The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1800 pounds and a standard deviation of 60 pounds. The company believes that, due to an improvement in the manufacturing process, the mean breaking strength, μ, of the cables is now greater than 1800 pounds. To see if this is the case, 80 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1818 pounds. Can we support, at the 0.01 level of significance, the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1800 pounds? Assume that the population standard deviation has not changed. Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis Ho and the alternative hypothesis H₁. Ho :O H₁ :0 (b) Determine the type of test statistic to use. (Choose one) ▼ (c) Find the value of the test statistic. (Round to three or more decimal places.) I |x 딤 a S 00 ロ=ロ OSO 0#0 O
The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1800 pounds and a standard deviation of 60 pounds. The company believes that, due to an improvement in the manufacturing process, the mean breaking strength, μ, of the cables is now greater than 1800 pounds. To see if this is the case, 80 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1818 pounds. Can we support, at the 0.01 level of significance, the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1800 pounds? Assume that the population standard deviation has not changed. Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis Ho and the alternative hypothesis H₁. Ho :O H₁ :0 (b) Determine the type of test statistic to use. (Choose one) ▼ (c) Find the value of the test statistic. (Round to three or more decimal places.) I |x 딤 a S 00 ロ=ロ OSO 0#0 O
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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![The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1800 pounds and a standard deviation of 60 pounds. The company believes that, due to an improvement in the manufacturing process, the mean breaking strength, μ, of the cables is now greater than 1800 pounds. To see if this is the case, 80 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1818 pounds. Can we support, at the 0.01 level of significance, the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1800 pounds? Assume that the population standard deviation has not changed.
Perform a one-tailed test. Then complete the parts below.
Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.)
(a) State the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \).
\[ H_0: \]
\[ H_1: \]
(b) Determine the type of test statistic to use.
(Choose one) ▼
(c) Find the value of the test statistic. (Round to three or more decimal places.)
\[ \]
(d) Find the critical value. (Round to three or more decimal places.)
\[ \]
(e) Can we support the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1800 pounds?
○ Yes ○ No
The image also includes a symbol and formula menu with symbols \( \mu, \sigma, p, \bar{x}, s, \hat{p} \) and relational operators such as \( <, \leq, =, \neq, \geq, > \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c10e8ad-3433-42a7-ad9a-20db80d2b0e4%2F7b29b581-af06-453d-a53f-480bd48950e3%2Fkq7pqqi_processed.png&w=3840&q=75)
Transcribed Image Text:The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1800 pounds and a standard deviation of 60 pounds. The company believes that, due to an improvement in the manufacturing process, the mean breaking strength, μ, of the cables is now greater than 1800 pounds. To see if this is the case, 80 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1818 pounds. Can we support, at the 0.01 level of significance, the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1800 pounds? Assume that the population standard deviation has not changed.
Perform a one-tailed test. Then complete the parts below.
Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.)
(a) State the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \).
\[ H_0: \]
\[ H_1: \]
(b) Determine the type of test statistic to use.
(Choose one) ▼
(c) Find the value of the test statistic. (Round to three or more decimal places.)
\[ \]
(d) Find the critical value. (Round to three or more decimal places.)
\[ \]
(e) Can we support the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1800 pounds?
○ Yes ○ No
The image also includes a symbol and formula menu with symbols \( \mu, \sigma, p, \bar{x}, s, \hat{p} \) and relational operators such as \( <, \leq, =, \neq, \geq, > \).
![According to previous studies, the mean distance each visitor in Greenspan National Park hikes during their visit is 27 kilometers. The park recently closed its shuttle system, which used to transport hikers to many of the park’s most popular hiking trails. Because of this, an administrator at the park suspects the mean distance, \(\mu\), is now less than 27 kilometers. The administrator chooses a random sample of 78 visitors. The mean distance hiked for the sample is 25.6 kilometers. Assume the population standard deviation is 7.1 kilometers.
Can the administrator conclude that the mean distance hiked by each visitor is now less than 27 kilometers? Perform a hypothesis test, using the 0.05 level of significance.
**(a) State the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \).**
- \( H_0: \) [Insert the mathematical expression here]
- \( H_1: \) [Insert the mathematical expression here]
**(b) Perform a Z-test and find the p-value.**
Here is some information to help you with your Z-test.
- The value of the test statistic is given by
\[
\frac{\overline{x} - \mu}{\sigma / \sqrt{n}}
\]
- The p-value is the area under the curve to the left of the value of the test statistic.
**Standard Normal Distribution Instructions:**
1. Select one-tailed or two-tailed.
- One-tailed
- Two-tailed
2. Enter the test statistic. (Round to 3 decimal places.)
3. Shade the area represented by the p-value.
4. Enter the p-value. (Round to 3 decimal places.)
**Diagram Description:**
A bell curve graph represents the standard normal distribution, with the x-axis labeled from -3 to 3. Areas under the curve are marked with values 0.1, 0.2, and 0.3 to illustrate the cumulative distribution.
**(c) Based on your answer to part (b), choose what the administrator can conclude, at the 0.05 level of significance.**
- 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 mean distance hiked by each visitor is now less than 27 kilometers.
- Since the p-value is less](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c10e8ad-3433-42a7-ad9a-20db80d2b0e4%2F7b29b581-af06-453d-a53f-480bd48950e3%2Fuqeh3ba_processed.png&w=3840&q=75)
Transcribed Image Text:According to previous studies, the mean distance each visitor in Greenspan National Park hikes during their visit is 27 kilometers. The park recently closed its shuttle system, which used to transport hikers to many of the park’s most popular hiking trails. Because of this, an administrator at the park suspects the mean distance, \(\mu\), is now less than 27 kilometers. The administrator chooses a random sample of 78 visitors. The mean distance hiked for the sample is 25.6 kilometers. Assume the population standard deviation is 7.1 kilometers.
Can the administrator conclude that the mean distance hiked by each visitor is now less than 27 kilometers? Perform a hypothesis test, using the 0.05 level of significance.
**(a) State the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \).**
- \( H_0: \) [Insert the mathematical expression here]
- \( H_1: \) [Insert the mathematical expression here]
**(b) Perform a Z-test and find the p-value.**
Here is some information to help you with your Z-test.
- The value of the test statistic is given by
\[
\frac{\overline{x} - \mu}{\sigma / \sqrt{n}}
\]
- The p-value is the area under the curve to the left of the value of the test statistic.
**Standard Normal Distribution Instructions:**
1. Select one-tailed or two-tailed.
- One-tailed
- Two-tailed
2. Enter the test statistic. (Round to 3 decimal places.)
3. Shade the area represented by the p-value.
4. Enter the p-value. (Round to 3 decimal places.)
**Diagram Description:**
A bell curve graph represents the standard normal distribution, with the x-axis labeled from -3 to 3. Areas under the curve are marked with values 0.1, 0.2, and 0.3 to illustrate the cumulative distribution.
**(c) Based on your answer to part (b), choose what the administrator can conclude, at the 0.05 level of significance.**
- 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 mean distance hiked by each visitor is now less than 27 kilometers.
- Since the p-value is less
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