Find the standard score (round to two decimal places as needed)  z=  Find the P-value (round to four decimal places as needed) P-value= Make an conclusion

MATLAB: An Introduction with Applications
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Chapter1: Starting With Matlab
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Please don't forget to read this! There are 3 more parts to the question. 

Find the standard score (round to two decimal places as needed) 

z= 

Find the P-value (round to four decimal places as needed)

P-value=

Make an conclusion

### Hypothesis Testing for Compulsive Buyers Questionnaire Scores

#### Problem Statement
A random sample of 36 subjects who identified themselves as compulsive buyers was obtained and given a questionnaire. They had a mean questionnaire score of 0.59 with a standard deviation of 0.18. Test the claim that the population of self-identified compulsive buyers has a mean greater than the mean of 0.51 for the general population. Use a 0.05 significance level.

---

#### Step 1: State the Null and Alternative Hypotheses
Select the correct null and alternative hypotheses for the test:

**A.** \( H_0: \mu = 0.51 \)
\[ H_a: \mu < 0.51 \]

**B.** (Correct)
\[ H_0: \mu = 0.51 \]
\[ H_a: \mu > 0.51 \]

**C.**
\[ H_0: \mu \neq 0.51 \]
\[ H_a: \mu = 0.51 \]

**D.**
\[ H_0: \mu = 0.51 \]
\[ H_a: \mu \neq 0.51 \]

---

#### Step 2: Calculate the Z-Score

Using the provided data:
- Sample Mean (\( \bar{x} \)) = 0.59
- Population Mean (\( \mu \)) = 0.51
- Standard Deviation (\( \sigma \)) = 0.18
- Sample Size (\( n \)) = 36

First, find the Standard Error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{0.18}{\sqrt{36}} = \frac{0.18}{6} = 0.03 \]

Then, calculate the Z-score:
\[ Z = \frac{\bar{x} - \mu}{SE} = \frac{0.59 - 0.51}{0.03} = \frac{0.08}{0.03} = 2.67 \]

So, the Z-score is:
\[ Z = \boxed{2.67} \]

*(Round to two decimal places as needed.)*

---
This Z-score can be used in further steps to determine the corresponding p-value, and to conclude whether there is sufficient evidence to reject the null hypothesis in favor of
Transcribed Image Text:### Hypothesis Testing for Compulsive Buyers Questionnaire Scores #### Problem Statement A random sample of 36 subjects who identified themselves as compulsive buyers was obtained and given a questionnaire. They had a mean questionnaire score of 0.59 with a standard deviation of 0.18. Test the claim that the population of self-identified compulsive buyers has a mean greater than the mean of 0.51 for the general population. Use a 0.05 significance level. --- #### Step 1: State the Null and Alternative Hypotheses Select the correct null and alternative hypotheses for the test: **A.** \( H_0: \mu = 0.51 \) \[ H_a: \mu < 0.51 \] **B.** (Correct) \[ H_0: \mu = 0.51 \] \[ H_a: \mu > 0.51 \] **C.** \[ H_0: \mu \neq 0.51 \] \[ H_a: \mu = 0.51 \] **D.** \[ H_0: \mu = 0.51 \] \[ H_a: \mu \neq 0.51 \] --- #### Step 2: Calculate the Z-Score Using the provided data: - Sample Mean (\( \bar{x} \)) = 0.59 - Population Mean (\( \mu \)) = 0.51 - Standard Deviation (\( \sigma \)) = 0.18 - Sample Size (\( n \)) = 36 First, find the Standard Error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{0.18}{\sqrt{36}} = \frac{0.18}{6} = 0.03 \] Then, calculate the Z-score: \[ Z = \frac{\bar{x} - \mu}{SE} = \frac{0.59 - 0.51}{0.03} = \frac{0.08}{0.03} = 2.67 \] So, the Z-score is: \[ Z = \boxed{2.67} \] *(Round to two decimal places as needed.)* --- This Z-score can be used in further steps to determine the corresponding p-value, and to conclude whether there is sufficient evidence to reject the null hypothesis in favor of
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