Posted by **Gummy** on Saturday, April 6, 2013 at 1:08pm.

2. In 2009 a random sample of 70 unemployed people in Alabama showed an average weekly benefit of $199.65. In Mississippi, for a random sample of 65 the number was $187.93. Assume population standard deviations of $32.48 and $26.15 respectively.

a. Using the 5% level of significance, test whether

the two means are different.

b. Assume the p-value for this test is .0439.

c. If the level of significant used was 10% (or 1%),

would the Ho be rejected?

- Statistics -
**PsyDAG**, Saturday, April 6, 2013 at 2:54pm
Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

## Answer This Question

## Related Questions

- Statistics - The null and alternate hypotheses are: H0: μ1 = μ2 H1: &#...
- statistics - A specific study found that the average number of doctor visits per...
- statistics - A specific study found that the average number of doctor visits per...
- Statistics - 1. A study of 40 people found that they could do on the average15 ...
- Statistics - 1. A study of 40 people found that they could do on the average15 ...
- basic statistics for the behavioral sciences - on a standard test of motor ...
- probs & stats - 1. A study of 40 people found that they could do on the ...
- statistics - Based on a random sample of 25 units of lawnmowers, the average ...
- statistics - a random sample of 100 college students is taken from the student ...
- Statistics - A random sample of 14 evenings at the O'Sullivan house hold showed ...

More Related Questions