Answer:
![(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 10.862](https://tex.z-dn.net/?f=%20%28242-220%29%20-1.988%20%5Csqrt%7B%5Cfrac%7B20%5E2%7D%7B47%7D%20%2B%5Cfrac%7B31%5E2%7D%7B42%7D%7D%3D%2010.862)
![(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 33.138](https://tex.z-dn.net/?f=%20%28242-220%29%20%2B1.988%20%5Csqrt%7B%5Cfrac%7B20%5E2%7D%7B47%7D%20%2B%5Cfrac%7B31%5E2%7D%7B42%7D%7D%3D%2033.138)
And the 95% confidence interval for the difference in the means is given by: ![10.862 \leq \mu_A -\mu_B \leq 33.138](https://tex.z-dn.net/?f=%2010.862%20%5Cleq%20%5Cmu_A%20-%5Cmu_B%20%5Cleq%2033.138)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
sample mean for inhibitor A
sample standard deviation for inhibitor A
sample size for A
sample mean for inhibitor B
sample standard deviation for inhibitor B
sample size for A
Solution to the problem
For this case the confidence interval for the difference of means is given by:
![(\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_A}{n_A} +\frac{s^2_B}{n_B}}](https://tex.z-dn.net/?f=%20%28%5Cbar%20X_A%20-%5Cbar%20X_B%29%20%5Cpm%20t_%7B%5Calpha%2F2%7D%20%5Csqrt%7B%5Cfrac%7Bs%5E2_A%7D%7Bn_A%7D%20%2B%5Cfrac%7Bs%5E2_B%7D%7Bn_B%7D%7D)
The degrees of freedom are given by:
![df = n_A +n_B -2 = 47+42-2= 87](https://tex.z-dn.net/?f=%20df%20%3D%20n_A%20%2Bn_B%20-2%20%3D%2047%2B42-2%3D%2087)
Since the Confidence is 0.95 or 95%, the value of
and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,87)".And we see that ![t_{\alpha/2}=1.988](https://tex.z-dn.net/?f=t_%7B%5Calpha%2F2%7D%3D1.988)
And replacing we got:
![(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 10.862](https://tex.z-dn.net/?f=%20%28242-220%29%20-1.988%20%5Csqrt%7B%5Cfrac%7B20%5E2%7D%7B47%7D%20%2B%5Cfrac%7B31%5E2%7D%7B42%7D%7D%3D%2010.862)
![(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}}= 33.138](https://tex.z-dn.net/?f=%20%28242-220%29%20%2B1.988%20%5Csqrt%7B%5Cfrac%7B20%5E2%7D%7B47%7D%20%2B%5Cfrac%7B31%5E2%7D%7B42%7D%7D%3D%2033.138)
And the 95% confidence interval for the difference in the means is given by: ![10.862 \leq \mu_A -\mu_B \leq 33.138](https://tex.z-dn.net/?f=%2010.862%20%5Cleq%20%5Cmu_A%20-%5Cmu_B%20%5Cleq%2033.138)