Question

the following data show the number of hours per day 12 adults spent in front of screens watching television - related content. complete parts a and b below. 1.7 4.7 3.7 5.2 7.6 7.3 5.3 2.2 5.1 1.7 2.3 8.5 a. construct a 99% confidence interval to estimate the average number of hours per day adults spend in front of screens watching television - related content. the 99% confidence interval to estimate the average number of hours per day adults spend in front of screens watching television - related content is from hours to hours. (round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample mean

The sample data is \(x = \{1.7,4.7,3.7,5.2,7.6,7.3,5.3,2.2,5.1,1.7,2.3,8.5\}\). The sample size \(n = 12\). The sample - mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
\(\sum_{i=1}^{12}x_{i}=1.7 + 4.7+3.7+5.2+7.6+7.3+5.3+2.2+5.1+1.7+2.3+8.5 = 55.3\)
\(\bar{x}=\frac{55.3}{12}\approx4.61\)

Step2: Calculate the sample standard deviation

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\).
First, calculate \((x_{i}-\bar{x})^{2}\) for each \(i\):
\((1.7 - 4.61)^{2}=(-2.91)^{2}=8.4681\), \((4.7 - 4.61)^{2}=(0.09)^{2}=0.0081\), \((3.7 - 4.61)^{2}=(-0.91)^{2}=0.8281\), \((5.2 - 4.61)^{2}=(0.59)^{2}=0.3481\), \((7.6 - 4.61)^{2}=(2.99)^{2}=8.9401\), \((7.3 - 4.61)^{2}=(2.69)^{2}=7.2361\), \((5.3 - 4.61)^{2}=(0.69)^{2}=0.4761\), \((2.2 - 4.61)^{2}=(-2.41)^{2}=5.8081\), \((5.1 - 4.61)^{2}=(0.49)^{2}=0.2401\), \((1.7 - 4.61)^{2}=(-2.91)^{2}=8.4681\), \((2.3 - 4.61)^{2}=(-2.31)^{2}=5.3361\), \((8.5 - 4.61)^{2}=(3.89)^{2}=15.1321\)
\(\sum_{i = 1}^{12}(x_{i}-\bar{x})^{2}=8.4681+0.0081 + 0.8281+0.3481+8.9401+7.2361+0.4761+5.8081+0.2401+8.4681+5.3361+15.1321 = 66.299\)
\(s=\sqrt{\frac{66.299}{11}}\approx2.45\)

Step3: Determine the critical - value

Since the sample size \(n = 12\) (small sample, \(n<30\)) and we want a 99% confidence interval, the degrees of freedom \(df=n - 1=11\). Looking up in the \(t\) - distribution table, the critical value \(t_{\alpha/2}\) for a 99% confidence interval (\(\alpha=1 - 0.99 = 0.01\), \(\alpha/2=0.005\)) is \(t_{0.005,11}=3.106\).

Step4: Calculate the margin of error

The formula for the margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\).
\(E = 3.106\times\frac{2.45}{\sqrt{12}}\approx3.106\times\frac{2.45}{3.464}\approx2.19\)

Step5: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\).
\(4.61-2.19<\mu<4.61 + 2.19\)
\(2.42<\mu<6.80\)

Answer:

from \(2.42\) hours to \(6.80\) hours