Question

The owner of a factory wants to estimate the mean cost of manufacturing a product. She takes a random sample of 36 products and finds the sample mean cost to be $90. Suppose the population standard deviation is known to be σ = $10.

A) Create a 90% confidence interval for the population mean.

B) Interpret your confidence interval from part (a).

C) Without calculating another confidence interval, if the sample of 36 products instead had a mean of $200, would a 90% confidence interval be narrower, wider or the same width as your confidence interval from part a)? Briefly explain.

Answer 1

Solution:
Given that, σ=$10, n= 36
x̄= $90
(1–α)%=90%
α=0.10
α/2 =0.05
Zα/2 =1.645 ....from standard normal table.
Margin of error=E=Zα/2 ×(σ/✓n)
=1.645 ×(10/✓36)
=1.645× 1.6667
=2.7417
Margin of error=E=2.7417
90% confidence interval for population mean given as,
x̄± Margin of error=(90 - 2.6667,90+2.7417)
=(87.2583,92.7417)

Lower limit =87.2583
Upper limit=92.7417
B)90% confidence interval for population mean lies between (87.2583,92.7417)
C)As we know the factor affecting the confidence interval are,
I) confidence level
II) standard deviation
III) sample size
Here, all these parameter are the same so the margin of error for both confidence interval are same and the sample mean does not affect the width of confidence interval so both confidence interval have the same width.