00:01
Hello students in first question given x bar is equal to 153, mu is equal to 150, s is equal to 7 .5 and n is equal to 14.
00:09
Here you want to test the null hypothesis h naught such that mu is equal to 150 against the alternative hypothesis h a such that mu greater than 150.
00:21
The test statistic is t is equal to x bar minus mu divided by s by root n.
00:29
On substituting the values we will get the test statistic as 2 .53.
00:41
The p -value corresponding to the test statistic is 0 .0078.
00:50
Here p -value is less than the significance level alpha.
01:03
Therefore, here we reject the null hypothesis.
01:13
There is sufficient evidence to conclude that the mean time is greater than 150 minutes.
01:19
In second question given that x bar is equal to 6 .8, mu naught is equal to 6, sigma is equal to 1 .82, n is equal to 28.
01:29
Here we want to test the null hypothesis h naught such that mu is equal to 6 against the alternative hypothesis h1 such that mu naught equal to 6.
01:41
Under h naught the test statistic is z naught is equal to x bar minus mu naught divided by sigma by root n.
01:50
Now on substituting the values we will get the test statistic as 6 .8 minus 6 divided by 1 .82 by root 28.
02:02
Therefore, the test statistic is 2 .3259.
02:07
Now table value z alpha by 2 is 1 .96.
02:11
Here the test statistic z naught 2 .3259 is greater than table value z alpha by 2 1 .96.
02:23
Hence we reject the null hypothesis at 5 percent level of significance.
02:35
Therefore, we conclude that the mean waiting time during lunch hour is not exact 6 minutes.
02:43
Given mu is equal to 200, x bar is equal to 182 .9, sigma is equal to 121 .8, n is equal to 54, alpha is equal to 0 .10...