No.PZ2022071202000090 (选择题)

问题如下：

Question Assuming that the variances of the underlying populations are equal, the pooled estimate of the common variance is 2,678.05. The t-test statistic appropriate to test the hypothesis that the two population means are equal is closest to:

选项：

A.1.90. B.0.29. C.0.94.

解释：

Solution

C is correct.

$t=\frac{\left({\bar{x}}_1?{\bar{x}}_2\right)?\left({\mu }_1?{\mu }_2\right)}{{\left(\frac{s_p^2}{n_1}+\frac{s_p^2}{n_2}\right)}^{0.5}}$

In this case we have:

$s_p^2$ = (200 - 185)/(2678.05/25+2678.05/18)0.5

$t=\frac{\left(200?185\right)?\left(0\right)}{{\left(\frac{2678.05}{25}+\frac{2678.05}{18}\right)}^{0.5}}$ = 0.93768 ~ 0.94

A is incorrect. The mistake is to divide 2678.05 by 43 (25 + 18) and use the square root of the result as the divisor for the t-statistic: (200 - 185)/(2678.05/43)^0.5 = 1.90.

B is incorrect. The mistake is to take an equally weighted average of the given standard deviations and use this as the divisor for the t-statistic: (200 - 185)/[(45 + 60)/2] = 0.28571.