Question

In: Statistics and Probability

Everyone knows that exercise is important. Recently, employees of one large international corporation were surveyed and...

Everyone knows that exercise is important. Recently, employees of one large international corporation were surveyed and asked, How many minutes do you spend daily on some form of rigorous exercise? From a random sample of 35 employees, the mean time spent on vigorous daily exercise was 28.5 minutes. Assume the population standard deviation is 6.8 minutes.

a. Develop a 95% confidence interval estimate of the population mean. Show your work.

b. Develop a 99% confidence interval estimate of the population mean. Show your work.

c. Test if the population mean is less than 30 minutes with the 5% significance level.

d. Test if the population mean is different from 30 minutes with the 5% significance level.

Solutions

Expert Solution

Given, = 28.5, = 6.8

95% confidence interval for is

- Z/2 * / sqrt(n ) < < + Z/2 * / sqrt(n )

28.5 - 1.96 * 6.8 / sqrt(35) < < 28.5 + 1.96 * 6.8 / sqrt(35)

26.2472 < < 30.7528

95% CI is ( 26.2472, 30.7528)

99% confidence interval for is

- Z/2 * / sqrt(n ) < < + Z/2 * / sqrt(n )

28.5 - 2.5758 * 6.8 / sqrt(35) < < 28.5 + 2.5758 * 6.8 / sqrt(35)

25.5394 < < 31.4606

99% CI is ( 25.5394 , 31.4606)

H0: >= 30

Ha: < 30

Test statistics

z = - / / sqrt(n)

= 28.5 - 30 / 6.8 / sqrt(35)

= -1.305

This is test statistics value.

Critical value at 0.05 significance level is -1.645

Since test statistics value > -1.645, we do not have sufficient evidence to reject H0.

p-value = P( Z < z)

= P( Z < -1.305)

= 1 - P( Z < 1.305)

= 1 - 0.9041

= 0.0959

p-value = 0.0959

We conclude at 0.05 level that we fail to support the claim that population mean is less than 30

minutes.

H0: = 30

Ha: 30 (Two tailed)

Test statistics

z = - / / sqrt(n)

= 28.5 - 30 / 6.8 / sqrt(35)

= -1.305

This is test statistics value.

Criitcal value at 0.05 level is -1.96, 1.96

Since test statistics value lie in between -1.96 and 1.96, we do not have sufficient evidence

to reject H0.

p-value = 2 * P(Z < z) ( 2 is multiplied to probability since this is two tailed test)

= 2 * P( Z < -1.305)

= 2 * 0.0959

= 0.1918

p-value = 0.1918

We conclude at 0.05 level that we fail to support the claim that population mean is different

from 30 minutes.

orchestra answered 1 year ago

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