00:01
We are given in the question that sample mean let it be denoted by x -war is 16 .805 and sample size let it be denoted by n is 29 and sample standard deviation let it be denoted by s is 2 .738.
00:21
And we have to test the hypothesis at population mean let it be denoted by mu is equal to 5338.
00:31
So let us consider the null and alternative hypothesis.
00:37
Let it be denoted by h node, that is null hypothesis as mu is equal to 15 .8.
00:47
And alternative hypothesis, let it be denoted by h1 as mu greater than 15 .8.
00:59
So we know by test statistics that t is equal to x bar minus mu note divide s divide under root n.
01:17
So putting the values, we get t value for degree of freedom that is 29 minus 1 is equal to x bar that is 16 .805 minus mu note that is the value at which we have to test the hypothesis which is 15 .8 divide s which is 2 .738 divide under root n, that is under root 29.
01:46
So it will be, that is the t value at 28 is 1 .005 divide 0 .5084.
01:59
So the t value will be 1 .9768.
02:09
Now the p value, that is probability where t value is greater than 1 .9768 is equal to 0 .029...