## SOLUTION

In this question, they are asking us to find the 99% confidence interval estimate of the difference in the two population means. There are two different formulas that could be used here, but only one is correct: the **Equal Variances t Estimate**, and the *Unequal Variances t Estimate*. In order to choose the correct formula, we must first conduct an F test to check if equal variances can be assumed…

## STEP 1 | F-test to check for equal variances:

### Given

### Hypotheses

#### Null Hypothesis

(Variances are equal)

#### Alternative Hypothesis

(Variances are not equal)

### Test Statistic

The F-statistic is calculated as:

### Rejection Region

#### Degrees of Freedom

#### Critical Value

The significance level is 0.01 (1%) and it is a two-tailed test.

The upper critical value for and at this level is 22.0.

The lower critical value is 1/12.0 = 0.0833.

#### Decision Rule

Therefore, the null hypothesis will be rejected if or if .

### Conclusion

Since the calculated F-value of F = 0.7534 is greater than the lower critical value (0.0833) and less than the upper critical value (22.0), we **fail to reject the null hypothesis**.

**Therefore, we can assume that the variances of the two populations are equal.**

## STEP 2 | Confidence Interval Estimator for When

### Given

### Confidence Interval

The 99% confidence interval is calculated as:

#### Degrees of Freedom

#### Critical Value

From the t-distribution table, for 10 degrees of freedom, the critical t-value for a 99% confidence level (two-tailed) is approximately 3.1693.

#### Pooled Variance

Since we have assumed equal variances, we first calculate the pooled variance:

#### Calculation

#### Conclusion

The 99% confidence interval for the difference in the average hours of watching the sport channel between Toronto and Windsor falls between -187.4533 and 197.8135 hours.