Question

1. A process is known to produce bricks whose weights are normally distributed with standard deviation 0.12 pounds. A random sample of sixteen bricks from today's output had a mean weight of 4.07 pounds.
   1. Find a 99% confidence interval for the mean weight of all bricks produced today. (6)

2. Without doing the calculations, state whether a 95% confidence interval for the population mean would be wider than, narrower than, or the same width as that found in (a). (2)

3. It is decided that tomorrow a sample of twenty bricks will be taken. Without doing the calculations, state whether a correctly calculated 99% confidence interval for the mean weight of tomorrow's output will be wider than, narrower than, or the same width as that found in (a). (2)

Answer 1

Question (a)

Given Sample Mean = 4.07

Stamdard Deviation = 0.13

Sample size, n = 16

Given the population is normally distributed, So

99% confidence interval fort the mean weight = Z/2 * ( / )

= 1- confidence level

= 1 - 0.99

= 0.01

/2 = 0.005

So we need to find a z-score that has an area of 0.995 to its left

The z-score of 2.576 has an area of 0.995 to its left

99% confidence interval fort the mean weight = 4.07 2.576 * (0.13 / )

= 4.07 2.576 * (0.13 /4)

= 4.07 2.576 *0.035

= 4.07 0.08372

= (3.98628, 4.15372)

So the 99% confidence interval for the mean weight of all bricks produced today is (3.98628, 4.15372)

Question (b)

We have decreased the confidecne level from 99% to 95% which implies the Z/2 will decrease from 2.576 to 1.96

So  the Margin of error Z/2 * ( / ) decreases as we decrease the confidence level from 99% to 95% since the Standard deviation S and sample size remains same in both the cases

The confidence interval Z/2 * ( / ) becomes narrow as we decreae the confidence level from 99% to 95% since the Sample Mean , Standard deviation S and sample size remains same in both the cases

A 95% confidence interval for the population mean would be narrower than a 99% confidence interval for the population mean

Question (c)

The sample size is increased from 16 to 20

So  Z/2 * ( / ) decreases as weincrease the sample size from 16 to 20, since the Sample size is in denominator, increasing it would decrease the Maegin of error Z/2 * ( / ) as Standard deviation S and confidecne level reamain the same in the both the cases

The confidence interval Z/2 * ( / ) becomes narrow as we increase the sample size from 16 to 20 since the Sample Mean , Standard deviation S and Confidecne level remain the same in both the cases

correctly calculated 99% confidence interval for the mean weight of tomorrow's output with a sample size of 20 will be narrower than  as that found in the case with a sample size of 16