A data set about speed dating includes "like" ratings of male dates made by the female dates. The summary statistics are [tex]n=188, \bar{x}=7.87, s=1.92[/tex]. Use a 0.01 significance level to test the claim that the population mean of such ratings is less than 8.00. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, [tex]P[/tex]-value, and state the final conclusion that addresses the original claim.

### What are the null and alternative hypotheses?

A. [tex]H_0: \mu\ \textless \ 8.00[/tex]

B.
[tex]\[
\begin{array}{l}
H_0: \mu=8.00 \\
H_1: \mu\ \textgreater \ 8.00
\end{array}
\][/tex]

C.
[tex]\[
\begin{array}{l}
H_0: \mu=8.00 \\
H_1: \mu \neq 8.00
\end{array}
\][/tex]

D.
[tex]\[
\begin{array}{l}
H_0: \mu=8.00 \\
H_1: \mu\ \textless \ 8.00
\end{array}
\][/tex]

### Determine the test statistic.

[tex]\[ \square \][/tex] (Round to two decimal places as needed.)



Answer :

To address the claim that the population mean of the "like" ratings is less than 8.00 with the given data, let's follow the hypothesis testing steps.

### Step 1: State the Hypotheses

The null hypothesis ([tex]\(H_0\)[/tex]) and the alternative hypothesis ([tex]\(H_1\)[/tex]) are formulated based on the claim that the population mean is less than 8.00.

- The null hypothesis ([tex]\(H_0\)[/tex]): The population mean is equal to 8.00.
- The alternative hypothesis ([tex]\(H_1\)[/tex]): The population mean is less than 8.00.

So the correct hypotheses are:

[tex]\[ \begin{aligned} H_0: \mu = 8.00 \\ H_1: \mu < 8.00 \end{aligned} \][/tex]

This matches option D.

### Step 2: Calculate the Test Statistic

The test statistic for a t-test is calculated using the formula:

[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \][/tex]

Given data:
- [tex]\(\bar{x} = 7.87\)[/tex] (sample mean)
- [tex]\(n = 188\)[/tex] (sample size)
- [tex]\(s = 1.92\)[/tex] (sample standard deviation)
- [tex]\(\mu_0 = 8.00\)[/tex] (mean under the null hypothesis)

Plugging these values into the formula, we have:

[tex]\[ t = \frac{7.87 - 8.00}{\frac{1.92}{\sqrt{188}}} = \frac{-0.13}{\frac{1.92}{\sqrt{188}}} \approx -0.93 \][/tex]

So, the test statistic is approximatively:

[tex]\[ t \approx -0.93 \][/tex]

### Step 3: Calculate the P-Value

Using the t-distribution with [tex]\(df = n - 1 = 188 - 1 = 187\)[/tex] degrees of freedom, you can find the p-value associated with the test statistic. For a left-tailed test, the p-value is the area to the left of [tex]\(t = -0.93\)[/tex].

The p-value can be found using a t-distribution table or computational tools. In this case, the p-value is:

[tex]\[ p \approx 0.17721 \][/tex]

### Step 4: Compare P-Value with Significance Level and Make a Conclusion

The significance level ([tex]\(\alpha\)[/tex]) is given as 0.01. To make a decision, we compare the p-value to the significance level:

- If [tex]\(p\)[/tex]-value < [tex]\(\alpha\)[/tex]: Reject the null hypothesis.
- If [tex]\(p\)[/tex]-value >= [tex]\(\alpha\)[/tex]: Fail to reject the null hypothesis.

In this case:

[tex]\[ 0.17721 > 0.01 \][/tex]

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

### Conclusion

Based on the test statistic of approximately -0.93 and a p-value of approximately 0.17721, we do not have sufficient evidence to reject the null hypothesis at the 0.01 significance level. Thus, we conclude that there is not enough evidence to support the claim that the population mean of "like" ratings is less than 8.00.