Question

The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken from anesthetized bears. The standard deviation of weights of bears is known to be 10 lb. A researcher takes a sample of 37 bears and based on this sample, they test the hypothesis that the average weight of a bear in Yellowstone National Park is different from 180 lb. The p-value was calculated to be 0.0226. What was the average weight of bears in the sample?

answer correct to 2 decimal places

Answer 1

Answer)

Null hypothesis Ho : u = 180

Alternate hypothesis Ha : u not equal to 180

As the population standard deviation is given

We can use standard normal z table to estimate the answer

Given test is two tailed

So, we will divide the p-value in to two equal parts to find the area in one tail

= 0.0226/2

= 0.0113

From z table, P(z<-2.28) = 0.0113 = P(z>2.28)

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

-2.28 = (sample mean - 180)/(10/√37)

Sample mean = 176.25

When z = 2.28

2.28 = (sample mean - 180)/(10/√37)

Sample mean = 183.75