Question

Question 6. (10 Marks) A factory hiring people to work on an assembly line gives job applicants a test of manual agility. This test counts how many strangely shaped pegs the applicants can fit into matching holes in a one-minute period. The table below summarizes the data by gender of the job applicant. Male Female Sample size Sample Mean Sample Standard Deviato2.52 3.39 50 65 19.39 18.55 At a 0.05, is there significant evidence that male applicants perform better at this test?

Answer 1

Answer

We are assuming that the population variances are unequal for male and female

Given that

x1(bar)= 19.39, s1 = 2.52, n1 = 65

x2(bar)= 18.55, s2 = 3.39, n2 = 50

$H_o: \mu_1 = \mu_2 \\ H_a : \mu_1 > \mu_2$

test statistic t =

-(19.39 -18.55)/V(2.522/65) (3.392/50) 0.84/0.5723 = 1.47

degree of freedom = minimum of (n1-1) or (n2-1) = 50 -1 = 49 (because 50 is minimum out of 50 and 65)

using t distribution table, with t value (1.47) and df(49) for a one sided test, we get

p value = 0.0740

p value is more than 0.05 significance level, failing to reject the null hypothesis.

Therefore, we can conclude that male applicants are not performing better this test as compared to female

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