Question

A student wanted to estimate the number of chocolate chips in a commercial brand of cookie. He sampled 100 cookies and found an average of 10.5 chips per cookie. If we assume the standard deviation is 8, what is a 99% confidence interval for the average number of chips per cookie?
A. (8.4,12.6)
B. (8.9,12.1)
C. (5.3,10.7)

Answer 1

Answer: A. (8.4,12.6)

Step-by-step explanation:

Confidence interval$(\mu)$ is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

, where n is the sample size

$\sigma$ = Population standard deviation.

$\overline{x}$= Sample mean

$z_{\alpha/2}$ = Two tailed z-value for significance level of $\alpha$.

Given : Confidence level = 99% = 0.99

Significance level = $\alpha=1-0.99=0.01$

By standard normal z-value table ,

Two tailed z-value for Significance level of 0.01 :

$z_{\alpha/2}=z_{0.005}=2.576$

Also,

n=100

$\sigma= 8$

$\overline{x}=10.5$

Then, the required 99% confidence interval for the average number $(\mu)$ of chips per cookie :-

$10.5\pm (2.576)\dfrac{8}{\sqrt{100}}\\\\ =10.5\pm 2.0608\\\\=(10.5-2.0608,\ 10.5+2.0608)=(8.4392,\ 12.5608)\approx(8.4,\ 12.6)$

Hence, the 99% confidence interval for the average number of chips per cookie = (8.4,12.6)