00:01
Hello students, in this question we have 5 subpar.
00:02
So the first one is we have to find out the statistic critical value, p -value and also the decision.
00:08
So here the first one is h1.
00:10
First one is h1 alternate hypothesis given mu less than or equal to 100, x bar equal to 97 .6.
00:17
So 97 .6 and sigma equal to 6 .5 and n value is equal to 24.
00:22
So here t equal to x bar minus mu naught by h by root n.
00:25
So it is equal to 97 .6 minus 100 divided by 6 .5 by root 24.
00:33
It is equal to minus 1 .8089.
00:35
Therefore, t value equal to minus 1 .8089 and here the p -value with respect to t statistic is equal to 0 .04178.
00:43
Since here p -value is less than alpha, therefore h naught is rejected.
00:49
Group 1 population average is considered.
00:54
Therefore, group 1 population average is considered to be less than 100 and here the critical value is equal to minus 1 .7139 and here the degrees of freedom is equal to 23.
01:12
And coming to second part, this is a one sample t -test and the second part is one sample proportion z -test.
01:21
So here h1 p is not equal to 0 .5 and p -cap equal to 0 .51 and n equal to 1021.
01:27
So here z equal to p -cap minus p naught by square root of p naught into 1 minus p naught by n.
01:32
Z equal to, after plugging the values, we get 0 .6077 and here the critical value is equal to 0 .5433.
01:40
And here the p -value is equal to 0 .54 with respect to z statistic and since p -value is greater than alpha, h naught cannot be rejected.
01:52
So population proportion p -cap of group 1 population is assumed to be equal to the p -naught value.
02:07
Thank you.
02:08
And for the third one, alternative hypothesis mu1 not equal to mu2 and here x1 bar equal to 3 .23, x2 bar equal to 3 .15 and s1 equal to 0 .58, s2 equal to 0 .65, n1 equal to 15, n2 equal to 16 and sp value is equal to 0 .617.
02:27
So it is a two sample t -test.
02:31
So we know the formula that is t equal to x1 bar minus x2 bar by sp into square root of 1 by n1 plus 1 by n2...