An assembly line has a machine that welds two parts together. When the machine is properly calibrated, its welds have a mean length of 24.00 mm with a standard deviation of 0.27 mm. It has been a while since the machine has been serviced. Because of this, the foreman has told you to perform a hypothesis test to determine if the standard deviation, o, of the weld lengths is greater than when this machine is properly calibrated. To do so, you choose at random a sample of size 13 and examine the weld lengths. You find they have a sample standard deviation of 0.32 mm. Assume the weld lengths follow a normal distribution. Is there enough evidence to conclude that the population standard deviation, o, is greater than 0.27 mm? To answer, complete the parts below to perform a hypothesis test. Use the 0.10 level of significance. (a) State the null hypothesis Ho and the alternative hypothesis H₁ that you would use for the test.
An assembly line has a machine that welds two parts together. When the machine is properly calibrated, its welds have a mean length of 24.00 mm with a standard deviation of 0.27 mm. It has been a while since the machine has been serviced. Because of this, the foreman has told you to perform a hypothesis test to determine if the standard deviation, o, of the weld lengths is greater than when this machine is properly calibrated. To do so, you choose at random a sample of size 13 and examine the weld lengths. You find they have a sample standard deviation of 0.32 mm. Assume the weld lengths follow a normal distribution. Is there enough evidence to conclude that the population standard deviation, o, is greater than 0.27 mm? To answer, complete the parts below to perform a hypothesis test. Use the 0.10 level of significance. (a) State the null hypothesis Ho and the alternative hypothesis H₁ that you would use for the test.
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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![**Hypothesis Testing for Standard Deviation of Weld Lengths**
An assembly line has a machine that welds two parts together. When properly calibrated, its welds have a mean length of 24.00 mm with a standard deviation of 0.27 mm. Since it has been a while since the machine was serviced, a hypothesis test is required to determine if the standard deviation, σ, of the weld lengths is greater than 0.27 mm when properly calibrated. A sample of size 13 with a sample standard deviation of 0.32 mm is used, assuming a normal distribution of weld lengths.
**Objective:**
Determine if there’s enough evidence to conclude that the population standard deviation, σ, is greater than 0.27 mm using a 0.10 level of significance.
**Steps:**
**(a) State the Hypotheses:**
- Null Hypothesis (\( H_0 \)): \( \sigma \leq 0.27 \) mm
- Alternative Hypothesis (\( H_1 \)): \( \sigma > 0.27 \) mm
**(b) Perform a Chi-Square Test:**
- Use the test statistic:
\[
\chi^2 = \frac{{(n-1)s^2}}{\sigma^2}
\]
where \( n = 13 \) is the sample size, \( s = 0.32 \) mm is the sample standard deviation, and \( \sigma = 0.27 \) mm.
- The p-value is the area under the chi-square distribution curve to the right of the test statistic value.
**Chi-Square Distribution Diagram:**
1. **Step 1:** Enter number of degrees of freedom (n-1).
2. **Step 2:** Select one-tailed or two-tailed – One-tailed is selected.
3. **Step 3:** Enter the test statistic (round to 3 decimal places).
4. **Step 4:** Shade the area represented by the p-value on a chi-square graph.
5. **Step 5:** Enter the p-value (round to 3 decimal places).
**(c) Conclusion Based on Significance Level:**
Evaluate the p-value against the level of significance (0.10):
- If the p-value is less than or equal to the level of significance, reject \( H_0 \). Conclude that there is enough evidence](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c10e8ad-3433-42a7-ad9a-20db80d2b0e4%2F67abe6fe-4c9d-4cea-ae71-e83dd7978994%2F5sryv76_processed.png&w=3840&q=75)
Transcribed Image Text:**Hypothesis Testing for Standard Deviation of Weld Lengths**
An assembly line has a machine that welds two parts together. When properly calibrated, its welds have a mean length of 24.00 mm with a standard deviation of 0.27 mm. Since it has been a while since the machine was serviced, a hypothesis test is required to determine if the standard deviation, σ, of the weld lengths is greater than 0.27 mm when properly calibrated. A sample of size 13 with a sample standard deviation of 0.32 mm is used, assuming a normal distribution of weld lengths.
**Objective:**
Determine if there’s enough evidence to conclude that the population standard deviation, σ, is greater than 0.27 mm using a 0.10 level of significance.
**Steps:**
**(a) State the Hypotheses:**
- Null Hypothesis (\( H_0 \)): \( \sigma \leq 0.27 \) mm
- Alternative Hypothesis (\( H_1 \)): \( \sigma > 0.27 \) mm
**(b) Perform a Chi-Square Test:**
- Use the test statistic:
\[
\chi^2 = \frac{{(n-1)s^2}}{\sigma^2}
\]
where \( n = 13 \) is the sample size, \( s = 0.32 \) mm is the sample standard deviation, and \( \sigma = 0.27 \) mm.
- The p-value is the area under the chi-square distribution curve to the right of the test statistic value.
**Chi-Square Distribution Diagram:**
1. **Step 1:** Enter number of degrees of freedom (n-1).
2. **Step 2:** Select one-tailed or two-tailed – One-tailed is selected.
3. **Step 3:** Enter the test statistic (round to 3 decimal places).
4. **Step 4:** Shade the area represented by the p-value on a chi-square graph.
5. **Step 5:** Enter the p-value (round to 3 decimal places).
**(c) Conclusion Based on Significance Level:**
Evaluate the p-value against the level of significance (0.10):
- If the p-value is less than or equal to the level of significance, reject \( H_0 \). Conclude that there is enough evidence
![Over the years, a sushi restaurant has had a mean customer satisfaction rating of 71.2 with a variance of 26.8. The owner wants to see if using a new menu will have any effect on the variance, \(\sigma^2\). The owner surveys a random sample of 18 customers who ordered from the new menu. For the customers surveyed, the variance of ratings is 45.1. Assume the ratings for customers who order from the new menu follow a normal distribution.
Is there enough evidence to conclude that the population variance, \(\sigma^2\), differs from 26.8? To answer, complete the parts below to perform a hypothesis test. Use the 0.05 level of significance.
**(a)** State the null hypothesis \(H_0\) and the alternative hypothesis \(H_1\) that you would use for the test.
\(H_0:\ \) [ ]
\(H_1:\ \) [ ]
![Diagram of hypothesis choice description between sigma and s]
**(b)** Perform a chi-square test and find the p-value.
Here is some information to help you with your chi-square test.
- The value of the test statistic is given by \(\chi^2 = \frac{{(n-1)s^2}}{{\sigma^2}}\).
- The p-value is two times the area under the curve to the right of the value of the test statistic.
**Chi-square Distribution:**
- **Step 1:** Enter the number of degrees of freedom.
- **Step 2:** Select one-tailed or two-tailed.
- One-tailed ⃝
- Two-tailed ⃝
- **Step 3:** Enter the test statistic. (Round to 3 decimal places.)
- **Step 4:** Shade the area represented by the p-value.
- **Step 5:** Enter the p-value. (Round to 3 decimal places.)
(Graph showing chi-square distribution curve with x-axis ranging from 0 to 35)
**(c)** Based on your answer to part (b), choose what can be concluded, at the 0.05 level of significance, about the population variance of ratings for customers who order from the new menu.
- ○ Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to conclude that the variance differs from](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c10e8ad-3433-42a7-ad9a-20db80d2b0e4%2F67abe6fe-4c9d-4cea-ae71-e83dd7978994%2F9d028om_processed.png&w=3840&q=75)
Transcribed Image Text:Over the years, a sushi restaurant has had a mean customer satisfaction rating of 71.2 with a variance of 26.8. The owner wants to see if using a new menu will have any effect on the variance, \(\sigma^2\). The owner surveys a random sample of 18 customers who ordered from the new menu. For the customers surveyed, the variance of ratings is 45.1. Assume the ratings for customers who order from the new menu follow a normal distribution.
Is there enough evidence to conclude that the population variance, \(\sigma^2\), differs from 26.8? To answer, complete the parts below to perform a hypothesis test. Use the 0.05 level of significance.
**(a)** State the null hypothesis \(H_0\) and the alternative hypothesis \(H_1\) that you would use for the test.
\(H_0:\ \) [ ]
\(H_1:\ \) [ ]
![Diagram of hypothesis choice description between sigma and s]
**(b)** Perform a chi-square test and find the p-value.
Here is some information to help you with your chi-square test.
- The value of the test statistic is given by \(\chi^2 = \frac{{(n-1)s^2}}{{\sigma^2}}\).
- The p-value is two times the area under the curve to the right of the value of the test statistic.
**Chi-square Distribution:**
- **Step 1:** Enter the number of degrees of freedom.
- **Step 2:** Select one-tailed or two-tailed.
- One-tailed ⃝
- Two-tailed ⃝
- **Step 3:** Enter the test statistic. (Round to 3 decimal places.)
- **Step 4:** Shade the area represented by the p-value.
- **Step 5:** Enter the p-value. (Round to 3 decimal places.)
(Graph showing chi-square distribution curve with x-axis ranging from 0 to 35)
**(c)** Based on your answer to part (b), choose what can be concluded, at the 0.05 level of significance, about the population variance of ratings for customers who order from the new menu.
- ○ Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to conclude that the variance differs from
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