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 NO?). A random sample of n = 10 observations of X yield the following data: 55.95, 56.54, 57.58, 55.13, 57.48, 56.06,69.93, 18.3, 52.57, 58.46. (a) Give a point estimate of pand o based on the data (3) (b) Find the endpoints of a 95% confidence interval for f. [2] (e) Suppose now that X follows a N, 2). Repeat part b) and interpret the interval (2)

Answer 1

a) ∑x = 568

∑x² = 32300.29

n = 10

Mean , x̅ = Ʃx/n = 568/10 = 56.8

Standard deviation, s = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(32300.2928-(568)²/10)/(10-1)] = 2.0519

Point Estimate of µ = 56.8

Point Estimate of σ = √(s²/n) = √(4.2103/10) = 0.6489

b)

95% Confidence interval :

At α = 0.05 and df = n-1 = 9, two tailed critical value, t-crit = T.INV.2T(0.05, 9) = 2.262

Lower Bound = x̅ - t-crit*s/√n = 56.8 - 2.262 * 2.0519/√10 = 55.3322

Upper Bound = x̅ + t-crit*s/√n = 56.8 + 2.262 * 2.0519/√10 = 58.2678

55.332 < µ < 58.268

c)

σ = 2

95% Confidence interval :

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.960

Lower Bound = x̅ - z_c*σ/√n = 56.8 - 1.96 * 2/√10 = 55.560

Upper Bound = x̅ + z_c*σ/√n = 56.8 + 1.96 * 2/√10 = 58.040

55.560 < µ < 58.040

We can be 95% confident that the mean is between 55.560 and 58.040