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许晏宁 · 2022年10月09日

No.PZ2022071202000090 (选择题)

NO.PZ2022071202000090

问题如下:

Question
Independent samples drawn from normally distributed populations exhibit the following characteristics:

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= x ˉ 1 ? x ˉ 2 ? μ 1 ? μ 2 s p 2 n 1 + s p 2 n 2 0.5

In this case we have:

s p 2 = (200 - 185)/(2678.05/25+2678.05/18)0.5

t= 200?185 ? 0 2678.05 25 + 2678.05 18 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.

这道题涉及的是哪个知识点,common variance在这里有什么意义?

1 个答案

星星_品职助教 · 2022年10月09日

同学你好,

本题考察特定检验下的t统计量计算,实际上就是直接代数进公式。两个sample mean表格中直接给出;“... to test the hypothesis that the two population means are equal ”代表μ1-μ2=0。Sp即pooled estimator给出后就不需要自己另行计算了。

代入n1=25和n2=18,即可得到答案。

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