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a data set about speed dating includes \like\ ratings of male dates mad…

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.
c. ( h_0:mu = 6.00 )
( h_1:mult6.00 )
d. ( h_0:mult6.00 )
( h_1:mugt6.00 )
determine the test statistic.

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

determine the p - value.
0.001 (round to three decimal places as needed.)
state the final conclusion that addresses the original claim.
( h_0 ). there is evidence to conclude that the mean of the population of ratings is 6.00.

Explanation:

Step1: Determine the null and alternative hypotheses

The null hypothesis \(H_0\) is a statement of equality. The alternative hypothesis \(H_1\) is the claim we are testing. Here, the claim is that the population mean \(\mu<6.00\), so \(H_0:\mu = 6.00\) and \(H_1:\mu<6.00\) (already given as option C).

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 \(\bar{x} = 5.55\), \(\mu = 6.00\), \(s = 1.87\), \(n=184\).

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Step3: Find the P - value

For a one - tailed \(t\) - test with \(n - 1=184 - 1 = 183\) degrees of freedom (using technology or a \(t\) - distribution table approximation), the P - value is the probability of getting a \(t\) - statistic as extreme or more extreme than the calculated value. For \(t=-3.27\) and \(df = 183\), the P - value is approximately \(0.001\).

Step4: Make a decision

We compare the P - value with the significance level \(\alpha = 0.01\). Since \(P - value=0.001<\alpha = 0.01\), we reject the null hypothesis \(H_0\).

Answer:

Reject \(H_0\). There is sufficient evidence to conclude that the mean of the population of ratings is less than \(6.00\).