College

**Thesis Testing - Means Quiz**

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**Question 8 of 8**

This quiz: 21 points possible
This question: 4 points possible

A data set about speed dating includes "like" ratings of male dates made by the female dates. The summary statistics are: [tex]n=186, \bar{x}=5.82, s=2.08[/tex]. Use a 0.05 significance level to test the claim that the population mean of such ratings is less than 6.00. Assume that a simple random sample has been selected.

1. **Identify the null and alternative hypotheses:**

A. [tex]H_0: \mu < 6.00[/tex], [tex]H_1: \mu > 6.00[/tex]
B. [tex]H_0: \mu = 6.00[/tex], [tex]H_1: \mu < 6.00[/tex]
C. [tex]H_0: \mu = 6.00[/tex], [tex]H_1: \mu \neq 6.00[/tex]
D. [tex]H_0: \mu = 6.00[/tex], [tex]H_1: \mu > 6.00[/tex]

2. **Determine the test statistic:**
[tex]\square[/tex] (Round to two decimal places as needed.)

3. **Determine the P-value:**
[tex]\square[/tex] (Round to three decimal places as needed.)

4. **State the final conclusion that addresses the original claim:**
Choose the correct statement from the options above.

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**Note:** Please calculate the test statistic and P-value using the appropriate statistical methods or software to complete the analysis.

Answer :

To test the claim that the population mean of "like" ratings is less than 6.00, we'll use hypothesis testing with the following steps:

Step 1: Define the Hypotheses

- Null Hypothesis ([tex]\(H_0\)[/tex]): The population mean [tex]\(\mu\)[/tex] is equal to 6.00. (i.e., [tex]\(\mu = 6.00\)[/tex])
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The population mean [tex]\(\mu\)[/tex] is less than 6.00. (i.e., [tex]\(\mu < 6.00\)[/tex])

Step 2: Gather the Given Data

- Sample size ([tex]\(n\)[/tex]) = 186
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 5.82
- Sample standard deviation ([tex]\(s\)[/tex]) = 2.08
- Population mean under the null hypothesis = 6.00
- Significance level ([tex]\(\alpha\)[/tex]) = 0.05

Step 3: Calculate the Test Statistic

We use a t-test since the sample size is relatively large and the population standard deviation is unknown. The test statistic [tex]\(t\)[/tex] is calculated using the formula:

[tex]\[
t = \frac{\bar{x} - \text{Population mean}}{s / \sqrt{n}}
\][/tex]

Substituting the given values:

[tex]\[
t = \frac{5.82 - 6.00}{2.08 / \sqrt{186}}
\][/tex]

This calculation leads to a test statistic of approximately [tex]\(t = -1.18\)[/tex].

Step 4: Determine the P-value

The p-value is the probability of obtaining a test statistic at least as extreme as the one computed, assuming the null hypothesis is true. For a one-tailed t-test with 185 degrees of freedom (since [tex]\(n-1=185\)[/tex]), calculate the p-value corresponding to the test statistic [tex]\(t = -1.18\)[/tex].

The p-value is approximately 0.12.

Step 5: Make a Decision

Compare the p-value to the significance level [tex]\(\alpha = 0.05\)[/tex]:

- If the p-value is less than [tex]\(\alpha\)[/tex], reject the null hypothesis.
- If the p-value is greater than or equal to [tex]\(\alpha\)[/tex], fail to reject the null hypothesis.

Since the p-value (0.12) is greater than the significance level (0.05), we fail to reject the null hypothesis.

Step 6: Conclusion

Based on the hypothesis test, we do not have sufficient evidence to support the claim that the population mean of "like" ratings is less than 6.00. Therefore, we conclude that there is not enough evidence to support that the population mean is less than 6.00.