Question

Historically, the population of wolves in Canada have been found to have an average adult weight of 38.9 kilograms. The Canadian Wildlife Service believes that due to climate change, this may have increased. A random sample of 12 wolves is captured and weighed and the following results are obtained. Assume that the weights are normally distributed:

44.8, 40.7, 42.6, 46.8, 48.7, 35.4, 38.9, 39.5, 38.3, 48.9, 48.4, 38.4.

At a 0.05 significance level, test the hypothesis that the average weight of a wolf is more than 38.9.

a)	Identify the null hypothesis and alternative hypothesis. H0: μ ≤ 38.9 H1: μ > 38.9. H0: μ = 38.9 H1: μ < 38.9. H0: μ = 38.9 H1: μ ≠ 38.9. None of the above

b) State the critical value.
For full marks your answer should be accurate to at least three decimal places.

Critical value: ____

c) Determine the value of the test statistic.
For full marks your answer should be accurate to at least three decimal places.

Test statistic:___

d)

Identify the correct interpretation. Since the test statistic is in the rejection region, reject H0 and accept H1. Since the test statistic is in the rejection region, do not reject H0. Since the test statistic is not in the rejection region, reject H0 and accept H1. Since the test statistic is not in the rejection region, do not reject H0.

e)

What conclusion can be drawn from the interpretation at 0.05 significance level? There is not enough statistical evidence that the average weight is different than 38.9. There is significant evidence that the average weight is different than 38.9. There is not enough statistical evidence that the average weight is more than 38.9. There is significant evidence that the average weight is more than 38.9. There is not enough statistical evidence that the average weight is less than 38.9. There is significant evidence that the average weight is less than 38.9.

Answer 1

Answer)

A)

Ho : u <= 38.9

Ha : u > 38.9

B)

As the population s.d is unknown here we will use t distribution to conduct the test

N = 12

Degrees of freedom is = n-1 = 11

For df 11 and alpha 0.05,critical t value from t table is = 1.796

Rejection region is if t > 1.796

Reject Ho

C)

First we need to find the sample mean and s.d for the given data

Sample mean = 42.6167

S.d = 4.7472

Test statistics = (sample mean - claimed mean)/(s.d/√n)

= (42.6167 - 38.9)/(4.7472/√12)

= 2.712

D)

Since test statistics is in the rejection region (as 2.712 > 1.796) reject Ho and accept H1

E)

There is significant evidence that the average weight is more than 38.9.