00:01
So a magazine is claiming that the amount of leisure time spent per week is 40 hours.
00:06
And you believe that that is too high.
00:12
And so you take a sample of 60 men and find that the mean amount of leisure time is 37 .8 hours on the average for those men with a sample standard deviation pretty big of 12 .2 hours.
00:26
And we want to use a 5 % significance level to decide.
00:31
Whether we have evidence that it is actually lower than 60 and say that the magazine is exaggerating, basically.
00:39
Now, we can use a z value because our sample size is greater than or equal to 30.
00:45
So if we find that test statistic, we would be taking that 37 .8 hours minus the 40, divided by the sample standard deviation, divided by that square root of 60.
00:58
And that gives us a test statistic of negative 1 .3968.
01:04
Using software, i've used that entire value, and that tells me that test statistic being less than or equal to that test statistics, since we're doing a less than test, gives me a p value of 0 .081 .2...