A study of 50 mathematics professors showed that they spend, on average,12.4 minutes correcting a student's final examinations. Find a 90% confidence interval of the mean time for all mathematics peofessors spend on conrrecting final examinations when standard deviation = 2.2 minutes.
CI90 = mean ± 1.645 (sd/√n)
With your data:
CI90 = 12.4 ± 1.645 (2.2/√50)
I'll let you finish the calculation.
To find the 90% confidence interval for the mean time spent by mathematics professors on correcting final examinations, we can use the formula:
Confidence Interval = [sample mean - margin of error, sample mean + margin of error]
Let's calculate the margin of error using the following steps:
Step 1: Identify the given information:
Sample Size (n) = 50
Sample Mean (x̄) = 12.4 minutes
Standard Deviation (σ ) = 2.2 minutes
Confidence Level (CL) = 90% (which implies α = 1 - CL = 0.10)
Step 2: Determine the critical value (z-value) corresponding to the confidence level.
Since we have a large sample size (n > 30) and know the population standard deviation, we can use the Z-distribution. The z-value is obtained from the standard normal distribution table or calculator. For a 90% confidence level, the critical z-value is approximately 1.645.
Step 3: Calculate the margin of error:
Margin of Error (E) = z-value * (standard deviation / √n)
E = 1.645 * (2.2 / √50)
Step 4: Compute the Confidence Interval:
Confidence Interval = [sample mean - margin of error, sample mean + margin of error]
CI = [12.4 - (1.645 * (2.2 / √50)), 12.4 + (1.645 * (2.2 / √50))]
Now, let's calculate the confidence interval:
CI = [12.4 - (1.645 * (2.2 / √50)), 12.4 + (1.645 * (2.2 / √50))]
Simplifying the equation:
CI = [12.4 - 0.631, 12.4 + 0.631]
CI = [11.769, 13.031]
Therefore, the 90% confidence interval for the mean time spent by mathematics professors on correcting final examinations, when the standard deviation is 2.2 minutes, is [11.769, 13.031] minutes.