00:01
Hello, so here we have a sample size and is equal to 36.
00:03
The probability of a type 1 error, alpha, is equal to 0 .05, and then our null hypothesis, mu is equal to 100 versus our alternative hypothesis that mu is going to be less than 100.
00:15
Now, our population mean is mu not equal to 100, so the alternative hypothesis is left -tailed.
00:22
From the given information, the population variance is unknown.
00:25
So we apply a one sample t test.
00:28
And then in part a here, we have our sample mean x bar is going to be equal to 106.
00:36
Sample standard deviation is equal to 15.
00:40
The degrees of freedom is n minus 1.
00:43
So we have 35 degrees of freedom.
00:48
And then using the excel function for the critical value, we get our t critical value being equal to negative 1 .6896.
00:57
So our decision is to reject the null hypothesis if t is less than negative 1 .6896.
01:06
So t here is equal to x bar minus mu not divided by s over the square root of n.
01:11
So we get that our test statistic is going to be equal to 106 minus 100 divided by 15 divided by 6 or the square root of 36, which is equal to 2 .4, which we observe is going to be greater.
01:28
Then negative 1 .6896.
01:30
So therefore we fail to reject the null hypothesis.
01:39
Therefore, it can be concluded that the population mean is not less than 100.
01:45
And then in part b, we have the sample mean x bar is equal to 104.
01:52
Standard deviation, s is equal to 10.
01:55
Again, 35 degrees of freedom now.
01:57
So we can again use the excel function to find the t critical value, which is going to be given at 0 .05 with 35 degrees of freedom, giving us negative 1 .6896.
02:14
And then our decision is to reject h -0 if t is less than negative 1 .6896...