Astudy of one thousand teens found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 30 hours. the population standard deviation is 2 hours. what is the 95% confidence interval for the mean?

Given:

Sample Mean = 30
Sample size = 1000
Population Standard deviation or σ=2
Confidence interval = 95%

to compute for the confidence interval

Population Mean or μ= sample mean ± (z×SE)

where:

SE→ Standard Error

SE=σ√n= 30√1000=0.9486

Critical Value of z for 95% confidence interval =1.96

μ=30±(1.96×0.9486)
μ=30±1.8594

Upper Limit

μ = 30 + 1.8594 = 31.8594

Lower Limit

μ = 30 − 1.8594 = 28.1406

28.1406<u<31.8594

idk

step-by-step explanation:

maybe try   if it is work from a book, if not try

560

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