<h2>
Answer with explanation:</h2>
Let
be the population mean.
By observing the given information, we have :-
![H_0:\mu\leq4\\\\H_a:\mu>4](https://tex.z-dn.net/?f=H_0%3A%5Cmu%5Cleq4%5C%5C%5C%5CH_a%3A%5Cmu%3E4)
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 : ![\overline{x}=3.15](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D3.15)
Standard deviation : ![\sigma=1.2](https://tex.z-dn.net/?f=%5Csigma%3D1.2)
Test statistic for population mean :-
![z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B%5Coverline%7Bx%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![\Rightarrow\ z=\dfrac{3.15-4}{\dfrac{1.2}{\sqrt{121}}}\\\\\Rightarrow\ z=-7.79166666667\approx-7.79](https://tex.z-dn.net/?f=%5CRightarrow%5C%20z%3D%5Cdfrac%7B3.15-4%7D%7B%5Cdfrac%7B1.2%7D%7B%5Csqrt%7B121%7D%7D%7D%5C%5C%5C%5C%5CRightarrow%5C%20z%3D-7.79166666667%5Capprox-7.79)
Critical value (one-tailed) corresponds to the given significance level :-
![z_{\alpha}=z_{0.1}=1.2816](https://tex.z-dn.net/?f=z_%7B%5Calpha%7D%3Dz_%7B0.1%7D%3D1.2816)
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.