Respuesta :
Answer with explanation:
Let [tex]\mu[/tex] be the population mean.
By observing the given information, we have :-
[tex]H_0:\mu\leq4\\\\H_a:\mu>4[/tex]
Since the alternative hypotheses is left tailed so the test is a right-tailed test.
We assume that the time spend by students per day is normally distributed.
Given : Sample size : n=121 , since n>30 so we use z-test.
Sample mean : [tex]\overline{x}=3.15[/tex]
Standard deviation : [tex]\sigma=1.2[/tex]
Test statistic for population mean :-
[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]\Rightarrow\ z=\dfrac{3.15-4}{\dfrac{1.2}{\sqrt{121}}}\\\\\Rightarrow\ z=-7.79166666667\approx-7.79[/tex]
Critical value (one-tailed) corresponds to the given significance level :-
[tex]z_{\alpha}=z_{0.1}=1.2816[/tex]
Since the observed value of z (-7.79) is less than the critical value (1.2816) , so we do not reject the null hypothesis.
Hence, we conclude that we have enough evidence to accept that the college students spend an average of 4 hours or less studying per day.