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A data set about speed dating includes "like" ratings of male dates made by the female dates. The summary statistics are [tex]$n=190, \bar{x}=5.54, s=1.93$[/tex]. Use a 0.10 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. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.

1. What are the null and alternative hypotheses?
A. [tex]$H_0: \mu=6.00$[/tex]
B. [tex]$H_0: \mu\ \textless \ 6.00$[/tex], [tex]$H_1: \mu\ \textgreater \ 6.00$[/tex]
C. [tex]$H_0: \mu=6.00$[/tex], [tex]$H_1: \mu\ \textless \ 6.00$[/tex]
D. [tex]$H_0: \mu=6.00$[/tex], [tex]$H_1: \mu \neq 6.00$[/tex]

2. Determine the test statistic.
(Round to two decimal places as needed.)

Answer :

Let's solve the problem step-by-step to determine the test statistic, identify the null and alternative hypotheses, calculate the p-value, and make a conclusion to address the claim.

### Step 1: Identify the Null and Alternative Hypotheses

We are testing the claim that the population mean is less than 6.00. Therefore, the hypotheses are:

- Null Hypothesis (H₀): The population mean is equal to 6.00. [tex]\( H_0: \mu = 6.00 \)[/tex]
- Alternative Hypothesis (H₁): The population mean is less than 6.00. [tex]\( H_1: \mu < 6.00 \)[/tex]

So, the correct choice is C: [tex]\( H_0: \mu = 6.00 \)[/tex] and [tex]\( H_1: \mu < 6.00 \)[/tex].

### Step 2: Calculate the Test Statistic

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

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

Where:
- [tex]\(\bar{x} = 5.54\)[/tex] is the sample mean.
- [tex]\(\mu_0 = 6.00\)[/tex] is the hypothesized population mean under the null hypothesis.
- [tex]\(s = 1.93\)[/tex] is the sample standard deviation.
- [tex]\(n = 190\)[/tex] is the sample size.

Substituting these values into the formula gives us:

[tex]\[
t = \frac{5.54 - 6.00}{1.93 / \sqrt{190}} \approx -3.29
\][/tex]

### Step 3: Determine the P-value

For a one-tailed t-test, the p-value is the probability that the t-statistic is less than the observed value, given that the null hypothesis is true. Using the t-distribution with [tex]\(n - 1 = 189\)[/tex] degrees of freedom, we find:

The p-value is approximately 0.0006.

### Step 4: Make the Conclusion

With a significance level [tex]\(\alpha = 0.10\)[/tex], we compare the p-value to [tex]\(\alpha\)[/tex]:

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

Since the p-value [tex]\(0.0006\)[/tex] is less than the significance level [tex]\(0.10\)[/tex], we reject the null hypothesis.

### Final Conclusion

There is sufficient evidence to support the claim that the population mean of the "like" ratings is less than 6.00.

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