The dean at a local university is concerned about what affects the GPAs of students. A study is done to

Question

see if students who use social media extensively have lower GPAs than students who do not use social media extensively. Two random samples of students were taken from the university and the sample statistics are below:

Population Sample Size Sample Mean Sample Standard Deviation
Students use social media extensively n1= 135 x ̅1= 3.01 s1= 0.98
Students do not use social media extensively n2= 72 x ̅2= 3.89 s2= 0.38

In order to help the dean to see the effect of extensive use of social media on GPA, create a 95% confidence interval for the difference between the mean GPA of students who use and do not use social media extensively. Interpret this interval in context of the study above. (Hint: Does 0 lie in the interval? What does it mean?) Show calculator command and/or formula used to get answer in order to receive full credit. Use full sentences to interpret your results.

Answer 1

To construct a 95% confidence interval for the difference in mean GPAs between students who use social media extensively (Group 1) and those who do not (Group 2), we can use the following formula for the confidence interval of the difference between two means:

\[ \text{CI} = (\bar{x}_1 - \bar{x}_2) \pm t^* \times SE \]

Where:

• \(\bar{x}_1\) is the mean GPA of students using social media extensively.
• \(\bar{x}_2\) is the mean GPA of students not using social media extensively.
• \(t^*\) is the critical t-value from the t-distribution for the desired confidence level and degrees of freedom.
• \(SE\) is the standard error of the difference in means.

Step 1: Calculate the difference in sample means.

\[ \bar{x}_1 - \bar{x}_2 = 3.01 - 3.89 = -0.88 \]

Step 2: Calculate the standard error (SE) of the difference in means.

The formula for standard error of the difference between two sample means is:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]

Substituting the values:

\[ s_1 = 0.98, \quad n_1 = 135, \quad s_2 = 0.38, \quad n_2 = 72 \]

\[ SE = \sqrt{\frac{0.98^2}{135} + \frac{0.38^2}{72}} \]

Calculating \(s_1^2\) and \(s_2^2\):

\[ 0.98^2 = 0.9604 \quad \text{and} \quad 0.38^2 = 0.1444 \]

Now calculate \(SE\):

\[ SE = \sqrt{\frac{0.9604}{135} + \frac{0.1444}{72}} = \sqrt{0.007117 + 0.0020056} = \sqrt{0.0091226} \approx 0.0957 \]

Step 3: Determine the degrees of freedom using the formula for two samples:

Using the Welch-Satterthwaite equation:

\[ df \approx \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \]

Calculating:

\[ \frac{s_1^2}{n_1} = \frac{0.9604}{135} = 0.0071163 \] \[ \frac{s_2^2}{n_2} = \frac{0.1444}{72} = 0.0020056 \]

Thus,

\[ df \approx \frac{(0.0071163 + 0.0020056)^2}{\frac{(0.0071163^2)}{134} + \frac{(0.0020056^2)}{71}} \approx \frac{0.0091219^2}{0.0000002279 + 0.00005780} \approx \frac{0.00008314}{0.00005782} \approx 143.63 \]

Using a t-table or calculator, find \(t^*\) for \(df \approx 144\) at the 95% confidence level, which is approximately 1.976.

Step 4: Construct the confidence interval.

\[ \text{Margin of Error} = t^* \times SE = 1.976 \times 0.0957 \approx 0.189 \]

Thus, the confidence interval is:

\[ (-0.88 - 0.189, -0.88 + 0.189) = (-1.069, -0.691) \]

Interpretation:

The 95% confidence interval for the difference in mean GPAs between students who use social media extensively and those who do not is (-1.069, -0.691). This interval does not include 0, which indicates that we can be 95% confident that students who use social media extensively have lower GPAs compared to those who do not use social media extensively. In practical terms, this suggests a significant negative impact of extensive social media use on student GPAs at this university.