Question

A psychologist wants to test whether there is any difference in puzzle-solving abilities between boys and girls. Independent samples of twelve boys and ten girls were chosen at random. The boys took a mean of

42

minutes to solve a certain puzzle, with a standard deviation of

4.8

minutes. The girls took a mean of

37

minutes to solve the same puzzle, with a standard deviation of

5.9

minutes. Assume that the two populations of completion times are normally distributed, and that the population variances are equal. Can we conclude, at the

0.05

level of significance, that the mean puzzle-solving times for boys,

μ1

is greater than the mean puzzle-solving times for girls,

μ2

?

Perform a one-tailed test. Then fill in the table below.

Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)

The null hypothesis:	H0:
The alternative hypothesis:	H1:
The type of test statistic:	(Choose one)ZtChi squareF
The value of the test statistic: (Round to at least three decimal places.)
The p-value: (Round to at least three decimal places.)
Can we conclude that the mean puzzle-solving times for boys is greater than the mean puzzle-solving times for girls?		Yes	No

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ1 = μ2
Alternative Hypothesis, Ha: μ1 > μ2

Pooled Variance
sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((12 - 1)*4.8^2 + (10 - 1)*5.9^2)/(12 + 10 - 2))*(1/12 + 1/10))
sp = 2.2793

Test statistic,
t = (x1bar - x2bar)/sp
t = (42 - 37)/2.2793
t = 2.194

P-value Approach
P-value = 0.020
As P-value < 0.05, reject the null hypothesis.

yes, we can conclude that the mean puzzle-solving times for boys is greater than the mean puzzle-solving times for girls