Question

We need to compare the mean student debt of students from two states. The debes can bo assumed to be Normally datrbuted, but we cannot assume that they have equal variances: We took a random sampin trom each stale. In the fist state, the sample mean was $17175 and the sample standard deviacion was $1356, for a sampie of sice 20 . hn fre aecond siate. the sampie mean was $21660 and the nampio standard deviabion was $1363, for a sample of size 20. Find the lest starintic (Give tavo decimal places)

Answer 1

In this case, the test statistic to be approximately -11.61 (rounded to two decimal places).

Explanation:

To compare the mean student debt of students from two states, we can use a two-sample t-test since we don't assume equal variances. The t-test compares the means of two samples to determine if there is a significant difference between them.

Given the following information:

• - First state sample: Sample mean = $17175, Sample standard deviation = $1356, Sample size = 20.
• - Second state sample: Sample mean = $21660, Sample standard deviation = $1363, Sample size = 20.
• - Critical value = -2.024 (assuming a two-tailed test)

To calculate the test statistic for the t-test, we use the formula:

t = (x₁ - x₂) / √[(s₁² / n₁) + (s₁² / n₁)]

where x₁ and x₁ are the sample means, s₁ and s₂ are the sample standard deviations, and n₁ and n₁ are the sample sizes.

Plugging in the values:

t = ($17175 - $21660) / √[(($1356)² / 20) + (($1363)² / 20)]

Calculating this expression, we find the test statistic to be approximately -11.61 (rounded to two decimal places).

Since the critical value provided is -2.024 and the test statistic is much smaller than the critical value, we can conclude that there is a significant difference between the mean student debts of the two states.