Question

A polling company conducted a survey asking 40 fourth year students from Ohio and 33 fourth-year students from Pennsylvania about on average, how many hours they spend on the computer per week. Ohio had an average of 52, while Pennsylvania had an average of 33. Ohio and Pennsylvania have standard deviations of 7.22 and 10.11 respectively. Construct a 99.9% confidence interval for the difference between these results.

Answer 1

Answer:

The 99.9% confidence interval for the difference between these results is between 12.09 and 25.91 hours per week.

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction of normal variables:

When two normal variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of variances.

Ohio had an average of 52, standard deviation of 7.22, sample of 40.

This means that $\mu_O = 52, s_O = \frac{7.22}{\sqrt{40}} = 1.1416$

Pennsylvania had an average of 33, standard deviation of 10.11, sample of 33.

This means that $\mu_P = 33, s_P = \frac{10.11}{\sqrt{33}} = 1.76$

Distribution of the difference:

$\mu = \mu_O - \mu_P = 52 - 33 = 19$

$s = \sqrt{s_O^2 + s_P^2} = \sqrt{1.1416^2 + 1.76^2} = 2.1$

Confidence interval:

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

$\alpha = \frac{1 - 0.999}{2} = 0.0005$

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 $1 - 0.0005 = 0.9995$, so Z = 3.29.

Now, find the margin of error M as such

$M = zs$

$M = 3.29*2.1 = 6.91$

The lower end of the interval is the sample mean subtracted by M. So it is 19 - 6.91 = 12.09 hours per week

The upper end of the interval is the sample mean added to M. So it is 19 + 6.91 = 25.91 hours per week.

The 99.9% confidence interval for the difference between these results is between 12.09 and 25.91 hours per week.