157$

Therefore, the standard error is obtained as 0.0157.

Answer 20

Section 3:

To determine

To find: The margin of error for 95% confidence level.

Section 3:

Expert Solution

Answer to Problem 15E

Solution: The margin of error for 95% confidence level is $m=0.0308_$ .

Explanation of Solution

Calculation: The formula for margin of error m is defined as:

$m=z*×SEp^$

Here, $z*$ is the critical value of the standard normal density curve.

The standard error is obtained as $SEp^=0.0157$ in previous part. The value of $z*$ for 95% confidence level is $z*=1.96$ from the standard normal table provided in the book.

So, the margin of error is obtained as:

$m=z*×SEp^=1.96×0.0157=0.0308$

Therefore, the margin of error is obtained as 0.0308.

Answer 21

Section 3:

To determine

To find: The margin of error for 95% confidence level.

Answer 22

Section 3:

Expert Solution

Answer to Problem 15E

Solution: The margin of error for 95% confidence level is $m=0.0308_$ .

Answer 23

Section 3:

Expert Solution

Answer 24

Section 3:

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation: The formula for margin of error m is defined as:

$m=z*×SEp^$

Here, $z*$ is the critical value of the standard normal density curve.

The standard error is obtained as $SEp^=0.0157$ in previous part. The value of $z*$ for 95% confidence level is $z*=1.96$ from the standard normal table provided in the book.

So, the margin of error is obtained as:

$m=z*×SEp^=1.96×0.0157=0.0308$

Therefore, the margin of error is obtained as 0.0308.

Answer 27

(b)

To determine

To explain: Whether the guidelines to use the large-sample confidence interval for population proportion are satisfied.

(b)

Expert Solution

Answer to Problem 15E

Solution: Yes, the guidelines are satisfied to use the large-sample confidence interval for the population proportion.

Explanation of Solution

The guidelines required to be satisfied to use large-sample confidence interval for the population proportion is that the count of successes and the count of failures both should be at least 10.

In the provided problem of cell phone survey, the number of successes is defined as the number of respondents who used their cell phone while in the store within the last 30 days to make call So, the number of successes is 462.

The number of failures is obtained as,

$Number of failures=1003−462=541$

The obtained number of successes and failures shows that they are more than 10.

Therefore, the guidelines to use the large-sample confidence interval are satisfied for a population proportion.

Answer 28

(b)

To determine

To explain: Whether the guidelines to use the large-sample confidence interval for population proportion are satisfied.

Answer 29

(b)

Expert Solution

Answer to Problem 15E

Solution: Yes, the guidelines are satisfied to use the large-sample confidence interval for the population proportion.

Answer 30

(b)

Expert Solution

Answer 31

(b)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The guidelines required to be satisfied to use large-sample confidence interval for the population proportion is that the count of successes and the count of failures both should be at least 10.

In the provided problem of cell phone survey, the number of successes is defined as the number of respondents who used their cell phone while in the store within the last 30 days to make call So, the number of successes is 462.

The number of failures is obtained as,

$Number of failures=1003−462=541$

The obtained number of successes and failures shows that they are more than 10.

Therefore, the guidelines to use the large-sample confidence interval are satisfied for a population proportion.

Answer 34

(c)

To determine

To find: The 95% large-sample confidence interval for the population proportion.

(c)

Expert Solution

Answer to Problem 15E

Solution: The 95% large-sample confidence interval is $(0.4302,0.4918)_$ .

Explanation of Solution

Calculation: The formula for large-sample confidence interval for population proportion p is defined as:

$p^±m$

Here, $p^$ is the sample proportion and m is is the margin of error.

The sample proportion $p^$ is obtained as 0.461 and the margin of error is obtained as 0.0308 in part (a).

Substitute the values of margin of error and sample proportion in the formula for confidence interval. Therefore, the large-sample confidence interval is obtained as:

$p^±m=0.461±0.0308=(0.461−0.0308,0.461+0.0308)=(0.4302,0.4918)$

Therefore, the large-sample confidence interval for the population proportion is obtained as $(0.4302,0.4918)$ .

Answer 35

(c)

To determine

To find: The 95% large-sample confidence interval for the population proportion.

Answer 36

(c)

Expert Solution

Answer to Problem 15E

Solution: The 95% large-sample confidence interval is $(0.4302,0.4918)_$ .

Answer 37

(c)

Expert Solution

Answer 38

(c)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Calculation: The formula for large-sample confidence interval for population proportion p is defined as:

$p^±m$

Here, $p^$ is the sample proportion and m is is the margin of error.

The sample proportion $p^$ is obtained as 0.461 and the margin of error is obtained as 0.0308 in part (a).

Substitute the values of margin of error and sample proportion in the formula for confidence interval. Therefore, the large-sample confidence interval is obtained as:

$p^±m=0.461±0.0308=(0.461−0.0308,0.461+0.0308)=(0.4302,0.4918)$

Therefore, the large-sample confidence interval for the population proportion is obtained as $(0.4302,0.4918)$ .

Answer 41

(d)

To determine

To explain: A short statement on the meaning of the obtained confidence interval.

(d)

Expert Solution

Answer to Problem 15E

Solution: The obtained confidence interval shows that it is 95% confident that between 43.02% an 49.18% of cell phone owners used their cell phone while in a store within last 30 days to make call to any friend or family member for advice on their purchase.

Explanation of Solution

The 95% confidence interval is obtained as

$(0.4302,0.4918)$ in the previous part. So, the 95% confidence interval in the form of percents is calculated as:

$(0.4302,0.4918)=(43.02%,49.18%)$

This shows that there is 95% confidence that the percentage of people who used their cell phone while in a store within the last 30 days to make call to any friend or family member for advice on their purchase will lie between 43.02% and 49.18%.

Answer 42

(d)

To determine

To explain: A short statement on the meaning of the obtained confidence interval.

Answer 43

(d)

Expert Solution

Answer to Problem 15E

Solution: The obtained confidence interval shows that it is 95% confident that between 43.02% an 49.18% of cell phone owners used their cell phone while in a store within last 30 days to make call to any friend or family member for advice on their purchase.

Answer 44

(d)

Expert Solution

Answer 45

(d)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

The 95% confidence interval is obtained as

$(0.4302,0.4918)$ in the previous part. So, the 95% confidence interval in the form of percents is calculated as:

$(0.4302,0.4918)=(43.02%,49.18%)$

This shows that there is 95% confidence that the percentage of people who used their cell phone while in a store within the last 30 days to make call to any friend or family member for advice on their purchase will lie between 43.02% and 49.18%.