Construct a $99 \%$ confidence interval for $n=42, \bar{x}=18.2$, and $\sigma=3.96$.
A) $16.63

Question

Construct a $99 \%$ confidence interval for $n=42, \bar{x}=18.2$, and $\sigma=3.96$.
A) $16.63<\mu<19.77$
B) $17.00<\mu<19.40$
C) $17.00<\mu<19.77$
D) $16.63<\mu<19.40$

Quizplus · Accepted Answer

To construct a 99% confidence interval for the mean when the population standard deviation ($\sigma$) is known, we use the formula: $\bar{x} \pm Z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $n$ is the sample size, and $Z$ is the Z-score corresponding to the desired level of confidence. For a 99% confidence level, the Z-score is approximately 2.576.Given $n=42$, $\bar{x}=18.2$, and $\sigma=3.96$, we calculate the margin of error (ME) as follows:$$ME = Z\frac{\sigma}{\sqrt{n}} = 2.576 \times \frac{3.96}{\sqrt{42}} \approx 1.57$$Thus, the confidence interval is:$$\bar{x} \pm ME = 18.2 \pm 1.57 = (16.63, 19.77)$$Therefore, the correct choice is $16.63<\mu<19.77$.

Construct A 99%99 \%99% Confidence Interval For n=42,xˉ=18.2n=42, \bar{x}=18.2n=42,xˉ=18.2 , And σ=3.96\sigma=3.96σ=3.96 A)

Construct A $99 \%$ Confidence Interval For $n=42, \bar{x}=18.2$ , And $\sigma=3.96$
A)