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.
what are the null and alternative hypotheses?
a. ( h_0:mu = 7.00 )
( h_1:mu
eq7.00 )
b. ( h_0:mu = 7.00 )
( h_1:mult7.00 )
c. ( h_0:mu = 7.00 )
( h_1:mugt7.00 )
d. ( h_0:mult7.00 )
( h_1:mugt7.00 )

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 \(7.00\). So, \(H_0:\mu = 7.00\) (the status - quo assumption) and \(H_1:\mu<7.00\) (the claim we are testing).

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 \(n = 191\), \(\bar{x}=6.51\), \(s = 1.96\), and \(\mu = 7.00\)

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

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

\(\sqrt{191}\approx13.82\), \(1.96/\sqrt{191}\approx1.96/13.82\approx0.142\)

$$t=\frac{-0.49}{0.142}\approx - 3.45$$

Step3: Find the P - value

Since this is a left - tailed test (\(H_1:\mu<7.00\)) with \(n-1=191 - 1=190\) degrees of freedom. Using a \(t\) - distribution table or technology (e.g., a TI - 84 Plus: tcdf(-100,-3.45,190)), the P - value is approximately \(0.0003\)

Step4: Make a conclusion

Since the P - value (\(0.0003\)) is less than the significance level \(\alpha = 0.10\), we reject the null hypothesis.

Answer:

B. \(H_0:\mu = 7.00\), \(H_1:\mu<7.00\)

Test statistic \(t\approx - 3.45\)

P - value \(\approx0.0003\)

Final conclusion: Reject \(H_0\). There is sufficient evidence to support the claim that the population mean of such ratings is less than \(7.00\)