Question

a data set about speed dating includes \like\ ratings of male dates made by the female dates. the summary statistics are ( n = 184 ), ( overline{x}=5.55 ), ( s = 1.87 ). use a 0.01 significance level to test the claim that the population mean of such ratings is less than 6.00. 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 = 6.00 )( h_1:mu>6.00 )
b. ( h_0:mu = 6.00 )( h_1:mu
eq6.00 )
c. ( h_0:mu = 6.00 )( h_1:mu<6.00 )
d. ( h_0:mu<6.00 )( h_1:mu>6.00 )
determine the test statistic.
(round to two decimal places as needed.)
determine the p - value.
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the test statistic

The formula for the \(t\)-test statistic in a one - sample \(t\) - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\).
Here, \(\bar{x} = 5.55\), \(\mu=6.00\), \(s = 1.87\), and \(n = 184\).

$$t=\frac{5.55 - 6.00}{1.87/\sqrt{184}}$$

$$t=\frac{- 0.45}{1.87/13.56465}$$

$$t=\frac{-0.45}{0.1378}$$

\(t\approx - 3.27\)

Step2: Calculate the degrees of freedom and find the P - value

The degrees of freedom is \(df=n - 1=184-1 = 183\).
Since this is a left - tailed test (\(H_1:\mu<6.00\)), the P - value is the probability of getting a \(t\) - value less than \(-3.27\) with \(df = 183\).
Using a \(t\) - distribution table or a calculator, the P - value is approximately \(0.001\)

Answer:

• Test statistic: \(-3.27\)
• P - value: \(0.001\)

Since the P - value (\(0.001\)) is less than the significance level (\(\alpha = 0.01\)), we reject the null hypothesis \(H_0:\mu = 6.00\). There is sufficient evidence at the \(0.01\) significance level to support the claim that the population mean of such ratings is less than \(6.00\).