Question

20. A medical center conducted a study to investigate cholesterol levels in people who have had heart attacks. A random sample of 16 people was obtained from the names of all patients of the medical center who had a heart attack in the previous year. Of the people in the sample, the mean cholesterol level was 264.70 milligrams per deciliter (mg/dL) with standard deviation 42.12 mg/dL. Assuming all conditions for inference were met, whic of the following is a 90 percent confidence interval for the mean ch the medical center who had a heart attack in the previous year? olesterol level, in mg/dL, of all patients of (A) (242.26, 287.14) (B) (244.06, 285.34) (C) (246.24, 283.16) (D) (247.38, 282.02) (E) (260.09, 269.31)

Answer 1

Solution

Answer: (C) (246.24, 283.16)

given : sample size (n) = 16, sample mean (T) = 264.70, sample standard deviaton (s) = 42.12

Confidence level (C) = 0.90

Since population standard deviation is not known, Use t-distribution

degrees of freedom (df) = n-1 = 16 – 1= 15

a=1-C=1-0.90 = 0.10

Refer t-table or use excel function "=T.INV.2T(0.10, 15)" to find the value of t.

download for excel to find ta

https://drive.google.com/open?id=1Mh6HZEk7JzR6LFr9GZuOKXMBI7s1ppXv

..ta = 1.753

formula : Confidence interval for the population mean, u

= ī tq*{}

42.12 ) = 264.70 +1.753 * V16

= 264.70 +18.46

= (264.70 - 18.46.264.70 + 18.46)

= (246.24,283.16)