A psychologist is collecting data on the time it takes to learn a certain task.For 50 randomly selected adult subjects,the sample mean is 16.40 minutes and the sample standard deviation is 4.00 minutes.Construct the 95 percent confidence interval for the mean time required by all adults to learn the task.
A) $\left[ \begin{array} { l l }
8.56 & 24.24
\end{array} \right]$
B) $[ 15.47 \quad 17.33 ]$
C) $[ 16.24 \quad 16.56 ]$
D) $\left[ \begin{array} { l l }
15.26 & 17.54
\end{array} \right]$
E) $[ 17.12 \quad 48.48 ]$

Question

A psychologist is collecting data on the time it takes to learn a certain task.For 50 randomly selected adult subjects,the sample mean is 16.40 minutes and the sample standard deviation is 4.00 minutes.Construct the 95 percent confidence interval for the mean time required by all adults to learn the task.
A) $\left[ \begin{array} { l l } 
8.56 & 24.24
\end{array} \right]$
B) $[ 15.47 \quad 17.33 ]$
C) $[ 16.24 \quad 16.56 ]$
D) $\left[ \begin{array} { l l } 
15.26 & 17.54
\end{array} \right]$
E) $[ 17.12 \quad 48.48 ]$

Quizplus · Accepted Answer

We can use the formula for a confidence interval for the population mean:

$$\text{CI} = \bar{x} \pm t_{\alpha/2,n-1} \frac{s}{\sqrt{n}}$$

where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, $t_{\alpha/2,n-1}$ is the t-score with $\alpha/2$ area in the upper tail and $n-1$ degrees of freedom.

Substituting the given values, we get:

$$\text{CI} = 16.40 \pm t_{0.025,49} \frac{4.00}{\sqrt{50}}$$

Using a t-table or calculator, we find that $t_{0.025,49} \approx 2.009$. Substituting this value, we get:

$$\text{CI} = 16.40 \pm 2.009 \frac{4.00}{\sqrt{50}} \approx [15.26, 17.54]$$

Therefore, the best choice is D, which matches this interval.

A Psychologist Is Collecting Data on the Time It Takes [8.5624.24]\left[ \begin{array} { l l } 8.56 & 24.24\end{array} \right][8.56​24.24​]

A Psychologist Is Collecting Data on the Time It Takes $\left[ \begin{array} { l l } 8.56 & 24.24\end{array} \right]$