Question

A researcher collated data on Americans’ leisure time activities. She found the mean number of hours spent watching television each weekday to be 2.7 hours with a standard deviation of 0.4 hours. Jonathan believes that his football team buddies watch less television than the average American. He gathered data from 15 football teammates and found the mean to be 2.3. Which of the following shows the correct z-statistic for this situation?

Answer 1

Answer:

-3.87...just took the test and got this question right

Answer 2

Answer:

Z Test = -3.87298

p= .0001

Then the results are significant at p values < 0.1

Step-by-step explanation:

To answer this question we have to consider some things.

1) The Mean of Population is given by this formula:

$\mu =\frac{\sum X_{i}}{N}\; \mu=2.8$

The Population Standard Deviation is given by this formula:

$\sigma =\frac{\sqrt{(x_{i}-\mu)^{2}}}{N}\: \sigma=0.4$

On the other hand, the Team buddies are a sample of this population, whose mean is:

$\bar{x} = \frac{\sum x_{i} }{n}=2.3$

The Sample Standard Deviation is given by this formula:

$\S =\frac{\sqrt{(x_{i}-\mu)^{2}}}{N-1}\: \s=$

The Z test shows us the validity of the results.

2)

For the Z test we need the Variance $\sigma ^{2}$

Z = (M - μ) / √(σ2 / n)

$Z_{score}=\frac{2.3-2.7}{\sqrt{\frac{0.16}{0.15}}}\\Z_{score}= -3.87298$

3)

Z Test = -3.87298 standard deviation units, since it's a negative value is 3.897 below the mean.

p= .0001 the p value.

Then the results are significant at p values < 0.1