A sample of size $n=54$ has sample mean $\bar{x}=54.8$ and sample standard deviation $s=9.5$.
Part: $0 / 2$
Part 1 of 2
Construct a $98 \%$ confidence interval for the population mean $\mu$. Round the answers to one decimal place. A $98 \%$ confidence interval for the population mean is $\square< \mu< \square$.
Thus, a 98% confidence interval for the population mean is \(\boxed{51.8}<\mu<\boxed{57.8}\).
Step 1 :We are given a sample of size \(n=54\) with sample mean \(\bar{x}=54.8\) and sample standard deviation \(s=9.5\).
Step 2 :We are asked to construct a 98% confidence interval for the population mean \(\mu\).
Step 3 :The formula for a confidence interval is \(\bar{x} \pm Z_{\alpha/2} \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z_{\alpha/2}\) is the z-score corresponding to the desired confidence level, \(s\) is the sample standard deviation, and \(n\) is the sample size.
Step 4 :The z-score for a 98% confidence interval is approximately 2.33.
Step 5 :Substituting the given values into the formula, we get \(\bar{x} \pm Z_{\alpha/2} \frac{s}{\sqrt{n}} = 54.8 \pm 2.33 \frac{9.5}{\sqrt{54}}\).
Step 6 :Calculating the margin of error, we get approximately 3.0121919698058695.
Step 7 :Subtracting the margin of error from the sample mean, we get the lower bound of the confidence interval, which is approximately 51.78780803019413.
Step 8 :Adding the margin of error to the sample mean, we get the upper bound of the confidence interval, which is approximately 57.812191969805866.
Step 9 :Rounding the lower and upper bounds to one decimal place, we get 51.8 and 57.8 respectively.
Step 10 :Thus, a 98% confidence interval for the population mean is \(\boxed{51.8}<\mu<\boxed{57.8}\).