z = -1.61 (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. OB. O A. Fail to reject Ho. There is not sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams. 23 O C. Fail to reject Ho. There is sufficient evidence to reject the claim that mean sodium content is no more than 91.6 milligrams. Reject Ho. There is not sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams. 00 O D. Reject Ho. There is sufficient M evidence to reject the claim that mean sodium content is no A more than 916 milligrams.
z = -1.61 (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. OB. O A. Fail to reject Ho. There is not sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams. 23 O C. Fail to reject Ho. There is sufficient evidence to reject the claim that mean sodium content is no more than 91.6 milligrams. Reject Ho. There is not sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams. 00 O D. Reject Ho. There is sufficient M evidence to reject the claim that mean sodium content is no A more than 916 milligrams.
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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![**Hypothesis Testing in a Fast Food Scenario: Understanding Sodium Content Claims**
**Problem Statement:**
A fast food restaurant estimates that the mean sodium content in one of its breakfast sandwiches is no more than 916 milligrams. A random sample of 45 breakfast sandwiches has a mean sodium content of 910 milligrams. Assume the population standard deviation is 25 milligrams. At $\alpha = 0.10$, do you have enough evidence to reject the restaurant's claim? Complete parts (a) through (e).
**Step-by-Step Solution:**
**Z-Score Calculation:**
\[ z = \frac{\bar{X} - \mu}{(\sigma / \sqrt{n})} \]
Given:
- Sample mean (\(\bar{X}\)) = 910 mg
- Population mean (\(\mu\)) = 916 mg
- Population standard deviation (\(\sigma\)) = 25 mg
- Sample size (n) = 45
- Significance level (\(\alpha\)) = 0.10
Using the above information, we calculate:
\[ z = \frac{910 - 916}{(25 / \sqrt{45})} \]
\[ z = \frac{-6}{(25 / 6.708)} \]
\[ z = \frac{-6}{3.725} \]
\[ z = -1.61 \]
**Decision Making:**
(d) Decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim.
**Options Provided:**
A) Fail to reject \( H_0 \). There is not sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams.
B) Reject \( H_0 \). There is not sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams.
C) Fail to reject \( H_0 \). There is sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams.
D) Reject \( H_0 \). There is sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams.
**Interpreting the Result:**
Based on the calculated z-score of -1.61 and comparing it against the critical value for \( \alpha = 0.10 \) (approximately ±1.645 for a two-tailed test), the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F952bcd3e-4a7c-474a-8dd6-ddd08471460a%2F9a644283-8d49-4680-9049-aa7613e764f9%2Ftracjhr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Hypothesis Testing in a Fast Food Scenario: Understanding Sodium Content Claims**
**Problem Statement:**
A fast food restaurant estimates that the mean sodium content in one of its breakfast sandwiches is no more than 916 milligrams. A random sample of 45 breakfast sandwiches has a mean sodium content of 910 milligrams. Assume the population standard deviation is 25 milligrams. At $\alpha = 0.10$, do you have enough evidence to reject the restaurant's claim? Complete parts (a) through (e).
**Step-by-Step Solution:**
**Z-Score Calculation:**
\[ z = \frac{\bar{X} - \mu}{(\sigma / \sqrt{n})} \]
Given:
- Sample mean (\(\bar{X}\)) = 910 mg
- Population mean (\(\mu\)) = 916 mg
- Population standard deviation (\(\sigma\)) = 25 mg
- Sample size (n) = 45
- Significance level (\(\alpha\)) = 0.10
Using the above information, we calculate:
\[ z = \frac{910 - 916}{(25 / \sqrt{45})} \]
\[ z = \frac{-6}{(25 / 6.708)} \]
\[ z = \frac{-6}{3.725} \]
\[ z = -1.61 \]
**Decision Making:**
(d) Decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim.
**Options Provided:**
A) Fail to reject \( H_0 \). There is not sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams.
B) Reject \( H_0 \). There is not sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams.
C) Fail to reject \( H_0 \). There is sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams.
D) Reject \( H_0 \). There is sufficient evidence to reject the claim that mean sodium content is no more than 916 milligrams.
**Interpreting the Result:**
Based on the calculated z-score of -1.61 and comparing it against the critical value for \( \alpha = 0.10 \) (approximately ±1.645 for a two-tailed test), the
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