Question

a data set about speed dating includes like ratings of male dates made by the female dates. the summary statistics are ( n = 186,overline{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
eq7.00 )
d. ( h_{0}:mu<7.00 )( h_{1}:mu>7.00 )

Explanation:

Step1: Determine the null hypothesis

The null hypothesis \(H_0\) is a statement of equality. Here, it is \(H_0:\mu = 7.00\) as we are comparing the population mean \(\mu\) to the value \(7.00\)

Step2: Determine the alternative hypothesis

The claim is that the population mean of such ratings is less than \(7.00\). So the alternative hypothesis \(H_1\) is \(H_1:\mu<7.00\) (a left - tailed test).

For a left - tailed \(t\) - test (since the population standard deviation \(\sigma\) is unknown and we use \(s\) instead), the test statistic \(t\) is given by the formula \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)

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

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The degrees of freedom \(df=n - 1=186-1 = 185\)

Using technology (or a \(t\) - table approximation), for \(t=-0.675\) and \(df = 185\), the \(P\) - value is the probability of getting a \(t\) - value less than \(-0.675\) in a \(t\) - distribution with \(df = 185\). Using a calculator or statistical software, \(P\) - value\(\approx0.25\)

Since the \(P\) - value (\(0.25\))> significance level (\(\alpha = 0.05\)), we fail to reject the null hypothesis.

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