Question

Austin hypothesizes that water consumption of captive elephants at his San Diego zoo differs significantly from the average captive elephant population. In the population, the mean amount of water drunk per day by elephants is 50 gallons. Austin records the gallons of water drunk by 25 elephants in captivity at the San Diego Zoo. The sample mean of water drunk by Austin’s sample is 57 with a sample variance of s2 = 25. Set alpha equal to 0.01.

a. State the null and alternative hypotheses in symbols.

b. Set up the criteria for making a decision. That is, find the critical value.

c. Compute the appropriate test statistic. Show your work.

d. Based on your answers above, evaluate the null hypothesis.

REJECT FAIL TO REJECT (circle one)

e. State your conclusion in words.

f. Compute the 99% confidence interval.

h. In words, interpret the confidence interval you calculated above.

Answer 1

Solution:

Given that

n= 25 sample size

population mean of water drunk by Austin elephants.

. Sample mean of water drunk by Austin elephants.

sample variance of water drunk by Austin elephants.

s = 5 sample standard deviations water drunk by Austin elephants.

level of significance

a) To test the hypothesis

. Vs.   

b) The t critical value at is

. From t table

The t critical value = 2.797

c) Test statistic

t = 7

Test statistic t = 7

d) Decision :

t > t critical value

7 > 2.797

Reject Ho.

e) Conclusion : Reject Ho , there is sufficient evidence to conclude that the water consumption of captive elephants at his San Diego Zoo differ significantly from the average captive elephants population.

f) The 99% confidence interval for population mean is

At

from t table

( 54.203 , 59.797)

The 99% confidence interval for population mean is

( 54.203, 59.797)

h) Interpretation of confidence interval.

We can be 99% confident that the true population mean lies in the confidence interval.