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A study of 40 English professors showed that they spend, on average, 12.6 minutes correcting a student's term paper. The standard deviation was 2.5 minutes. Construct the 90% confidence interval of the mean for all term papers.

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Final Answer:

The 90% confidence interval for the mean time spent correcting a student's term paper by all English professors is approximately 12.0 to 13.2 minutes.

Step-by-step explanation:

To construct a confidence interval for the mean time spent correcting a student's term paper by all English professors, we can use the t-distribution and the sample data provided in the study. The sample size (n) is 40, the mean (μ) is 12.6 minutes, and the standard deviation (σ) is 2.5 minutes. To find the confidence interval, we need to calculate the degrees of freedom (df), which is n - 1 = 39, and then use the t-distribution to find the critical value (t*). The confidence level is 90%, which corresponds to a significance level of α = 0.10.

The formula for calculating the confidence interval is:

μ ± t* (σ / sqrt(n))

Substituting the given values into this formula, we get:

12.6 ± t*(2.5 / sqrt(40))

To find t*, we can use a t-distribution table or a statistical software program to determine the value that corresponds to a significance level of α = 0.10 and degrees of freedom of df = 39. The critical value is approximately -1.715 and 1.715, which we round to two decimal places for simplicity.

Using these values, we can calculate the confidence interval:

12.6 - 1.715(2.5 / sqrt(40)) < μ < 12.6 + 1.715(2.5 / sqrt(40))

Simplifying this expression, we get:

μ approximately between 12.0 and 13.2 minutes, with a confidence level of 90%. This means that if we were to repeat this study multiple times under similar conditions, we would expect the true mean time spent correcting a student's term paper by all English professors to fall within this interval approximately 90% of the time.

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User Ezrqn Kemboi
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