A recent study found that 61 children who watched a commercial for potato chips featuring a celebrity endorser ate a mean of 38 grams of potato chips as compared to a mean of 27 grams for 51 children who watched a commercial for an alternative food snack. Suppose that the sample standard deviation for the children who watched the celebrity-endorsed commercial was 21.1 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 12.8 grams. Complete parts (a) through (d) below. tSTAT = 3.25 (Round to two decimal places as needed.) What is the corresponding p-value? p-value = 0.001 (Round to three decimal places as needed.) What is the correct conclusion? A. Reject Ho. There is sufficient evidence that the mean amount of potato chips eaten was significantly higher for children who watched the celebrity-endorsed commercial. O B. Do not reject Ho. There is insufficient evidence that the mean amount of potato chips eaten was significantly higher for children who watched the celebrity-endorsed commercial. ⒸC. Reject Ho. There is insufficient evidence that the mean amount of potato chips eaten was significantly higher for children who watched the celebrity-endorsed commercial. O D. Do not reject Ho. There is sufficient evidence that the mean amount of potato chips eaten was significantly higher for children who watched the celebrity-endorsed commercial. b. Assuming that the population variances are equal, construct a 95% confidence interval estimate of the difference μ₁-₂ between the mean amount of potato chips eaten by the children who watched the celebrity-endorsed commercial and children who watched the alternative food snack commercial. ☐Sµ₁ ≤H1-H₂= (Type integers or decimals rounded to two decimal places as needed.)
A recent study found that 61 children who watched a commercial for potato chips featuring a celebrity endorser ate a mean of 38 grams of potato chips as compared to a mean of 27 grams for 51 children who watched a commercial for an alternative food snack. Suppose that the sample standard deviation for the children who watched the celebrity-endorsed commercial was 21.1 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 12.8 grams. Complete parts (a) through (d) below. tSTAT = 3.25 (Round to two decimal places as needed.) What is the corresponding p-value? p-value = 0.001 (Round to three decimal places as needed.) What is the correct conclusion? A. Reject Ho. There is sufficient evidence that the mean amount of potato chips eaten was significantly higher for children who watched the celebrity-endorsed commercial. O B. Do not reject Ho. There is insufficient evidence that the mean amount of potato chips eaten was significantly higher for children who watched the celebrity-endorsed commercial. ⒸC. Reject Ho. There is insufficient evidence that the mean amount of potato chips eaten was significantly higher for children who watched the celebrity-endorsed commercial. O D. Do not reject Ho. There is sufficient evidence that the mean amount of potato chips eaten was significantly higher for children who watched the celebrity-endorsed commercial. b. Assuming that the population variances are equal, construct a 95% confidence interval estimate of the difference μ₁-₂ between the mean amount of potato chips eaten by the children who watched the celebrity-endorsed commercial and children who watched the alternative food snack commercial. ☐Sµ₁ ≤H1-H₂= (Type integers or decimals rounded to two decimal places as needed.)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Study Overview:
A recent study involving 61 children who watched a commercial for potato chips featuring a celebrity endorser found they consumed an average of 38 grams of potato chips. This contrasts with an average consumption of 27 grams among 51 children who watched a commercial for an alternative food snack. The sample standard deviation for those who watched the celebrity-endorsed commercial was 21.1 grams, while for the alternative food snack commercial, it was 12.8 grams.
### Statistical Analysis:
#### a. Test Statistic Calculation
- **t\(_{STAT}\)** = 3.25
(Rounded to two decimal places)
#### b. p-value Determination
- **p-value** = 0.001
(Rounded to three decimal places)
#### c. Conclusion from Statistical Test
- **Correct Conclusion:**
- **A.** Reject H\(_0\). There is sufficient evidence that the mean amount of potato chips eaten was significantly higher for children who watched the celebrity-endorsed commercial.
#### d. Confidence Interval Construction
- Assuming equal population variances, construct a 95% confidence interval for the difference (\(\mu_1 - \mu_2\)) in the mean amount of potato chips consumed by children who watched the celebrity-endorsed commercial versus the alternative food snack commercial.
#### Interval:
- \(\text{Enter values as needed: } [\text{lower bound}] ≤ \mu_1 - \mu_2 ≤ [\text{upper bound}]\)
- (Type integers or decimals rounded to two decimal places as needed.)
### Interpretation:
The study suggests that exposure to celebrity endorsements in commercials may influence higher potato chip consumption among children compared to commercials for alternative snacks. This is supported by statistical evidence and a significant p-value. The confidence interval provides a range of plausible values for the mean difference in consumption.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b6853aa-68ca-417e-9091-86ddfc5176c4%2F22a00f1c-4d99-4b6f-853f-f33621744b60%2Faevbjt_processed.png&w=3840&q=75)
Transcribed Image Text:### Study Overview:
A recent study involving 61 children who watched a commercial for potato chips featuring a celebrity endorser found they consumed an average of 38 grams of potato chips. This contrasts with an average consumption of 27 grams among 51 children who watched a commercial for an alternative food snack. The sample standard deviation for those who watched the celebrity-endorsed commercial was 21.1 grams, while for the alternative food snack commercial, it was 12.8 grams.
### Statistical Analysis:
#### a. Test Statistic Calculation
- **t\(_{STAT}\)** = 3.25
(Rounded to two decimal places)
#### b. p-value Determination
- **p-value** = 0.001
(Rounded to three decimal places)
#### c. Conclusion from Statistical Test
- **Correct Conclusion:**
- **A.** Reject H\(_0\). There is sufficient evidence that the mean amount of potato chips eaten was significantly higher for children who watched the celebrity-endorsed commercial.
#### d. Confidence Interval Construction
- Assuming equal population variances, construct a 95% confidence interval for the difference (\(\mu_1 - \mu_2\)) in the mean amount of potato chips consumed by children who watched the celebrity-endorsed commercial versus the alternative food snack commercial.
#### Interval:
- \(\text{Enter values as needed: } [\text{lower bound}] ≤ \mu_1 - \mu_2 ≤ [\text{upper bound}]\)
- (Type integers or decimals rounded to two decimal places as needed.)
### Interpretation:
The study suggests that exposure to celebrity endorsements in commercials may influence higher potato chip consumption among children compared to commercials for alternative snacks. This is supported by statistical evidence and a significant p-value. The confidence interval provides a range of plausible values for the mean difference in consumption.
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