Question

a data set about speed dating includes like ratings of male dates made by the female dates. the summary statistics are ( n = 180 ), ( \bar{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.
(round to two decimal places as needed.)

Explanation:

Step1: State 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 or no - difference hypothesis) 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 \(n = 180\), \(\bar{x}=6.89\), \(s = 2.22\), \(\mu = 7.00\)

$$
LATEXBLOCK0
$$

Step3: Find the P - value

For a one - tailed \(t\) - test with \(n-1=180 - 1=179\) degrees of freedom and \(t=-0.67\). Using a \(t\) - distribution table or technology (e.g., a TI - 84 Plus: tcdf(-1000,-0.67,179)), the P - value is approximately \(0.2514\)

Step4: Make a decision

Since the significance level \(\alpha = 0.05\) and \(P-\text{value}=0.2514>0.05\), we fail to reject the null hypothesis.

Step5: State the conclusion

There is not sufficient evidence at the \(0.05\) significance level to support the claim that the population mean of such ratings is less than \(7.00\)

Answer:

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