The hypotheses are; Null hypothesis: >= $13.04 vs alternative hypothesis: < $13.04 x = $12.75 = $2 n = 64 alpha = 0.05 By use of z-score; the test statistic is; z =( x - )/(/sqrt(n)) z = (12.75 - 13.04)/(2/sqrt(64)) z = -0.29/(2/8) z = -0.29/0.25 z = -1.16 From tables P-value = (z<-1.16) P-value = 0.123 Therefore; 0.123>0.05 Since the p-value is greater than alpha value, we fail to reject the null hypothesis and conclude that the mean cost of eating out is greater than or equal to $13.04 than that of eating in. 1