Question

(5 points) A milk processor monitors the number of bacteria per milliliter in raw milk received at the factory. A random sample of 10 one-milliliter specimens of milk supplied by one producer gives the following data (assume the population is Normal):

5370

4890

5100

4500

5260

5150

4900

4760

4700

4870

Construct and interpret a 90% confidence interval for the population mean μ.

Answer 1

calculate mean and standard deviation for the given sample data

Answer 2

calculate mean and standard deviation for the given sample data

Count, N: 10 Sum, Ex: 49500 Mean, X 4950 Variance, s2: 72066.666666667 Steps N 1 S = (di - T)? N-1 i=1 52 - E(X; - 8)2 N-1 (5370 - 4950)2 + ... + (4870 - 4950)2 10-1 648600 9 = 72066.666666667 S = 172066.666666667 = 268.45235455601

The number of degrees of freedom are df = 10 – 1=9, and the significance level is a = 0.1. Based on the provided information, the critical t-value for a = 0.1 and df = 9 degrees of freedom is te = 1.833 The 90% confidence for the population mean u is computed using the following expression te x 8 CI = X (1 X + te x 8 vn Therefore, based on the information provided, the 90 % confidence for the population mean u is CI = (4950 1.833 x 268.45235455601 1.833 x 268.45235455601 4950+ 10 V10 (4950 – 155.617, 4950 + 155.617) = (4794.383,5105.617)

What this means is if we repeated experiments, the proportion of such intervals that contain u would be 90%