Question

Suppose that the average time spent per day with digital media several years ago was 3 hours and 16 minutes. For last year, a random sample of 20 adults in a certain region spent the numbers of hours per day with digital media given in the accompanying table. Preliminary data analyses indicate that the t-interval procedure can reasonably be applied. Find and interpret a 95% confidence interval for last year's mean time spent per day with digital media by adults of the region. (Note: x=5.40 hr and s=2.28 hr.)Click here to view the digital media times. Click here to view page 1 of the t-table. Click here to view page 2 of the t-table1. The 95% confidence interval is from ___hour(s) ___to hour(s).(Round to two decimal places as needed.)2. Interpret the 95% confidence interval. Select all that apply.A. 95% of all adults in the region spent amounts of time per day on digital media last year that are between the interval's bounds.B. There is a 95% chance that the mean amount of time spent per day on digital media last year by all adults in the region is between the interval's bounds.C. With 95% confidence, the mean amount of time spent per day on digital media last year by all adults in the region is between the interval's bounds.D. 95% of all possible random samples of 20 adults in the region have mean amounts of time spent per day on digital media last year that are between the interval's bounds.

Answer 1

We have a sample with a size n = 20 with a mean of M = 5.40 hours and a standard deviation of s = 2.28 hours.

We have to find the 95% confidence interval from this sample.

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.

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

`s_M=(s)/(√(n))=(2.28)/(√(20))\approx(2.28)/(4.4721)\approx0.5098`

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\cdot0.5098=1.0670`

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

`\begin{gathered} LL=M-MOE=5.40-1.07=4.33 \\ UL=M+MOE=5.40+1.07=6.47 \end{gathered}`

The 95% confidence interval is from 4.33 hours to 6.47 hours.

We now have to find the correct interpretation of this confidence interval.

The confidence interval tell us that we expect 95% of the sample means drawn from this population will fall within this limits.

This correspond to the definition given in D: "95% of all possible random samples of 20 adults in the region have mean amounts of time spent per day on digital media last year that are between the interval's bounds."

We can also conclude that there is 95% confidence that the population mean will lie within the interval. This correspond to option C: "With 95% confidence, the mean amount of time spent per day on digital media last year by all adults in the region is between the interval's bounds."

Answer:

a) The 95% confidence interval is from 4.33 hours to 6.47 hours.

b) Options C and D