Question

we are performing a study about weekly per capita milk consumption. a previous study found weekly per capita milk consumption to be normally distributed, with a mean of 45.1 fluid ounces and a standard deviation of 8.8 fluid ounces. we randomly sample 30 people and record the weekly milk consumptions shown below. 38 55 33 39 44 49 30 64 37 48 41 23 39 54 45 41 41 50 48 47 36 29 23 41 61 42 50 46 53 39. the mean is 42.9 (round to one decimal place as needed.) find the standard deviation of your sample. the standard deviation is 9.8 (round to one decimal place as needed.) compare the mean and standard deviation of your sample with those of the previous study. discuss the differences. the sample mean is less than the previous mean, so, on average, consumption from the sample is greater than in the previous study. the sample standard deviation is less than the previous standard deviation by, so the milk consumption is more spread out in the sample.

Explanation:

Step1: Identify given values

Previous mean $\mu = 45.1$, sample mean $\bar{x}=42.9$, previous standard - deviation $\sigma = 8.8$, sample standard - deviation $s = 9.8$.

Step2: Compare means

Since $42.9<45.1$, the sample mean is less than the previous mean.

Step3: Compare standard deviations

Since $9.8>8.8$, the sample standard deviation is greater than the previous standard deviation. A larger standard deviation means the data is more spread out.

Answer:

The sample mean is less than the previous mean. The sample standard deviation is greater than the previous standard deviation, so the milk consumption is more spread out in the sample.