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The time spent, in hours, of teenagers on social media per year are normally distributed with a population standard deviation of 442 hours and an unknown population mean. If a random sample of 24 teenagers is taken and results in a sample mean of 1330 hours, find a 99% confidence interval for the population mean.

Respuesta :

Answer:

1329.85≤μ≤1330.14

Step-by-step explanation:

z score for 99% confidence interval

from the z table we can find that the score for 99% confidence is 2.58

formula

z=[tex]\frac{x-μ}{[tex]\frac{{σ}[tex]\sqrt{n}[/tex]}[/tex]}[/tex]

where

x-sample mean

μ-population mean

σ-population standard deviation

n-sample size

2.58=[tex]\frac{1330-μ}{[tex]\frac{{442}[tex]\sqrt{24}[/tex]}[/tex]}[/tex]

1330-2.58×24÷442≤μ≤1330+2.58×24÷442

1329.85≤μ≤1330.14

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