Question

a data set about speed dating includes \like\ ratings of male dates made by the female dates. the summary statistics are ( n = 192 ), ( overline{x}=6.52 ), ( s = 1.81 ). use a 0.05 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.

a. ( h_{0}:mu = 7.00 )
( h_{1}:mult7.00 )
b. ( h_{0}:mult7.00 )
( h_{1}:mugt7.00 )
c. ( h_{0}:mu = 7.00 )
( h_{1}:mugt7.00 )
d. ( h_{0}:mu = 7.00 )
( h_{1}:mu
eq7.00 )
determine the test statistic.

• 3.68 (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.
reject ( h_{0} ). there is sufficient evidence to conclude that the mean of the population of ratings is less than 7.00

Explanation:

Step1: Hypotheses

The null hypothesis \(H_0\) is a statement of equality. The claim is that the population mean \(\mu\) is less than \(7.00\). So, \(H_0:\mu = 7.00\) and \(H_1:\mu<7.00\) (Option A).

Step2: Test - statistic formula

For a one - sample \(t\) - test (since the population standard deviation \(\sigma\) is unknown), the formula is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\). Here, \(\bar{x} = 6.52\), \(\mu = 7.00\), \(s = 1.81\), and \(n = 192\).

$$t=\frac{6.52 - 7.00}{1.81/\sqrt{192}}$$

$$t=\frac{- 0.48}{1.81/13.8564}$$

$$t=\frac{-0.48}{0.1306}\approx - 3.68$$

Step3: P - value

Using a \(t\) - distribution table or technology (with degrees of freedom \(df=n - 1=192-1 = 191\)), for a one - tailed test (\(t=-3.68\)), the \(P\) - value is approximately \(0.000\) (using a \(t\) - calculator or software, since for a large \(df\), the \(t\) - distribution is close to the standard normal distribution and \(P(t<-3.68)\) is very small).

Step4: Conclusion

Since the \(P\) - value (\(0.000<0.05\)) (significance level \(\alpha = 0.05\)), we reject \(H_0\).

Answer:

• Hypotheses: \(H_0:\mu = 7.00\), \(H_1:\mu<7.00\) (Option A)
• Test statistic: \(-3.68\)
• P - value: \(0.000\)
• Conclusion: Reject \(H_0\). There is sufficient evidence to conclude that the mean of the population of ratings is less than \(7.00\)