One researcher wishes to estimate the mean number of hours that high school students spend watching TV on a weekday. A margin of error of 0.28 hour is desired. Past studies suggest that a population standard deviation of hours is reasonable. Estimate the minimum sample size required to estimate the population mean with the stated accuracy.

Complete question:

One researcher wishes to estimate the mean number of hours that high school students spend watching TV on a weekday. A margin of error of 0.28 hours is desired. Past studies suggest that a population standard deviation of 1.5 hours is reasonable. Estimate the minimum sample size required to estimate the population mean with the stated accuracy.

111 students

Step-by-step explanation:

Given the following :

Margin of Error (E) = 0.28

Population standard deviation (sd) = 1.5

Recall:

Margin of Error(E) = Z * (sd/√n)

Taking a confidence interval of 95%

The Z value at a 95% confidence interval is 1.96

Plugging our values, we have :

Margin of Error(E) = Z * (sd/√n)

0.28 = 1.96 * (1.5/√n)

0.28 = 2.94 / √n

√n × 0.28 = 2.94

√n = 2.94 / 0.28

√n = 10.5

Square both sides to obtain n

n = 10.5^2

n = 110.25

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

where are the choices?