Suppose that for years the mean of population 1 has been...

Suppose that for years the mean of population 1 has been accepted as the same as the mean of population 2, but that now population 1 is believed to have a greater mean than population 2. Letting α = 0.05 and assuming the populations have equal variances and x is approximately normally distributed, use the following data to test this belief. Sample 1 Sample 2 43.6 45.3 44.4 49.1 45.2 45.6 40.8 46.5 48.3 45.7 40.1 37.0 42.2 42.3 43.1 38.8 37.0 43.3 41.0 40.1 Sample 1 43.6 45.3 44.4 49.1 45.2 45.6 40.8 46.5 48.3 45.7 Sample 2 40.1 37.0 42.2 42.3 43.1 38.8 37.0 43.3 41.0 40.1 What is the alternative hypothesis? What is the observed t value? What is the p-value? What is the final decision about the null hypothesis and why?

Transcript

00:01 Okay, so we want to compare the means of two samples.

00:04 So our null hypothesis is going to be that the means of these samples are equal, and the alternative hypothesis is going to actually be that the mean of sample one is strictly bigger than the mean of sample two.

00:18 They want us to test whether population one now has a greater mean than population two.

00:25 The t statistic is going to be given by the difference in the sample means minus the difference in the sample means if the null hypothesis was true, which is zero because they'd be equal if the null hypothesis was true.

00:43 And then divided by the standard error if the null hypothesis was true.

00:49 Now, we're told to assume the populations have equal variances.

00:54 And so this standard error is going to be given by the square root of 1 over n1 plus 1 over n2 times the pool standard deviation, where the pool standard deviation is given by m1 minus 1, s1 squared plus n2 minus 1 s2 squared over m1 plus n2 minus 2 minus 2.

01:18 Now we've been given the data and we can compute these various quantities.

01:22 So the mean for sample one you can compute is given by 45 .45.

01:31 For sample two, it's given by 40 .49.

01:35 The standard deviation for sample one is given by 2 .33 and for sample two, the same to two decimal places.

01:43 And the number in each sample is 10.

01:49 So one can compute that sp is the full sum deviation is also 2 .33, which makes sense because both of the individual ones are.

01:57 And then that the standard error is therefore 1 .042.

02:03 And so here we have 45 .45 minus 40 .49 divided by 1 .042...