Question

1. Some colleges have pushed back the starting time of morning classes because students were not getting enough sleep. A random sample of 46 college students had an average of 6.54 hours of sleep with a standard deviation is 1.90 hours. Compute and interpret a 98% confidence interval for the population mean amount of sleep that college students get per night. Also determine the critical values and calculate the margin of error.

2.The Travel Channel wants to conduct a survey of adults who take weekend vacations to Las Vegas. If the channel wants to be 97% confident that the sample percentage is in error by no more than 1.75 percentage points, how many adults must be surveyed?

Answer 1

1)

sample mean, xbar = 6.54
sample standard deviation, s = 1.9
sample size, n = 46
degrees of freedom, df = n - 1 = 45

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.412

Critical value = 2.412

ME = tc * s/sqrt(n)
ME = 2.412 * 1.9/sqrt(46)
Margin of Error = 0.676

2)

The following information is provided,
Significance Level, α = 0.03, Margin of Error, E = 0.0175

The provided estimate of proportion p is, p = 0.5
The critical value for significance level, α = 0.03 is 2.17.

The following formula is used to compute the minimum sample size required to estimate the population proportion p within the required margin of error:
n >= p*(1-p)*(zc/E)^2
n = 0.5*(1 - 0.5)*(2.17/0.0175)^2
n = 3844

Therefore, the sample size needed to satisfy the condition n >= 3844 and it must be an integer number, we conclude that the minimum required sample size is n = 3844
Ans : Sample size, n = 3844

if you take z value upto 3 or 4 decimal answer would be change