posted by Anonymous .
we want to if men and women perform differently on statistics tests. The average of 15 men's scores was 76. The average of 10 women's scores was 83. The standard deviation in both cases is 10. Test at 5% significance level.
what is the null and alternate hypothese?
is it a one tail or two?
what is the value of statistic?
what is the decision rule?
what can we conclude?
Try a two-sample independent groups t-test since your sample sizes are small.
Ho: µ1 = µ2
Ha: µ1 does not equal µ2
This is a two-tailed test because the problem is just asking if there is a difference (no specific direction in the alternate hypothesis).
Degrees of freedom for this type of test is this:
df = n1 + n2 - 2
Find the critical or cutoff value using a t-table for degrees of freedom at .05 level of significance for a two-tailed test.
Use the correct formula for this type of test.
Substitute the values into the formula and calculate your test statistic. If your test statistic exceeds the critical value from the table, then the null will be rejected in favor of the alternative hypothesis and µ1 does not equal µ2. If your test statistic does not exceed the critical value from the table, then the null will not be rejected and you cannot conclude a difference.
Hope this will help get you started.