Math / Data & Statistics

QuestionSection 9.2 Homework Question 9, 9.2.28 HW Score: 47.14\%, 15.56 of 33 points Part 3 of 4 Points: 0 of 1 Save
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 951 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.32 hours with a standard deviation of 0.52 hour. Complete parts (a) through (d) below.
Click the icon to view the table of critical t-values. that both tails are accounted for in the confidence interval. D. Since the distribution of time spent eating and drinking each day is highly skewed right, a large sample size is needed to minimize the margin of error to ensure only the peak of the sampling distribution is captured in the confidence interval. (b) There are more than 200 million people nationally age 15 or older. Explain why this, along with the fact that the data were obtained using a random sample, satisfies the requirements for constructing a confidence interval. A. The sample size is greater than $10 \%$ of the population. B. The sample size is greater than $5 \%$ of the population. C. The sample size is less than $5 \%$ of the population. D. The sample size is less than $10 \%$ of the population. (c) Determine and interpret a 95\% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day.
Select the correct choice below and fill in the answer boxes, if applicable, in your choice. (Type integers or decimals rounded to three decimal places as needed. Use ascending order.) A. The nutritionist is $95 \%$ confident that the amount of time spent eating or drinking per day for any individual is between $\square$ and B. The nutritionist is $95 \%$ confident that the mean amount of time spent eating or drinking per day is between $\square$ and $\square$ hours. C. There is a $95 \%$ probability that the mean amount of time spent eating or drinking per day is between $\square$ and $\square$ hours.

Studdy Solution

STEP 1

1. The sample size is $n = 951$ .
2. The sample mean is $\bar{x} = 1.32$ hours.
3. The sample standard deviation is $s = 0.52$ hours.
4. The population size is more than 200 million people.
5. We are constructing a 95% confidence interval for the mean.

STEP 2

1. Verify the conditions for constructing a confidence interval.
2. Calculate the standard error of the mean.
3. Determine the critical t-value for a 95% confidence level.
4. Calculate the confidence interval.
5. Interpret the confidence interval.

STEP 3

Verify the conditions for constructing a confidence interval: - The sample size is $n = 951$ , which is large enough for the Central Limit Theorem to apply. - The sample is random. - The sample size is less than 5% of the population, satisfying the independence requirement.

STEP 4

Calculate the standard error of the mean (SEM):
$\text{SEM} = \frac{s}{\sqrt{n}} = \frac{0.52}{\sqrt{951}}$
$\text{SEM} \approx 0.0168$

STEP 5

Determine the critical t-value for a 95% confidence level with $n - 1 = 950$ degrees of freedom. Using a t-table or calculator, find:
$t^* \approx 1.96$ (assuming normal approximation due to large sample size)

STEP 6

Calculate the confidence interval:
$\text{Margin of Error} = t^* \times \text{SEM} = 1.96 \times 0.0168$
$\text{Margin of Error} \approx 0.0329$
$\text{Confidence Interval} = \left( \bar{x} - \text{Margin of Error}, \bar{x} + \text{Margin of Error} \right)$
$\text{Confidence Interval} = \left( 1.32 - 0.0329, 1.32 + 0.0329 \right)$
$\text{Confidence Interval} \approx (1.2871, 1.3529)$

STEP 7

Interpret the confidence interval: The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day for Americans age 15 or older is between 1.287 and 1.353 hours.

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Studdy Solution

STEP 1

1. The sample size is $n = 951$ .
2. The sample mean is $\bar{x} = 1.32$ hours.
3. The sample standard deviation is $s = 0.52$ hours.
4. The population size is more than 200 million people.
5. We are constructing a 95% confidence interval for the mean.

STEP 2

1. Verify the conditions for constructing a confidence interval.
2. Calculate the standard error of the mean.
3. Determine the critical t-value for a 95% confidence level.
4. Calculate the confidence interval.
5. Interpret the confidence interval.

STEP 3

Verify the conditions for constructing a confidence interval: - The sample size is $n = 951$ , which is large enough for the Central Limit Theorem to apply. - The sample is random. - The sample size is less than 5% of the population, satisfying the independence requirement.

STEP 4

Calculate the standard error of the mean (SEM):
$\text{SEM} = \frac{s}{\sqrt{n}} = \frac{0.52}{\sqrt{951}}$
$\text{SEM} \approx 0.0168$

STEP 5

Determine the critical t-value for a 95% confidence level with $n - 1 = 950$ degrees of freedom. Using a t-table or calculator, find:
$t^* \approx 1.96$ (assuming normal approximation due to large sample size)

STEP 6

STEP 7

Interpret the confidence interval: The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day for Americans age 15 or older is between 1.287 and 1.353 hours.

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Studdy Solution

STEP 1

1. The sample size is n=951 n = 951 n=951.2. The sample mean is xˉ=1.32 \bar{x} = 1.32 xˉ=1.32 hours.3. The sample standard deviation is s=0.52 s = 0.52 s=0.52 hours.4. The population size is more than 200 million people.5. We are constructing a 95% confidence interval for the mean.

STEP 2

1. Verify the conditions for constructing a confidence interval.2. Calculate the standard error of the mean.3. Determine the critical t-value for a 95% confidence level.4. Calculate the confidence interval.5. Interpret the confidence interval.

STEP 3

Verify the conditions for constructing a confidence interval: - The sample size is n=951 n = 951 n=951, which is large enough for the Central Limit Theorem to apply. - The sample is random. - The sample size is less than 5% of the population, satisfying the independence requirement.

STEP 4

Calculate the standard error of the mean (SEM):SEM=sn=0.52951 \text{SEM} = \frac{s}{\sqrt{n}} = \frac{0.52}{\sqrt{951}} SEM=n​s​=951​0.52​SEM≈0.0168 \text{SEM} \approx 0.0168 SEM≈0.0168

STEP 5

Determine the critical t-value for a 95% confidence level with n−1=950 n - 1 = 950 n−1=950 degrees of freedom. Using a t-table or calculator, find:t∗≈1.96 t^* \approx 1.96 t∗≈1.96 (assuming normal approximation due to large sample size)

STEP 6

STEP 7

Interpret the confidence interval: The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day for Americans age 15 or older is between 1.287 and 1.353 hours.

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1. The sample size is $n = 951$ .
2. The sample mean is $\bar{x} = 1.32$ hours.
3. The sample standard deviation is $s = 0.52$ hours.
4. The population size is more than 200 million people.
5. We are constructing a 95% confidence interval for the mean.

1. Verify the conditions for constructing a confidence interval.
2. Calculate the standard error of the mean.
3. Determine the critical t-value for a 95% confidence level.
4. Calculate the confidence interval.
5. Interpret the confidence interval.

Verify the conditions for constructing a confidence interval: - The sample size is $n = 951$ , which is large enough for the Central Limit Theorem to apply. - The sample is random. - The sample size is less than 5% of the population, satisfying the independence requirement.

Calculate the standard error of the mean (SEM):
$\text{SEM} = \frac{s}{\sqrt{n}} = \frac{0.52}{\sqrt{951}}$
$\text{SEM} \approx 0.0168$

Determine the critical t-value for a 95% confidence level with $n - 1 = 950$ degrees of freedom. Using a t-table or calculator, find:
$t^* \approx 1.96$ (assuming normal approximation due to large sample size)