Question

1. Assume that the weights of all girl scout cookie boxes are normally distributed with a mean of 9.0 ounces and a standard deviation of .24 ounces. If you were to generate a distribution of sample means for cookie box samples of size n=25 from this population, what would you expect the mean of sampling distribution to be? 8.75 oz, 0.05,9.0,225

2. a random sample of n=16 cookie boxes is obtained from the population of girl scout cookies (mean= 9.0 ounces, sigma= .24 ounces). What is the probability that this sample will have a mean weight greater than 9.06 ounces? 0.8413, 0.5000, 0.0228, 0.15787

3. Imagine that a new random sample of cookie boxes is obtained from the population of girl scout cookies, but this time the sample is much large n=100. compute the z score for a sample mean of 9.06? -2.50, 2.50, .25, -.25

4. What is the main problem with point estimates of population parameters? why can't we just use the sample mean as our estimate of the population mean without doing more computations? a) point estimates do not account for sampling error, they tell us nothing about the population parameters they are estimating, nothing, ping estimates are 100% accurate, we can't use them to estimate population means

Answer 1

here, s = sample standard deviation = sigma/sqrt(n)

4. When the sample mean is used as a point estimate of the population mean, some error can be expected owing to the fact that a sample is a subset of the population.

As they do not account for the sampling error