Question

a data set about speed dating includes \like\ ratings of male dates made by the female dates. the summary statistics are ( n = 191 ), ( overline{x}=6.51 ), ( s = 1.96 ). use a 0.10 significance level to test the claim that the population mean of such ratings is less than 7.00. 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.
( h_1:mu>7.00 )
( h_1:mu>7.00 )
determine the test statistic.

• 3.45 (round to two decimal places as needed.)

determine the p - value.
0.000 (round to three decimal places as needed.)
state the final conclusion that addresses the original claim.
( h_0 ). there is evidence to conclude that the mean of the population of ratings is 7.00.

Explanation:

Step1: Hypotheses

The null hypothesis \(H_0:\mu = 7.00\) (claim is about population mean). The alternative hypothesis \(H_1:\mu<7.00\) (since we are testing if population mean is less than \(7.00\)).

Step2: Test - statistic formula

The formula for the \(t\) - test statistic (when population standard deviation \(\sigma\) is unknown) is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\). Here, \(\bar{x} = 6.51\), \(\mu = 7.00\), \(s = 1.96\), \(n=191\).

$$t=\frac{6.51 - 7.00}{1.96/\sqrt{191}}$$

$$t=\frac{- 0.49}{1.96/\sqrt{191}}\approx\frac{-0.49}{0.142}\approx - 3.45$$

Step3: P - value

For a one - tailed \(t\) - test with \(n - 1=191-1 = 190\) degrees of freedom and \(t=-3.45\). Using a \(t\) - distribution table or technology (e.g., Excel: =T.DIST(-3.45,190,TRUE)), the \(P\) - value is approximately \(0.000\) (rounded to three decimal places).

Step4: Conclusion

Since the \(P\) - value (\(0.000\)) is less than the significance level \(\alpha = 0.10\), we reject \(H_0\).

Answer:

Reject \(H_0\). There is sufficient evidence to conclude that the mean of the population of ratings is less than \(7.00\).