Question

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization

thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they

spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.

The null and alternative hypotheses are:

$$\begin{gathered} a.Ho:\overline{x}=4.5,Ha:\overline{x}>4.5 \\ b.Ho:\mu \ge 4.5,Ha:\mu <4.5 \\ c.Ho:\mu =4.75,Ha:\mu >4.75 \\ d.Ho:\mu =4.5,Ha:\mu >4.5 \end{gathered}$$

Short answer

The correct option is d.$Ho:\mu =4.5,Ha:\mu >4.5$.

Step-by-step solution

Given Information

We are given,

Old mean = $4.5$

New mean = $4.75$

Standard deviation ($\sigma$)$=2$

n (Sample size) = $15$

Explanation Part a

The formula for calculating z is:

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

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.

Explanation Part b

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.

Explanation Part c

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.

Explanation Part d

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 $H_a:\mu >4.5$.

Therefore option d is correct.