We are given summary data from two states, Alaska and Hawaii, regarding people's awareness of environmental issues. We have the sample means, sample standard deviations, and sample sizes for each state. We want to test for a difference in the means. The given values are: $\bar{x}_H = 57.50$, $\bar{x}_A = 65.70$, $s_H = 18.65$, $s_A = 15.20$, $n_H = 145$, $n_A = 155$.

1. Problem Description

We are given summary data from two states, Alaska and Hawaii, regarding people's awareness of environmental issues. We have the sample means, sample standard deviations, and sample sizes for each state. We want to test for a difference in the means. The given values are:

$\bar{x}_H = 57.50$, $\bar{x}_A = 65.70$, $s_H = 18.65$, $s_A = 15.20$, $n_H = 145$, $n_A = 155$.

2. Solution Steps

We will perform a two-sample t-test to determine if there is a significant difference between the means of the two groups. The formula for the t-statistic is:

$t = \frac{(\bar{x}_A - \bar{x}_H)}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_H^2}{n_H}}}$

Plugging in the given values:

$t = \frac{(65.70 - 57.50)}{\sqrt{\frac{15.20^2}{155} + \frac{18.65^2}{145}}}$

$t = \frac{8.20}{\sqrt{\frac{231.04}{155} + \frac{347.8225}{145}}}$

$t = \frac{8.20}{\sqrt{1.4906 + 2.3988}}$

$t = \frac{8.20}{\sqrt{3.8894}}$

$t = \frac{8.20}{1.9721}$

$t \approx 4.158$

3. Final Answer

The calculated t-statistic is approximately 4.

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