Question

Confidence Intervals for Means

Complete each of the following calculations by providing the correct formula and values that you are using for each problem.

1) A sample size of n = 120 produced the sample mean of ? ̅ = 24.3. Assuming the population standard deviation ? = 5.2, compute a 99% confidence interval for the population mean. Interpret the confidence interval.


2) Assuming the population standard deviation ? = 4, how large should a sample be to estimate a population mean with a margin of error of 0.2 for a 95% confidence interval?


3) The manager of a plant would like to estimate the mean amount of time a worker takes to complete a specific task. Assume the population standard deviation for this task is 4.1 minutes.

a. After observing 70 workers completing the same type of task, the manager calculated the average time to be 12.7 minutes. Construct a 90% confidence interval for the mean task time. Interpret the confidence interval.

b. How large a sample size n should he observe to decrease the margin of error to 0.5 minutes for the 90% interval?

4) A sample of 25 was selected out of a specific population with mean equal to 18.4 and sample standard deviation of 3.6. Construct a 95% confidence interval for the mean of the population. Interpret the confidence interval.


5) A group of students were randomly selected to participate in a study that compared the female grades to the male grades for a specific test. There were 15 females with a mean grade of 94.3 and sample standard deviation of 3.6. There were 12 males with a mean grade of 90.6 and sample standard deviation of 5.1. Construct a 90% confidence interval for the difference between the females’ and males’ test grades. Interpret the confidence interval.

Answer 1

(1) Given : = 24.3, = 5.2, n = 120, = 0.01

Since population standard deviation is known and n > 30, the Zcritical (2 tail) for = 0.01, is 2.576

The Confidence Interval is given by ME, where

The Lower Limit = 24.3 - 1.22 = 23.08

The Upper Limit = 24.3 + 1.22 = 25.52

The Confidence Interval is (23.08 , 25.52)

We are 99% confident that the true mean lies within the limits of the confidence interval from 23.08 till 25.52.

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(2) Given ME = 0.2, = 4, = 0.05

The Zcritical at = 0.05 is 1.96

The ME is given by :

Squaring both sides we get: (ME)² = (Z critical)² * ²/n

Therefore n = (Zcritical * /ME)² = (1.96*4/0.2)² = 1536.64

Therefore n = 1537 (Taking it to the next whole number)

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(3) (a) Given : = 12.7, = 4.1, n = 70, = 0.10

Since population standard deviation is known and n <30, the Zcritical (2 tail) for = 0.10 is 1.645

The Confidence Interval is given by ME, where

The Lower Limit = 12.7 - 0.8 = 11.9

The Upper Limit = 12.7 + 0.8 = 13.5

The Confidence Interval is (11.9 , 13.5)

We are 90% confident that the true population mean lies within the limits of the confidence interval from 11.9 till 13.5.

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(b) Given ME = 0.5, = 4.1, = 0.05

The Zcritical at = 0.05 is 1.645

The ME is given by :

Squaring both sides we get: (ME)² = (Z critical)² * ²/n

Therefore n = (Zcritical * /ME)² = (1.96 * 4.1 / 0.5)² = 181.95

Therefore n = 182 (Taking it to the next whole number)

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(4) Given : = 18.4, S = 3.6, n = 25, = 0.05

Since population standard deviation is not known and n <30, the Zcritical (2 tail) for = 0.05, df = n - 1 = 24, is 2.064

The Confidence Interval is given by ME, where

The Lower Limit = 18.4 - 1.5 = 16.9

The Upper Limit = 18.4 + 1.5 = 19.9

The Confidence Interval is (16.9 , 19.9)

We are 95% confident that the true population mean lies within the limits of the confidence interval from 16.9 till 19.9.

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(5) Given:

= 94.3, s₁ = 3.6, n₁ = 15,

= 90.6, s₂ = 5.1, n₂ = 12,

Since s₁/s₂ = 3.6/5.1 = 0.7 (it lies between 0.5 and 2) we used the pooled standard deviation

The Pooled Variance is given by:

= 0.10

The tcritical (2 tail) for = , df = n₁ + n₂ – 2 = 15 + 12 – 2 = 25 is 1.708

The Confidence Interval is given by (- ) ME, where

(- ) = 94.3 - 90.6 = 3.7

The Lower Limit = 3.7 - 2.86 = 0.84

The Upper Limit = 3.7 + 2.86 = 6.56

The Confidence Interval is (0.84 , 6.56)

We are 90% confident that the true difference in population means lies within the limits of the confidence interval from 0.84 till 6.56.

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