Astudy of two hundred teens found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 12 hours. the population standard deviation is 4 hours. what is the margin of error for a 98% confidence interval?

Since we know the population standard deviation and the sample size is more than 30, therefore we use the z statistic test. The formula for calculating the z score is given as:

At 98% confidence level, the value of z is:

z = 2.33

Now we can calculate for the margin of error (MOE) using the formula:

MOE = z * σ / sqrt (n)

where,

σ = standard deviation = 4 hours

n = sample size = 200

Substituting the given values:

MOE = 2.33 * 4 / sqrt (200)

MOE = 0.66

Therefore the margin of error for 98% confidence level is 0.66

b.my answer is b.

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