00:01
It is given that to examine the redemption rate for the milkshake coupons, let us represent them as m and for the donut coupons let us represent them as d.
00:11
And it is given that the total number of coupons for milkshake, nm is given as 2247 and xm, which is the number of coupons redeemed, is given as 79.
00:23
And for donuts n d is given as double 619 and xd is given as 72 in part a we have to calculate the sample redemption rate for the milkshake that is p cap m it is calculated by xm by n m that is 79 divided by two double four seven simplifying this gives 0 .032 thus, the sample redemption rate for milkshake is 0 .032.
01:00
In part b, we have to calculate the sample redemption rate for the donuts and it is calculated as x d by nd, that is 72 divided by 261.
01:14
Simplifying this gives 0 .0 .1.
01:18
Thus, the sample redemption rate for the donut coupons is 0 .619.
01:24
In path c we have to calculate the point estimate for the true redemption difference of pm minus pd and it is calculated as difference of the sample redemption rate that is p cap m minus p cap d that is 0 .032 minus 0 .01.
01:50
1.
01:50
Simplifying this gives 0 .021.
01:54
Thus the point estimate for the true difference of the redemption rate is 0 .021.
02:03
In part d we have to calculate the 90 percentage confidence interval for the true difference of the proportions.
02:12
The 90 percentage confidence level implies 0 .10 significance level and the z critical value for the given significance level is that is that is 0 .1 by 2 is 1 .645 now the confidence interval is calculated as p cap m minus p p cap d into plus or minus z critical value into root of p cap m into 1.
02:46
Minus p cap n m divided by n m plus p cap d into 1 minus p cap d divided by n d simplifying this from part c this value is known as 0 .021 plus or minus 1 .645 into root of 0 .035 into root of 0 .030 into 1 minus 0 .032 divided by 2247 plus 0 .01 into 1 minus 0 .0 .1 divided by 2619.
03:39
Simplifying this values gives 0 .021 plus or minus 0 .012.
03:46
Now, simplifying further will give 0 .015...