Question

A local zoo wants to compare day versus evening visiting times to determine when is a more popular time for their polar bear 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. 4a. Which test will you use to determine significance? 4b. State the null and alternative hypotheses 4c. State your research criteria 4d. Calculate your test statistic 4e. State your decision regarding significance and interpret your results. 4f. Calculate the confidence interval, if applicable. 4g. Calculate the effect size and proportion of variance, if applicable

Answer 1

Answer 4a)

2 samples T test with unequal variances can be used to test the difference in mean values of the two independent samples

Answer 4b)

Null hypothesis: The average time spent in zoo during day and night time are same

Alternate hypothesis: The average time spent is zoo during day and night time are significantly different

Answer 4c)

The mean difference of the two independent samples is the criteria of research. Accoring to the null hypothesis this mean difference is not significant

Answer 4d)

t = (x1 - x2) / sqrt(s1^2/n1 + s2^2/n2)

t = (15-9) / sqrt((3.1^2/30)+(2^2/26))

t = 8.71

.

df =

.

On plugging the values we get:

df = ((3.1^2/30)+(2^2/26))^2/(((3.1^2/30)^2/29)+((2^2/26)^2/25))

df = 50.13~50

.

P value for t statistic = 8.71, df = 50 and two tailed test is: 0.0000

[use T table or excel function =t.dist.2t(8.71,50) to find the P value]

.

Answer 4e)

Decision: Since P value 0.000 < alpha 0.05, we reject the null hypothesis and conclude that the two different times have significantly different mean times of visitors