The human resources department of a consulting firm gives a standard creativity test to a randomly selected group of new hires every year. This year, 80 new hires took the test and scored a mean of 112.5 points with a standard deviation of 17.3. Last year, 55 new hires took the test and scored a mean of 117.5 points with a standard deviation of 17.3. Assume that the population standard deviations of the test scores of all new hires in the current year and the test scores of all new hires last year can be estimated by the sample standard deviations, as the samples used were quite large. Construct a 95% confidence interval for u1-u2, the difference between the mean test score u1 of new hires from the current year and the mean test score u2 of new hires from last year. Then complete the table below
Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.
What is the lower limit of the 95% confidence interval
What is the upper limit of the 95% confidence interval
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To construct a confidence interval for the difference between the mean test scores of new hires from the current year and the mean test scores of new hires from last year (u1 - u2), you can use the formula:
Confidence Interval = (x1 - x2) ± (t * sqrt(s1^2/n1 + s2^2/n2))
where:
x1 = mean test score of new hires from the current year
x2 = mean test score of new hires from last year
s1 = standard deviation of the test scores of new hires from the current year
s2 = standard deviation of the test scores of new hires from last year
n1 = number of new hires from the current year
n2 = number of new hires from last year
t = t-value based on the confidence level and degrees of freedom
In this case:
x1 = 112.5
x2 = 117.5
s1 = 17.3 (estimated from the sample)
s2 = 17.3 (estimated from the sample)
n1 = 80
n2 = 55
To find the t-value, you can use a t-distribution table or a statistical software. With a confidence level of 95% and degrees of freedom equal to the smaller sample size minus 1 (df = min(n1-1, n2-1)), the t-value is approximately 1.980.
Now, let's substitute the given values into the confidence interval formula:
Confidence Interval = (112.5 - 117.5) ± (1.980 * sqrt(17.3^2/80 + 17.3^2/55))
Note: The sqrt() represents the square root of its contents.
Calculating the square roots and performing the additions and subtractions:
Confidence Interval = -5 ± (1.980 * sqrt(0.036 + 0.055))
Calculating the values inside the square root:
Confidence Interval = -5 ± (1.980 * sqrt(0.091))
Calculating the square root:
Confidence Interval = -5 ± (1.980 * 0.302)
Calculating the value inside the parentheses:
Confidence Interval = -5 ± 0.599
Now, calculating the values of the lower and upper limits of the confidence interval:
Lower Limit = -5 - 0.599 = -5.599
Upper Limit = -5 + 0.599 = -4.401
Therefore, the lower limit of the 95% confidence interval for u1-u2 is -5.599, and the upper limit is -4.401.
To construct the confidence interval for the difference between the mean test scores of new hires from the current year and the mean test scores of new hires from last year (u1 - u2), you can use the formula:
CI = (x̄1 - x̄2) ± t * sqrt((s1^2/n1) + (s2^2/n2))
Where:
- x̄1 and x̄2 are the sample means
- s1 and s2 are the sample standard deviations
- n1 and n2 are the sample sizes
- t is the critical value from the t-distribution based on the desired confidence level (95% in this case), and the degrees of freedom (df = n1 + n2 - 2)
Given values:
x̄1 = 112.5
x̄2 = 117.5
s1 = s2 = 17.3
n1 = 80
n2 = 55
Now we'll calculate the confidence interval:
Degrees of freedom (df) = n1 + n2 - 2
df = 80 + 55 - 2
df = 133
t (critical value) can be determined using a t-table or a calculator with the given confidence level and degrees of freedom. For a 95% confidence level and df = 133, the t-value is approximately 1.9785.
Substituting the values into the formula:
CI = (112.5 - 117.5) ± 1.9785 * sqrt((17.3^2/80) + (17.3^2/55))
Now we can calculate the confidence interval:
CI = -5 ± 1.9785 * sqrt(301.19/80 + 301.19/55)
CI = -5 ± 1.9785 * sqrt(3.764875 + 5.474364)
Performing the calculations:
CI = -5 ± 1.9785 * sqrt(9.239239)
CI = -5 ± 1.9785 * 3.040646
Now we can find the lower and upper limits of the 95% confidence interval:
Lower limit = -5 - 1.9785 * 3.040646
Upper limit = -5 + 1.9785 * 3.040646
Lower limit ≈ -11.992
Upper limit ≈ 1.992
Therefore, the lower limit of the 95% confidence interval is approximately -11.992 and the upper limit is approximately 1.992.