00:01
Hello students, as per the given question, let us go through the situation step by step.
00:07
So, given that the null hypothesis h naught is mu e is equals to 75 which is the population mean which is 75 and given that the alternative hypothesis h a is for each situation is not equals to 75.
00:28
So, reported value p is equals to 0 .1032 when rounded to 4 decimals and we need calculate for bit a where we say that in the in the bit a the alternative hypothesis is different in each case.
00:48
So, it is given that alternative hypothesis is less than 75 and z is equals to minus 1 .63 and alpha is equals to 0 .05.
01:02
So, case since it is a two tailed test, we need to compare the p value for alpha by 2 which is 0 .025, if you can level of alpha which is equals to 0 .05.
01:22
So, we have a formula for that like if p is less than or equals to alpha by 2, then we reject h naught and if p value is greater than alpha by 2, then we fail to reject h naught.
01:40
So, in given that p is equals to 0 .1032 which is greater than alpha by 2 which is equals to 0 .025.
01:53
So, therefore, we say that we fail to reject the null hypothesis h naught and therefore, we say that we do not have enough evidence to conclude that the population proportion which is a greater than 75 and coming to the bit b where it is given that the alternative hypothesis h a is mu e which is less than 75 and z is equals to 1 .63 and alpha value is given as 0 .10.
02:30
So, using a sample proportion same process for bit as same as a, we have for p value if it is less than or equals to alpha by 2, then we reject h naught and if the p value is greater than alpha by 2, we fail to reject h naught.
02:51
So, as it is given that p value which is equals to 0 .1032 which is greater than the alpha by 2 value which is 0 .05...