Question

Problem 5: 17 points) Let X equal the weight in grams of a 52-gram snack pack of candies. Assume that the distribution of X is N(2,0?). A random sample of n = 10 observations of X yield the following data: 55.95, 56.54, 57,58, 55.13, 57.48, 56.06, 59.93, 58.3, 52.57, 58.46. (a) Give a point estimate of and a based on the data. [3] (b) Find the endpoints of a 95% confidence interval for . (2) (e) Suppose now that X follows a N(2). Repeat part b) and interpret the interval. 121

Answer 1

a) = (55.95 + 56.54 + 57.58 + 55.13 + 57.48 + 56.06 + 59.93 + 58.3 + 52.57 + 58.46)/10 = 56.8

T

s = sqrt(((55.95 - 56.8)^2 + (56.54 - 56.8)^2 + (57.58 - 56.8)^2 + (55.13 - 56.8)^2 + (57.48 - 56.8)^2 + (56.06 - 56.8)^2 + (59.93 - 56.8)^2 + (58.3 - 56.8)^2 + (52.57 - 56.8)^2 + (58.46 - 56.8)^2)/9) = 2.0519

b) df = 10 - 1 = 9

For 95% confidence interval, the critical value is t^* = 2.262

The 95% confidence interval is

S i Et** n

2.0519 = 56.8 + 2.262 * 10

55.33, 58.27

Lower limit = 55.33

Upper limit = 58.27

c) For 95% confidence interval, the critical value is z^* = 1.96

The 95% confidence interval is

i z* n

2 = 56.8±1.96 * 10

= 56.8 +1.24

55.56,58.04

We are 95% confident that the true population mean of weight lies in the above confidence interval.