91Ó°ÊÓ

We are given,

Old mean = $4.5$

New mean = $4.75$

Standard deviation ($\mathrm{σ}$)$=2$

n (Sample size) = $15$

The formula for calculating z is:

$$\begin{gathered} z=\frac{\overline{x}-{\mathrm{μ}}_x}{{\displaystyle \frac{\mathrm{σ}}{\sqrt{n}}}} \\ z=\frac{4.5-4.75}{{\displaystyle \frac{2}{\sqrt{15}}}} \\ z=5.39 \end{gathered}$$

We can see that it is much more than the expected time a teenager spent per week, therefore, we can say that option a is not correct.

We have obtained $z=5.39$which is much more than the expected time spent by a teenager on the phone, whereas, we are given null hypothesis as $4.5$which is the expected time spent on phone by teenager, therefore, we can conclude that the option b is incorrect.

we are given null hypothesis as $4.5$which is the expected time spent on phone by teenager, whereas, sample mean is given to us as $4.75$which does not fit in the scenario here.

Therefore, option c is incorrect.

Since we are given null hypothesis as $4.5$is the expected time spent on phone by teenager and alternative hypothesis mostly is what is being said by the researchers we conclude that $H_a:\mathrm{μ}>4.5$.

Therefore option d is correct.