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We are performing the independent means t-test. We do a t-test because we do not actually know what the population variances are. We are only given sample information - so we cannot do a z-test
MANUAL t-TEST:
We can also perform the test manually
The null and alternative hypotheses are :
Ho : x1 = x2 ( the sample means are equal )
Ha : x1 not equal x2 ...
l
We are performing the independent means t-test. We do a t-test because we do not actually know what the population variances are. We are only given sample information - so we cannot do a z-test
MANUAL t-TEST:
We can also perform the test manually
The null and alternative hypotheses are :
Ho : x1 = x2 ( the sample means are equal )
Ha : x1 not equal x2 ( the sample means are different)
The standard error is given by:
SE = sqrt[ (s12/n1) + (s22/n2) ]
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
SE = sqrt [11.05^2/22 + 12.34^2/20] = 3.63
the degrees of freedom can be approximated by the smaller of n1 - 1 and n2 - 1; So d.f =20-1 = 19
est statistic. The test statistic is a t-score (t) defined by the following equation.
t = [ (x1 - x2) - d ] / SE
where x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
So test statistic = (112.42 - 122.78)/3.63 = -2.85
t-critical score for two-tails, alpha =0.05 and d.f =19 is 2.093
Since our test statistic (ignoring sign) of 2.85 is greater than the t-critical value, our test statistic falls in the rejection region. We thus reject the null hypothesis that there is no difference in the population means. There is in fact a difference at the 95% confidence level.
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