Answer with explanation:
Let [tex]\mu[/tex] represents the population mean .
Claim : The population mean of of male dates made by the female dates < 7.00.
Hypothesis for the given situation :
[tex]H_0:\mu\geq7\\\\H_a:\mu<7[/tex]
As alternative hypothesis is left-tailed , so the test is left-tailed test.
We assume that is a normal distribution.
Since sample size is large (>30) , so we use z-test.
The z-test statistic will be :
[tex]z=\dfrac{\overlinw{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]z=\dfrac{6.56-7}{\dfrac{1.92}{\sqrt{184}}}\approx-3.10[/tex]
The p-value = [tex]P(z<-3.10)=0.0009676[/tex]
Since the p-value (0.0009676) is less than the significance level (0.10) , so we reject the null hypothesis.
Thus , we have enough evidence to support the alternative hypothesis.
Final conclusion : There is evidence to conclude that the population mean of of male dates made by the female dates is is less than 7.00 .
