Question

Consider the data below Descriptive Statistics: C1, C2 Variable N Mean SE Mean StDev Minimum Q1 C1 25 161.56 6.72 33.58 106.00 125.00 C2 17 185.59 6.32 26.05 181.00 197.50 Median Q3 Maximum 143.00 167.50 234.00 208.00 227.50 288.00 Test the hypothesis that the mean for population 1 is less than the mean for population 2, assuming the the population variances are equal. The test statistic is found to be −2.48 . Find the p-value. Assume that the populations are normal.

(A) .0132   (B) .005 < p-value < .01   (C) .025 < p-value < .05   (D) .01 < p-value < .025   (E) .0066

Answer 1

This is problem of two sample t test with equal variance.

Given that

Test statistics =-2.48

Degree of freedom = n1+n2-2 =25+17-2 = 40

P-value = P( t₄₀ < -2.48) =0.008724

Therefore correct answer is

B) 0.005 < p- value <0.01

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Thank You!