Question

The number of hours per day that Americans spend on social networking is approximately normally distributed. A random sample of 20 Americans who use social networks had

Answer 1

Answer:

B. (2.869, 3.411)

Step-by-step explanation:

The question is incomplete:

The number of hours per day that Americans spend on social networking is approximately normally distributed. A random sample of 20 Americans who use social networks had M = 3.14 hours and s = 0.58 hours. Find a 95% confidence interval for the actual mean number of hours that Americans spend on social networking.

A. (2.860, 3.420)

B. (2.869, 3.411)

C. (2.886, 3.394)

D. Not appropriate

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=3.14.

The sample size is N=20.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.58}{\sqrt{20}}=\dfrac{0.58}{4.472}=0.13$

The degrees of freedom for this sample size are:

$df=n-1=20-1=19$

The t-value for a 95% confidence interval and 19 degrees of freedom is t=2.093.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.093 \cdot 0.13=0.271$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 3.14-0.271=2.869\\\\UL=M+t \cdot s_M = 3.14+0.271=3.411$

The 95% confidence interval for the mean is (2.869, 3.411).