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. Your grade will be based on selecting the appropriate formula, substituting the appropriate values, and correctly calculating the answer. 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)
sample mean, xbar = 24.3
sample standard deviation, σ = 5.2
sample size, n = 120

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

ME = zc * σ/sqrt(n)
ME = 2.58 * 5.2/sqrt(120)
ME = 1.22

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (24.3 - 2.58 * 5.2/sqrt(120) , 24.3 + 2.58 * 5.2/sqrt(120))
CI = (23.0753 , 25.5247)
Therefore, based on the data provided, the 99% confidence interval for the population mean is 23.0753 < μ < 25.5247 which indicates that we are 99% confident that the true population mean μ is contained by the interval (23.0753 , 25.5247)

2)

The following information is provided,
Significance Level, α = 0.05, Margin or Error, E = 0.2, σ = 4

The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.96 * 4/0.2)^2
n = 1536.64

Therefore, the sample size needed to satisfy the condition n >= 1536.64 and it must be an integer number, we conclude that the minimum required sample size is n = 1537

3)

a)

sample mean, xbar = 12.7
sample standard deviation, σ = 4.1
sample size, n = 70

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

ME = zc * σ/sqrt(n)
ME = 1.64 * 4.1/sqrt(70)
ME = 0.8

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (12.7 - 1.64 * 4.1/sqrt(70) , 12.7 + 1.64 * 4.1/sqrt(70))
CI = (11.8963 , 13.5037)
Therefore, based on the data provided, the 90% confidence interval for the population mean is 11.8963 < μ < 13.5037 which indicates that we are 90% confident that the true population mean μ is contained by the interval (11.8963 , 13.5037)

b)

The following information is provided,
Significance Level, α = 0.1, Margin or Error, E = 0.5, σ = 4.1

The critical value for significance level, α = 0.1 is 1.64.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.64 * 4.1/0.5)^2
n = 180.85

Therefore, the sample size needed to satisfy the condition n >= 180.85 and it must be an integer number, we conclude that the minimum required sample size is n = 181
Ans : Sample size, n = 181

4)
sample size, n = 25
degrees of freedom, df = n - 1 = 24

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.06

ME = tc * s/sqrt(n)
ME = 2.06 * 3.6/sqrt(25)
ME = 1.483

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (18.4 - 2.06 * 3.6/sqrt(25) , 18.4 + 2.06 * 3.6/sqrt(25))
CI = (16.9168 , 19.8832)
Therefore, based on the data provided, the 95% confidence interval for the population mean is 16.9168 < μ < 19.8832 which indicates that we are 95% confident that the true population mean μ is contained by the interval (16.9168 , 19.8832)

5)

5)

x1 = 94.3 , s1 = 3.6 , n1 = 15
x2 = 90.6 , s2 = 5.1 , n2 = 12

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.711

CI = (x1- x2) +/- t *sqrt(s1^2/n1+s2^2/n2)
= ( 94.3 - 90.6) +/- 1.711 * sqrt(3.6^2/15 + 5.1^2/12)
= (0.7209 , 6.6791)