# In a recent study of 35 ninth-grade students, the mean number of hours per week that they played video games was 16.6. The standard deviatio

Question

In a recent study of 35 ninth-grade students, the mean number of hours per week that they played video games was 16.6. The standard deviation of the population was 2.8.
a. Find the best point of estimate of the population mean
b. Find the 95% confidence level of the mean of the time playing video games
c. Find the 99% confidence interval of the mean time playing video games
d. Which is larger? Explain why.

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1 year 2021-09-04T15:19:37+00:00 1 Answers 473 views 0

a) 16.6

b) The 95% confidence level of the mean of the time playing video games is between 15.7 and 17.5 hours.

c) The 99% confidence level of the mean of the time playing video games is between 15.4 and 17.8 hours.

d) The margin of error increases as the confidence level increases, due to the value of z, which means that the 99% confidence interval is larger.

Step-by-step explanation:

a. Find the best point of estimate of the population mean

The best estimate for the population mean is the sample mean, which is 16.6

b. Find the 95% confidence level of the mean of the time playing video games

We have that to find our $$\alpha$$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$$\alpha = \frac{1 – 0.95}{2} = 0.025$$

Now, we have to find z in the Ztable as such z has a pvalue of $$1 – \alpha$$.

That is z with a pvalue of $$1 – 0.025 = 0.975$$, so Z = 1.96.

Now, find the margin of error M as such

$$M = z\frac{\sigma}{\sqrt{n}}$$

In which $$\sigma$$ is the standard deviation of the population and n is the size of the sample.

$$M = 1.96\frac{2.8}{\sqrt{35}} = 0.9$$

The lower end of the interval is the sample mean subtracted by M. So it is 16.6 – 0.9 = 15.7 hours

The upper end of the interval is the sample mean added to M. So it is 16.6 – 0.9 = 17.5 hours

The 95% confidence level of the mean of the time playing video games is between 15.7 and 17.5 hours.

c. Find the 99% confidence interval of the mean time playing video games

Following the same logic as above, we have that $$Z = 2.575$$. So

$$M = 2.575\frac{2.8}{\sqrt{35}} = 0.9$$

The lower end of the interval is the sample mean subtracted by M. So it is 16.6 – 1.2 = 15.4 hours.

The upper end of the interval is the sample mean added to M. So it is 16.6 + 1.2 = 17.8 hours.

The 99% confidence level of the mean of the time playing video games is between 15.4 and 17.8 hours.

d. Which is larger? Explain why.

The margin of error increases as the confidence level increases, due to the value of z, which means that the 99% confidence interval is larger.