Scott X1 93430 S1 5602 n135 Ligonier X1 98043 S14731 n2 40So

Scott: X₁ 93430, S₁= 5602, n₁=35

Ligonier: X₁= 98043, S₁=4731 n₂= 40

Solution

Let mu1 be the mean for Scott

Let mu2 be the mean for Ligonier

Ho: mu1=mu2 (i.e. null hypothesis)

Ha: mu1<mu2 (i.e. alternative hypothesis)

The test statistic is

Z=(xbar1-xbar2)/sqrt(s1^2/n1+s2^2/n2)

=(93430-98043)/sqrt(5602^2/35+4731^2/40)

=-3.82

It is a left-tailed test.

Given a=0.01, the critical value is Z(0.01) =-2.33 (from standard normal table)

Since Z=-3.82 is less than -2.33, we reject the null hypothesis.

So we can conclude that Ligonier is significantly more expensive to live in as compared to Scott.