Question

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Speed Dating Data Set 18 “Speed Dating” in Appendix B includes “attractive” ratings of male dates made by the female dates. The summary statistics are n = 199, x = 6.19, s = 1.99. Use a 0.01 significance level to test the claim that the population mean of such ratings is less than 7.00.

Short answer

The hypotheses are as follows.

$$\begin{gathered} H_0:\mu ⩾7.00 \\ {H}_1:\mu <7.00 \end{gathered}$$

The test statistic is -5.742.

The critical value is -2.35.

The null hypothesis is rejected.

Thus, there is enough evidence to support the claim that the population mean of the rating of the dates is less than 7.00.

Step-by-step solution

Given information

A sample is taken from females about the ratings of male dates. Summary statistics:

$$\begin{gathered} n=199 \\ \overline{x}=6.19 \\ s=1.99 \end{gathered}$$

The significance level is 0.01 to test the claim that the mean of such ratings for the population is 7.00.

Check the requirements

The required conditions are verified as follows.

1. The sample, satisfies the condition of asimple random sampling, as the sample is randomly collected for the ratings of the dates..
2. The sample size of199 is greater than 30. Thus, there is no need to check for normality of the sample, as the condition for the sample size is satisfied.

dfsdfsdfsd

As the population standard deviation is unknown, the t-test applies.

State the hypotheses

The null hypothesis $H_0$represents the population mean of the rating for the dates, which is greater than or equal to 7. As equality is not present in the original claim, the alternate hypothesis $H_1$represents the population mean of the rating for the dates, which is less than 7.

Let $\mu$be the population mean of the rating of the male dates.

State the null and alternate hypotheses.

role="math" localid="1648806642242" $$\begin{gathered} H_0:\mu ⩾7 \\ {H}_1:\mu <7 \end{gathered}$$

Thus, the test is left-tailed.

State the critical value

The degree of freedom is obtained by using the formula $df=n-1$, where, $n=199$as shown below.

$$\begin{gathered} df=199-1 \\ =198 \end{gathered}$$

The critical value can be obtained using the t distribution table with 198 and the significance level 0.01 for a one-tailed test.

From the t-table, the value corresponding to row 198 and column 0.01 for a one-tailed test is 2.35.

Thus, $t_{0.01}=2.35$.

Compute the observed test statistic

Apply the t-test to compute the test statistic using the formula $t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$.

Substitute the respective value in the above formula and simplify.

$$\begin{gathered} t=\frac{6.19-7}{\frac{1.99}{\sqrt{199}}} \\ =-5.7419 \end{gathered}$$

State the decision 

Reject the null hypothesis when the absolute value of the observed test statistic is greater than the critical value. Otherwise, fail to reject the null hypothesis.

$$\begin{gathered} \left|t\right|=\left|-5.7419\right| \\ =5.7419 \\ >2.35\left(t_{0.01}\right) \end{gathered}$$

The absolute value of the observed test statistic is greater than the critical value. This implies that the null hypothesis is rejected.

Conclusion

As the null hypothesis is rejected at a 0.01 level of significance, it can be concluded that there is sufficient evidence to support the claim that the mean of ratings is lesser than 7.00.