Answer to Question #300467 in Statistics and Probability for john michael

Question

the average hour spent using the computer by 20 senior high school students during online class is 7 hour with a standard deviation of 2 hours. construct a 98% confidence interval of average hour spent by all senior high school students. identify the width of the interval

Answer 1

The critical value for $\alpha = 0.02$ and $df = n-1 = 19$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.539483.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times \dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(7-2.539483\times\dfrac{2}{\sqrt{20}},7+2.539483\times\dfrac{2}{\sqrt{20}})$$

$$=(5.8643, 8.1357)$$

Therefore, based on the data provided, the 98% confidence interval for the population mean is $5.8643<\mu<8.1357,$ which indicates that we are 98% confident that the true population mean $\mu$ is contained by the interval $(5.8643, 8.1357).$

The width of the interval is

$$W=2(t_c\times \dfrac{s}{\sqrt{n}})=2.2714$$