Question

a class survey in a large class for first-year college students asked, \about how many hours do you study during a typical week?\ the mean response of the 463 students was \\( \bar{x} = 13.7 \\) hours. suppose that we know that the study time follows a normal distribution with standard deviation \\( \sigma = 7.4 \\) hours in the population of all first-year students at this university. regard these students as an srs from the population of all first-year students at this university. does the study give good evidence that students claim to study more than 13 hours per week on the average?
the national survey of student engagement conducts an annual survey that includes hours spent preparing for class each week. the numbers in this exercise are based on the results of the 2015 survey. you can find survey results at nuse.indiana.edu/html/findings.cfm.
what is the value of the test statistic \\( z \\)? give your answer to two decimal places.

Explanation:

Step1: Recall z - statistic formula

The formula for the z - statistic (also known as the z - score) for a sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

From the problem, we assume that the population mean \(\mu = 13\) (since we are testing if students study more than 13 hours per week on average), \(\bar{x}=13.7\), \(\sigma = 7.4\), and \(n = 463\).

Step2: Calculate the standard error

First, we calculate the standard error \(SE=\frac{\sigma}{\sqrt{n}}\). Substitute \(\sigma = 7.4\) and \(n = 463\) into the formula:
\(\sqrt{463}\approx21.517\)
\(SE=\frac{7.4}{21.517}\approx0.344\)

Step3: Calculate the z - statistic

Now, use the z - statistic formula \(z=\frac{\bar{x}-\mu}{SE}\). Substitute \(\bar{x}=13.7\), \(\mu = 13\), and \(SE\approx0.344\) into the formula:
\(z=\frac{13.7 - 13}{0.344}=\frac{0.7}{0.344}\approx2.03\)

Answer:

\(2.03\)