A sample of size n=15 is drawn from an approximately normal population whose standard deviation is $\sigma = 5.5$ The sample mean is $\bar { x } = 40.8$ Construct a 90 % confidence interval for $\mu .$
A) (30.28,51.32)
B) (40.80,43.14)
C) (39.80,41.80)
D) 38.46,43.14

Question

A sample of size  n=15  is drawn from an approximately normal population whose standard deviation is $\sigma = 5.5$ The sample mean is $\bar { x } = 40.8$ Construct a  90 %  confidence interval for  $\mu .$
A)  (30.28,51.32) 
B)  (40.80,43.14) 
C)  (39.80,41.80) 
D)  38.46,43.14

Quizplus · Accepted Answer

To construct a 90% confidence interval for the population mean $\mu$, we use the formula: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\sigma$ is the population standard deviation, $n$ is the sample size, and $z$ is the z-score corresponding to the desired level of confidence. For a 90% confidence level, the z-score is approximately 1.645.Plugging in the given values: $\bar{x} = 40.8$, $\sigma = 5.5$, and $n = 15$, we get:$$40.8 \pm 1.645 \times \frac{5.5}{\sqrt{15}}$$Calculating the margin of error:$$1.645 \times \frac{5.5}{\sqrt{15}} \approx 2.34$$Thus, the confidence interval is:$$40.8 \pm 2.34 = (38.46, 43.14)$$Therefore, the correct answer is D, which corresponds to the calculated confidence interval.

A Sample of Size N=15 Is Drawn from an Approximately