Answer :
To solve this hypothesis testing problem, we need to identify the null and alternative hypotheses, calculate the test statistic, determine the P-value, and state the conclusion based on the significance level.
1. Identify the null and alternative hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): The population mean [tex]\(\mu\)[/tex] is equal to 6.00.
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The population mean [tex]\(\mu\)[/tex] is less than 6.00.
- Notation: [tex]\(H_0: \mu = 6.00\)[/tex], [tex]\(H_1: \mu < 6.00\)[/tex] (Choice B is correct).
2. Calculate the test statistic:
- Formula for the test statistic in a t-test is:
[tex]\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\][/tex]
where:
- [tex]\(\bar{x} = 5.82\)[/tex] is the sample mean.
- [tex]\(\mu_0 = 6.00\)[/tex] is the population mean under the null hypothesis.
- [tex]\(s = 2.08\)[/tex] is the sample standard deviation.
- [tex]\(n = 186\)[/tex] is the sample size.
- Substituting these values into the formula gives us a test statistic of [tex]\(-1.18\)[/tex] (rounded to two decimal places).
3. Determine the P-value:
- The P-value is found using the t-distribution and indicates the probability of observing a test statistic as extreme as, or more extreme than, the observed value, assuming the null hypothesis is true.
- For a left-tailed test (since [tex]\(H_1: \mu < 6.00\)[/tex]), the P-value is calculated from the t-distribution with [tex]\(n - 1 = 185\)[/tex] degrees of freedom.
- The P-value is 0.12 (rounded to three decimal places).
4. State the final conclusion:
- Compare the P-value to the significance level ([tex]\(\alpha = 0.06\)[/tex]).
- Since the P-value (0.12) is greater than the significance level (0.06), we do not reject the null hypothesis.
- Conclusion: There is not enough evidence to support the claim that the population mean of such ratings is less than 6.00.
In summary, the statistical analysis indicates that based on the available data and a significance level of 0.06, we do not have sufficient evidence to claim that the mean "Ike" ratings of male dates by female dates in speed dating is less than 6.00.
1. Identify the null and alternative hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): The population mean [tex]\(\mu\)[/tex] is equal to 6.00.
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The population mean [tex]\(\mu\)[/tex] is less than 6.00.
- Notation: [tex]\(H_0: \mu = 6.00\)[/tex], [tex]\(H_1: \mu < 6.00\)[/tex] (Choice B is correct).
2. Calculate the test statistic:
- Formula for the test statistic in a t-test is:
[tex]\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\][/tex]
where:
- [tex]\(\bar{x} = 5.82\)[/tex] is the sample mean.
- [tex]\(\mu_0 = 6.00\)[/tex] is the population mean under the null hypothesis.
- [tex]\(s = 2.08\)[/tex] is the sample standard deviation.
- [tex]\(n = 186\)[/tex] is the sample size.
- Substituting these values into the formula gives us a test statistic of [tex]\(-1.18\)[/tex] (rounded to two decimal places).
3. Determine the P-value:
- The P-value is found using the t-distribution and indicates the probability of observing a test statistic as extreme as, or more extreme than, the observed value, assuming the null hypothesis is true.
- For a left-tailed test (since [tex]\(H_1: \mu < 6.00\)[/tex]), the P-value is calculated from the t-distribution with [tex]\(n - 1 = 185\)[/tex] degrees of freedom.
- The P-value is 0.12 (rounded to three decimal places).
4. State the final conclusion:
- Compare the P-value to the significance level ([tex]\(\alpha = 0.06\)[/tex]).
- Since the P-value (0.12) is greater than the significance level (0.06), we do not reject the null hypothesis.
- Conclusion: There is not enough evidence to support the claim that the population mean of such ratings is less than 6.00.
In summary, the statistical analysis indicates that based on the available data and a significance level of 0.06, we do not have sufficient evidence to claim that the mean "Ike" ratings of male dates by female dates in speed dating is less than 6.00.
