Question

. Suppose that you own a movie theater and that you have purchased a filling machine for candy bags which, according to the manufacturer, is supposed to fill each bag with 16 oz. of candy. You want to find out if the (population) mean μ is actually 16 oz. so, you take a random sample of 16 bags which yields the following data (in oz.): 5.87 16.02 15.78 15.83 15.69 15.81 16.01 15.82 15.92 15.82 15.90 15.70 15.88 15.80 5.90 15.83 2. Assume that everything is as described in problem 1 except that you took a random sample of 36 bags. Assume that the sample mean and sample variance as the sample mean and sample variance you calculated for 16 bags in problem 1 (a) [25 points] Find the 95% confidence interval for the population mean μ。( Do you need to assume that the weights of filled bags are approximately normally distributed?) (b) [25 points] On the basis of these data, what do you conclude about the claim that the population mean fill weight is 16 oz. Give your reasoning. Begin your reasoning with whether or not 16 is in the 95% confidence interval.

I need the answer for Question 2 by taking the information from Question 1 mentioned above..

Answer 1

2. a. Here mean=15.849 and standard deviation =0.09

Now sample size is n=36>30 so we will use z distribution

For 95% z value is 1.96 as $P(-1.96<z<1.96)=0.95$

Margin of Error= $E=z*\frac{\sigma }{\sqrt{n}}=1.96*\frac{0.09}{6}=0.0294$

Hence CI= $\mu \pm E=15.8488 \pm 0.0294=(15.8194,15.8782)$

b. As 16 is not in the range of CI, we conclude that population mean is not equal to 16.