Question

Noise levels at 66 manufacturing plants were measured in decibels yielding the following data:

132,148,139,105,126,118

Construct the 95%confidence interval for the mean noise level at such locations. Assume the population is approximately normal.

1.Calculate the sample mean for the given sample data. Round your answer to one decimal place.

2.Calculate the sample standard deviation for the given sample data. Round your answer to one decimal place.

3.Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

4. Construct the 95%95% confidence interval. Round your answer to one decimal place.

Answer 1

Solution:

Given:  Noise levels at 6 manufacturing plants were measured in decibels yielding the following data:

132,148,139,105,126,118

We have to construct the 95%confidence interval for the mean noise level at such locations.

Part 1) Calculate the sample mean for the given sample data

T= n

x
132
148
139
105
126
118
Στ = 768

Thus

T= n

768 T =

T = 128.0

Part 2) Calculate the sample standard deviation for the given sample data.

8 = 14 12 – (Στ)2 η η - 1

x	x^2
132	17424
148	21904
139	19321
105	11025
126	15876
118	13924
Στ = 768	12 = 99474

Thus

8 = 14 12 – (Στ)2 η η - 1

199474-(76812/6 6-1

99474 - 98304 $= 15 S =

1170 s=15

s= V234

S= 15.3

Part 3) Find the critical value that should be used in constructing the confidence interval.

df = n -1 = 6 - 1 = 5

two tail area = 1 - c = 1 - 0.95 = 0.05

t Table cum. prob one-tail 0.50 1.00 0.25 0.50 0.20 0.40 ta 0.15 0.30 totes 0.10 0.05 0.20 0.10 376 0.025 0.05 two-tails 0.000 0.000 0.000 0.000 0.000 1.000 0.816 0.765 0.741 .72 1.376 1.963 1.061 1.386 0.978 1.250 0.941 1.190 00201156 3.078 1.886 1.638 1.533 1.476 6.314 2.920 2.353 2.132 2.0152.571

t_c = 2.571

Part 4) Construct the 95% confidence interval.

Formula;

(1 – E <μ< + Ε)

where

E = te x s/n

E = 2.571 x 15.3/6

E = 2.571 x 15.3/2.44949

E = 16.1

Thus

(1 – E <μ< + Ε)

(128.0 – 16.1 <μ< 128.0 + 16.1)

(111.9 <μ< 144.1)

Thus the 95%confidence interval for the mean noise level at such locations is :

(111.9 <μ< 144.1)