a study of 40 professors showed they spent on average 13.6 minutes correcting a student papers. Given a standard deviation of 17, the 90% confidence interval of the mean time for all the test papers is what?
Formula:
CI90 = mean + or - 1.645(sd/√n)
With your data:
CI90 = 13.6 + or - 1.645(17/√40)
I'll let you take it from here.
To calculate the 90% confidence interval for the mean time spent correcting student papers, we need to use the formula:
Confidence interval = X̄ ± Z * (σ / √n)
Where:
X̄ is the sample mean,
Z is the Z-score (corresponding to the desired confidence level),
σ is the population standard deviation, and
n is the sample size.
In this case, we are provided with:
Sample mean (X̄) = 13.6 minutes
Population standard deviation (σ) = 17 minutes
Sample size (n) = 40 professors
Confidence level = 90%
Step 1: Find the Z-score
The Z-score represents the number of standard deviations a particular value lies from the mean. For a confidence level of 90%, we need to find the Z-score that corresponds to a tail probability of (1 - 0.90) / 2 = 0.05. Looking up this value in the standard normal distribution table, the Z-score is approximately 1.645.
Step 2: Calculate the confidence interval
Plugging the values into the formula, we get:
Confidence interval = 13.6 ± 1.645 * (17 / √40)
To simplify the calculation, let's first calculate (17 / √40):
(17 / √40) ≈ 2.687
Now, substitute this value in the formula:
Confidence interval = 13.6 ± 1.645 * 2.687
Calculating further:
Confidence interval = 13.6 ± 4.424
Step 3: Finalize the confidence interval
Now, we can express the confidence interval as a range:
Lower limit = 13.6 - 4.424 ≈ 9.176
Upper limit = 13.6 + 4.424 ≈ 17.024
Therefore, the 90% confidence interval for the mean time spent correcting student papers is approximately (9.176, 17.024) minutes. This means we are 90% confident that the true population mean lies within this range.