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In order to estimate the average time spent on the computer terminals per student at a local university, data were collected from a sample of 64 business students over a one-week period. Assume the population standard deviation is 1.2 hours.a) With a 0.98 probability, the margin of error is approximately?b) If the sample mean is 10 hours, then the 98% confidence interval is approximately
?c) The standard error of the mean is?

Respuesta :

Answer:

a) 0.3495

b) (9.65, 10.35)

Step-by-step explanation:

We are given the following in the question:

Sample mean, [tex]\bar{x}[/tex] = 10 hours

Sample size, n = 64

Alpha, α = 0.02

Population standard deviation, σ = 1.2 hours

a) Margin of error

Formula

[tex]z_{critical}\dfrac{\sigma}{\sqrt{n}}[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.02} = \pm 2.33[/tex]

Putting values, we get,

[tex]M.E = (2.33)\dfrac{1.2}{\sqrt{64}} = 0.3495[/tex]

b)  the sample mean is 10 hours, then the 98% confidence interval

[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

[tex]\bar{x} \pm M.E[/tex]

Putting the values, we get,

[tex]10\pm 0.3495\\=(9.6505, 10.3495)\\\approx (9.65, 10.35)[/tex]