Students in a representative sample of 69 second-year students selected from a large university in England participated in a study of academic procrastination. Each student in the sample completed the Tuckman Procrastination Scale, which measures procrastination tendencies. Scores on this scale can range from 16 to 64, with scores over 40 indicating higher levels of procrastination. For the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.85.

Required:
a. Construct a 95% confidence interval estimate of μ, the population mean procrastination scale for second-year students at this college.
b. How does the confidence interval for the population mean score for second-year students compare to the confidence interval for first-year students of (35.447, 38.633)?

Answer:

a) The 95% confidence interval estimate of μ is (39.35, 42.5).

b) Second-year students tend to procrastinate more than first-year students

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 69 - 1 = 68

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 68 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 1.9955

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 1.9955\frac{6.85}{\sqrt{69}} = 1.65[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 41 - 1.65 = 39.35

The upper end of the interval is the sample mean added to M. So it is 41 + 1.65 = 42.65

The 95% confidence interval estimate of μ is (39.35, 42.5).

b. How does the confidence interval for the population mean score for second-year students compare to the confidence interval for first-year students of (35.447, 38.633)?

The confidence interval for second-year students had higher values both at the lower and the upper end of the interval, which means that they tend to procrastinate more than first-year students

The choises are

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