Question

a data set about speed dating includes \like\ ratings of male dates made by the female dates. the summary statistics are ( n = 193 ), ( overline{x}=7.87 ), ( s = 2.09 ). use a 0.01 significance level to test the claim that the population mean of such ratings is less than 8.00. 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. what are the null and alternative hypotheses? oa. ( h_0: mu<8.00 ) ( h_1: mu>8.00 ) ob. ( h_0: mu = 8.00 ) ( h_1: mu
eq8.00 ) oc. ( h_0: mu = 8.00 ) ( h_1: mu>8.00 ) od. ( h_0: mu = 8.00 ) ( h_1: mu<8.00 )

Explanation:

Step1: Determine the null and alternative hypotheses

The null hypothesis \(H_0\) is a statement of equality. The claim is that the population mean \(\mu\) is less than \(8.00\). So, \(H_0:\mu = 8.00\) (equality for the null) and \(H_1:\mu<8.00\) (the claim as the alternative).

Step2: Calculate the test - statistic

The formula for the \(t\) - test statistic (since the population standard deviation \(\sigma\) is unknown) is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\).
Given \(\bar{x} = 7.87\), \(\mu = 8.00\), \(s = 2.09\), \(n = 193\).

$$t=\frac{7.87 - 8.00}{2.09/\sqrt{193}}$$

$$t=\frac{- 0.13}{2.09/13.892}$$

$$t=\frac{-0.13}{0.1504}\approx - 0.864$$

Step3: Find the P - value

For a one - tailed \(t\) - test with \(n - 1=193 - 1 = 192\) degrees of freedom (using technology or a \(t\) - table approximation). Using a \(t\) - distribution calculator, for \(t=-0.864\) and \(df = 192\), the \(P\) - value is approximately \(P = 0.194\)

Step4: Make a decision

Since the significance level \(\alpha=0.01\) and \(P - value=0.194>0.01\), we fail to reject the null hypothesis.

Answer:

• Null and alternative hypotheses: \(H_0:\mu = 8.00\), \(H_1:\mu<8.00\) (Option D)
• Test statistic: \(t\approx - 0.864\)
• \(P\) - value: Approximately \(0.194\)
• Conclusion: Fail to reject \(H_0\). There is not sufficient evidence at the \(0.01\) significance level to support the claim that the population mean of such ratings is less than \(8.00\)