00:01
Hello, so here we're giving the sample size n is 65, so the degrees of freedom is then 64.
00:05
The sample mean x bar is 103, and the sample standard deviation is 11 .5.
00:11
So the hypotheses here to be tested would be h0, where we have mu is going to be equal to 100.
00:19
And then the alternative hypothesis would be that mu would not be equal to 100.
00:26
So here the test statistic for the hypothesis test about a population where sigma is known is given by t is equal to x bar minus mu not.
00:37
That's going to be 103 minus 100 and then divided by s divided by square root of n.
00:45
So 11 .5 divided by the square root of 64, which gives us 2 .09.
00:53
So then we get here that the area under the t distribution curve to the left of negative 2 .09 is the same as the area under the curve to the right of t equals 2 .09, which we obtain from the t table.
01:08
And we see that t equals 2 .098 is between 1 .998 and 2 .386.
01:13
We then see the area in the tail on the right of t equals 2 .09 is between 2 .025 and 0 .01.
01:21
We then double these amounts and get the p value then must be between 0 .05 and 0 .02.
01:32
So we have here that our p value then is going to be less than alpha, where alpha is 0 .05.
01:40
So therefore, we reject the null hypothesis at a 5 % level of significance and then conclude that mu is not equal.
01:52
Equal to 100.
01:54
So we can reject the null hypothesis and then we can accept the alternative hypothesis.
01:59
And then for part b, we are given that our sample size n is 65, degrees of freedom 64, sample mean 96 .5, and a standard deviation is 11.
02:11
So now we get that our t statistic is going to be equal to x bar minus mu not, which is going to be 96 .5 minus 100...