Question

(12 of 15)

You suspect that unemployment in Omaha is less than nationwide average. After collecting data each month for 12 years, you nd that the mean unemployment rate for Omaha is 3.6% with a sample standard deviation of 2.4%. The national average over the time period was 4.2%. The null hypothesis you wish to test is:

H0 : μ=4.2% H0 : μ=4.2%

Should you accept or reject the null hypothesis?

Reject; Omaha's rate is 3 standard errors from the national average.

Reject; Omaha's rate is 0.25 standard deviations from the national average.

Accept; Omaha's rate is 0.25 standard deviations from the national average.

Accept; Omaha's rate is 3 standard errors from the national average.

Answer 1

Option 4. We need to accept the null hypothesis

The calculations are:

sample mean	3.60%
sample SD	2.40%
National average	4.20%
n	144

1. We need to find t values at 5% significance = (sample mean - pop mean) / SD/Sqrt(n)

= (3.6%-4.2%)/(4.2%/sqrt(144)) = -1.71

2. Find the critical t value for 143 degrees of freedom and significance level of 5% = +/- 1.96

3. Since the t value -1.71 does not fall in rejection region we need to accept the null hypothesis at 5% significance

4. SE = SD/Sqrt(n) = 2.4%/sqrt(144) = 0.002

3*SE = 3*0.002 = 0.006= 0.6%

So, National average - 3*SE= Sample average

Sample average = 4.2%-0.6% = 3.2%