00:01
Hello students, here in the qstn we want to test the null hypothesis h0 such that p greater than or equal to 0 .90 against the alternative hypothesis ha such that p less than 0 .90.
00:14
Given x is equal to 1068, n is equal to 1200, p is equal to 0 .90, sample proportion of satisfied customers p cap is equal to 1068 divided by 1200 which is equal to 0 .89.
00:34
Now we have the test statistic z is equal to p cap minus p divided by square root of p into 1 minus p divided by n.
00:47
Now substitute the values into the test statistic equation, we will get the test statistic as 0 .89 minus 0 .90 divided by square root of 0 .90 into 1 minus 0 .90 divided by 1200.
01:10
Therefore, the test statistic z is equal to minus 1 .15.
01:17
Now the p value corresponding to the test statistic is p value is equal to z less than minus 1 .15.
01:32
Now from the table the corresponding p value is 0 .1251 given the significance level alpha is equal to 0 .05.
01:49
We have p value greater than alpha which is 0 .1251 is greater than 0 .05.
02:01
Therefore, we fail to reject the null hypothesis.
02:18
Based on the p value approach at a 5 percentage significance level, there is not enough evidence to suggest that the retailer needs to improve its services...