Question

A researcher was interested in the amount of time cats and dogs nap while their owners are away from home on a regular work day. The 30 cats napped an average of 7.0 hours with a standard deviation of 1.5 hours. The 30 dogs napped a mean of 5.0 hours with a standard deviation of 1.5 hours. Assume that the two populations are normally distributed with equal but unknown standard deviations. Test at the 5% significance level if the amount of time cats nap is longer than the amount of time dogs nap. What is the critical value and what is your conclusion?

Answer 1

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

Rejection Region
This is right tailed test, for α = 0.05 and df = n1 + n2 - 2 = 58
Critical value of t is 1.672.
Hence reject H0 if t > 1.672

Pooled Variance
sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((30 - 1)*1.5^2 + (30 - 1)*1.5^2)/(30 + 30 - 2))*(1/30 + 1/30))
sp = 0.3873

Test statistic,
t = (x1bar - x2bar)/sp
t = (7 - 5)/0.3873
t = 5.164

P-value Approach
P-value = 0
As P-value < 0.05, reject the null hypothesis.