A psychology researcher has a theory that predicts women will tend to carry more cash than men.A random sample of Ersatz University students revealed that 16 females had a mean of $22.30 in their wallets with a standard deviation of $3.20,while 16 males had a mean of $17.30 with a standard deviation of $9.60.The test statistic for the researcher's hypothesis is:

Question

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To test the hypothesis, we can use a two-sample t-test for the difference between means, assuming equal variances.

The test statistic can be calculated as follows:

t = (x1 - x2) / (s pooled * sqrt(2/n))

where x1 and x2 are the sample means, s pooled is the pooled standard deviation, and n is the sample size.

First, we need to calculate the pooled standard deviation:

s pooled = sqrt(((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / (n1 + n2 - 2))

where s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.

Plugging in the values, we get:

s pooled = sqrt(((16 - 1) * 3.20^2 + (16 - 1) * 9.60^2) / (16 + 16 - 2)) = 6.527

Next, we calculate the t-statistic:

t = (22.30 - 17.30) / (6.527 * sqrt(2/16)) = 1.976

The degrees of freedom for the t-test is (n1 + n2 - 2) = 30.

Using a two-tailed t-distribution with alpha = 0.05 and degrees of freedom = 30, the critical value is 2.042.

Since our calculated t-statistic (1.976) is less than the critical value (2.042), we fail to reject the null hypothesis that there is no significant difference between the mean cash carried by females and males.

Therefore, the best choice is D: 1.976.

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