Question

Can you help me

Answer 1

Investigators are interested in whether a new diet lowers total cholesterol in a group of individuals. They take a sample of 31 participants on the new diet. They know that the mean cholesterol level of the general population (µ) is 200 mg/dL, but they do not have information on the standard deviation of the population. The mean total cholesterol level in their sample of participants is 196 mg/dL and their sample has a standard deviation of 8 mg/dL.

You will use the information above for several of the following questions.

Answer 2

Use the data above examining the research question of interest: "Does the population of people on a new diet have a different mean cholesterol level than the general population?"

Which of the below represents a 95% confidence interval for the unknown population mean cholesterol level of people on a new diet.

(hint: since the population standard deviation (σ) is unknown, you must use the t-distribution for your confidence interval construction.)

Select one:
A. [193.24 – 200.31]
B. [198.32 – 202.31]
C. [194.29 – 204.31]
D. [190.73 – 198.04]
E. [193.07 – 198.93]

Answer 3

@kropot72

Answer 4

The confidence interval is found from:
\[\large C.I.=(\bar{x}-Z _{t}\frac{s}{\sqrt{n}},\ \bar{x}+Z _{t}\frac{s}{\sqrt{n}})\]
where Zt is found from a table of the t-distribution, using df = 31 -1 =30 and a two-tail probability value of 0.05 (for a 95% confidence interval).

Answer 5

You can find a t-distribution table here:
http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf