Step 1 :Given that the sample mean (\(\bar{x}\)) is 25.5, the sample standard deviation (s) is 22.1, and the sample size (n) is 131, we want to find a 95% confidence interval for the mean number of times per day that college students text.
Step 2 :We use the formula for a confidence interval for a population mean with a known standard deviation: \(\bar{x} \pm t_{\frac{\alpha}{2}, n-1} \cdot \frac{s}{\sqrt{n}}\)
Step 3 :In this formula, \(t_{\frac{\alpha}{2}, n-1}\) is the t-score for a two-tailed test with significance level \(\alpha\) and \(n-1\) degrees of freedom. Since we want a 95% confidence interval, \(\alpha = 0.05\), and we need to find the value of \(t_{0.025, 130}\).
Step 4 :Using a t-distribution table or a statistical software, we find that \(t_{0.025, 130} = 1.9783804054271528\).
Step 5 :Substituting the given values and the calculated t-score into the formula, we get the confidence interval: \((25.5 - 1.9783804054271528 \cdot \frac{22.1}{\sqrt{131}}, 25.5 + 1.9783804054271528 \cdot \frac{22.1}{\sqrt{131}})\)
Step 6 :Solving the above expression, we get the confidence interval: \((21.67997304129269, 29.32002695870731)\)
Step 7 :\(\boxed{\text{Final Answer: With 95% confidence, the population mean number of texts per day is between 21.680 and 29.320.}}\)