Answer:
Step-by-step explanation:
Given that :
Mean = 7.8
Standard deviation = 0.5
sample size = 30
Sample mean = 7.3 5.4772
The null and the alternative hypothesis is as follows;
[tex]\mathbf{ H_o: \mu \geq 7.8}[/tex]
[tex]\mathbf{ H_1: \mu < 7.8}[/tex]
The test statistics can be computed as :
[tex]z = \dfrac{X- \mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \dfrac{7.3- 7.8}{\dfrac{0.5}{\sqrt{30}}}[/tex]
[tex]z = \dfrac{-0.5}{\dfrac{0.5}{5.4772}}[/tex]
[tex]z = - 5.4772[/tex]
The p-value at 0.05 significance level is:
p-value = 1- P( Z < -5.4772)
p value = 0.00001
Decision Rule:
The decision rule is to reject the null hypothesis if p value is less than 0.05
Conclusion:
At the 0.05 significance level, there is sufficient information to reject the null hypothesis. Therefore ,we conclude that college students watch fewer movies a month than high school students.