A manufacturing process produces bags of cookies. The distribution of the weight

Question

A manufacturing process produces bags of cookies. The distribution of the weights of the contents of these bags is approximately Normal with a mean of 15.0 oz. and standard deviation 1.0 oz. Suppose that we take many samples of size n = 16, from this population, weigh each bag in the sample and then compute the mean weight for each sample of 16. The Central Limit Theorem tells us that the shape of the distribution of the means of those samples, as the number of samples taken increases, will look approximately ___ NORMAL, with a mean 15 and a standard deviation of .25 Suppose | draw one sample of size n = 16 and find the mean of the weights of the bags in that sample is 14.6 oz. A 98% Confidence interval to estimate the population mean u would be: suppose we draw random samples of size n from the initial population and want to construct a confidence interval with the same confidence level as in problem 10), BUT we want to reduce the Margin of Error for this confidence interval to a value of no more than .3 What sample size is needed to achieve this level of accuracy? If samples of size n=100 are selected randomly from the original population, the probability that the sample mean will be between 14.868 and 15.132 ounces is about: Suppose we want to take samples of size n from the initial population. If we want the distribution of the sample means from those samples to have a standard deviation of .03, what is the sample size n that we must use to achieve this goal?

Explanation / Answer

10) for 98% CI,z =2.326

std error =0.25

hence confidence interval =sample mean -/+ z*std error =14.0184 ; 15.1816

11)margin of errorE =0.3

hence sample size n=(z*std deviation/E)2 =~61

12)for n=100; std error =0.1

hence P(14.868<X<15.132)=P(-1.32<Z<1.32)=0.9066-0.0934=0.8132

13)sample size=(1/0.03)2 =~1112

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