Question

A pet store wants to compare day versus evening visiting times to determine when is a more popular time for their Siamese cat exhibit. They randomly observed people viewing the exhibit and noted 30 visitors stayed for an average of 15 minutes with a standard deviation of 3.1 during the day. During the evening, 26 visitors stayed for an average of 9 minutes with a standard deviation of 2.0. Assuming a normal distribution, determine if there is a significant difference in viewing time between day and evening hours.

1. Which test will you use to determine significance?

2. State the null and alternative hypotheses

3. State your research criteria

4.  Calculate your test statistic

5.  State your decision regarding the significance and interpret your results.

6.  Calculate the confidence interval, if applicable.

7.  Calculate the effect size and proportion of variance, if applicable.

Answer 1

(1)here we use t-test with

(2)null hypothesis H0:mean1=mean2 and alternate hypothesis H1:mean1≠mean2

(3) if t>t(0.05/2, 54) we reject the null hypothesis

(4)statistic t=|(mean1-mean2)|/((s_p*(1/n1 +1/n2)^1/2) with df is n=n1+n2-2=30+26-2=54 and s_p²=((n1-1)s1²+(n2-1)s2²)/n

t-test
	sample	mean	s	s2	n	(n-1)s2
	day	15.0000	3.1000	9.6100	30	278.6900
	evening	9.0000	2.0000	4.0000	26	100.0000
	difference=	6.0000	sum=	13.6100	56	378.6900
	sp2=	7.0128
	sp=	2.6482
	SE=	0.7096
	t=	8.4559
one tailed	p-value=	0.0000
two tailed	p-value=	0.0000
critical	t(0.05)	2.0049

(5) since calculated t=8.4559 is more than critical t=2.0049 at typical level of significance alpha=0.05, so we reject H0 and conclude that there is significant  difference between day time and evening time

(6) i think there is no need of calculation of confidence interval

(7) effect size=(difference of mean )/(standard deviation of difference of mean)=|(mean1-mean2)/((s_p)=6/2.6482=2.2657