Question

a data set about speed dating includes like ratings of male dates made by the female dates. the summary statistics are n = 186, x = 6.89, s = 2.22. 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.
what are the null and alternative hypotheses?
a. ( h_0: mu = 7.00 )
( h_1: mu < 7.00 )
b. ( h_0: mu = 7.00 )
( h_1: mu > 7.00 )
c. ( h_0: mu = 7.00 )
( h_1: mu
eq 7.00 )
d. ( h_0: mu < 7.00 )
( h_1: mu > 7.00 )
determine the test statistic.

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

determine the p - value.
0.252 (round to three decimal places as needed.)
state the final conclusion that addresses the original claim.
fail to reject ( h_0 ). there is not 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

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 \(n = 186\), \(\bar{x}=6.89\), \(s = 2.22\), \(\mu = 7.00\)

$$t=\frac{6.89 - 7.00}{2.22/\sqrt{186}}$$

$$t=\frac{- 0.11}{2.22/13.64}$$

$$t=\frac{-0.11}{0.163}\approx - 0.67$$

Step3: P - value

For a one - tailed \(t\) - test with \(n-1=186 - 1=185\) degrees of freedom and \(t=-0.67\). Using a \(t\) - distribution table or technology (e.g., TI - 84: tcdf(-100,-0.67,185)), the \(P\) - value is approximately \(0.252\)

Step4: Conclusion

Since the \(P\) - value (\(0.252\))> significance level (\(\alpha = 0.05\)), we fail to reject \(H_0\). There is not sufficient evidence to support the claim that the population mean of such ratings is less than \(7.00\)

Answer:

• Hypotheses: \(H_0:\mu = 7.00\), \(H_1:\mu<7.00\) (Option A)
• Test statistic: \(t\approx - 0.67\)
• \(P\) - value: \(P\approx0.252\)
• Conclusion: Fail to reject \(H_0\). There is not sufficient evidence to conclude that the mean of the population of ratings is less than \(7.00\)