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:

Answer 1

Answer:

The teenagers spent 4.5 hours per week, on average, on the phone.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 4.5

Sample mean, $\bar{x}$ = 4.75

Sample size, n = 15

Alpha, α = 0.05

Sample standard deviation, s = 2

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 4.5\text{ hours per week}\\H_A: \mu > 4.5\text{ hours per week}$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{4.75 - 4.5}{\frac{2}{\sqrt{15}} } = 0.4841$

Now,

$t_{critical} \text{ at 0.05 level of significance, 14 degree of freedom } = 1.76$

Since,                  

$t_{stat} < t_{critical}$

We accept the null hypothesis.  Thus, the teenagers spent 4.5 hours per week, on average, on the phone. The sample contradicted the organization's claim.