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 )
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
( h_{0} ). there is evidence to conclude that the mean of the population of ratings is 7.00

Explanation:

Step1: State null and alternative hypotheses

The claim is that the population mean is less than \(7.00\). So, \(H_0:\mu = 7.00\) (null hypothesis, no difference) and \(H_1:\mu<7.00\) (alternative hypothesis, claim we are testing).

Step2: Calculate test statistic

The formula for the \(t\) - test statistic (since population standard deviation \(\sigma\) is unknown) is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\). Given \(\bar{x} = 6.89\), \(\mu = 7.00\), \(s = 2.22\), \(n = 186\).

$$t=\frac{6.89 - 7.00}{2.22/\sqrt{186}}\approx\frac{- 0.11}{2.22/13.64}\approx\frac{-0.11}{0.163}\approx - 0.67$$

Step3: Find P - value

For a one - tailed \(t\) - test with \(n-1=186 - 1=185\) degrees of freedom (using technology or a \(t\) - table approximation). The \(P\) - value is the probability of getting a \(t\) - statistic less than \(-0.67\). Using a \(t\) - distribution calculator, \(P\approx0.252\)

Step4: Make a decision

Since the significance level \(\alpha = 0.05\) and \(P - value=0.252>0.05=\alpha\), we fail to reject \(H_0\).

Answer:

Fail to reject \(H_0\). There is not sufficient evidence to conclude that the mean of the population of ratings is less than \(7.00\)