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

\(\begin{array}{l}n = 199\\\bar x = 6.19\\s = 1.99\end{array}\)

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

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.

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

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.

\(\begin{array}{l}{H_0}:\mu \ge 7\\{H_1}:\mu < 7\end{array}\)

Thus, the test is left-tailed.

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

\(\begin{array}{c}df = 199 - 1\\ = 198\end{array}\)

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, \({t_{0.01}} = 2.35\).

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

Substitute the respective value in the above formula and simplify.

\(\begin{array}{c}t = \frac{{6.19 - 7}}{{\frac{{1.99}}{{\sqrt {199} }}}}\\ = - 5.7419\end{array}\)

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.

\(\begin{array}{c}\left| t \right| = \left| { - 5.7419} \right|\\ = 5.7419\\ > 2.35\left( {{t_{0.01}}} \right)\end{array}\).

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

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.