Question

A contributor for the local newspaper is writing an article for the weekly fitness section. To prepare for the story, she conducts a study to compare the exercise habits of people who exercise in the morning to the exercise habits of people who work out in the afternoon or evening. She selects three different health centers from which to draw her samples. The 57 people she sampled who work out in the morning have a mean of 5.2 hours of exercise each week. The 70 people surveyed who exercise in the afternoon or evening have a mean of 4.5 hours of exercise each week. Assume that the weekly exercise times have a population standard deviation of 0.6 hours for people who exercise in the morning and 0.4 hours for people who exercise in the afternoon or evening. Let Population 1 be people who exercise in the morning and Population 2 be people who exercise in the afternoon or evening.
Step 1 of 2: Construct a 95% confidence interval for the true difference between the mean amounts of time spent exercising each week by people who work out in the morning and those who work out in the afternoon or evening at the three health centers. Round the endpoints of the interval to one decimal place, if necessary.

Answer 1

Answer:

The 95% confidence interval for the true difference between the mean amounts of time spent exercising each week by people who work out in the morning and those who work out in the afternoon or evening at the three health centers is (0.5, 0.9).

Step-by-step explanation:

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

Central Limit Theorem

The Central Limit Theorem establishes 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.

Subtraction between normal variables:

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

In the morning:

Sample of 57, mean of 5.2, standard deviation of 0.6, so:

$\mu_1 = 5.2$

$s_1 = \frac{0.6}{\sqrt{57}} = 0.0795$

In the afternoon/evening:

Sample of 70, mean of 4.5, standard deviation of 0.4, so:

$\mu_2 = 4.5$

$s_2 = \frac{0.2}{\sqrt{70}} = 0.0239$

Distribution of the difference:

$\mu = \mu_1 - \mu_2 = 5.2 - 4.5 = 0.7$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0795^2 + 0.0239^2} = 0.083$

Confidence interval:

The confidence interval is:

$\mu \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower bound of the interval is:

$\mu - zs = 0.7 - 1.96*0.083 = 0.5$

The upper bound of the interval is:

$\mu + zs = 0.7 + 1.96*0.083 = 0.9$

The 95% confidence interval for the true difference between the mean amounts of time spent exercising each week by people who work out in the morning and those who work out in the afternoon or evening at the three health centers is (0.5, 0.9).