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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.
Option 4. We need to accept the null hypothesis
The calculations are:
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%
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