Some latest study shows that college students are increasingly skimping on their sleep. A researcher from a small community college wants to estimate the mean weekdays sleep time of its students. He takes a random sample of 87 students and records their weekday mean sleep time as 6.80 hours. Assume that the population standard deviation is fairly stable at 1.40 hours.
Some latest study shows that college students are increasingly skimping on their sleep. A researcher from a small community college wants to estimate the mean weekdays sleep time of its students. He takes a random sample of 87 students and records their weekday mean sleep time as 6.80 hours. Assume that the population standard deviation is fairly stable at 1.40 hours.
(a) Construct a 94% confidence interval for the population mean weekday sleep time of all students of this college.
Use the following table of zα2zα2 for most common αα values.
| αα | 0.005 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.1 |
|---|---|---|---|---|---|---|---|---|
| zαzα | 2.575 | 2.33 | 2.05 | 1.88 | 1.75 | 1.645 | 1.555 | 1.28 |
| zα2zα2 | 2.81 | 2.575 | 2.33 | 2.17 | 2.05 | 1.96 | 1.88 | 1.645 |
Find zα2=zα2=
Round the margin of error and confidence interval values to 2 decimal places.
Calculate the margin of error E=E=
What is the 94% confidence interval?
≤μ≤≤μ≤
(b) If the researcher wants to be 99% confident that the sample mean is within 0.30 hours (or 18 min) of the true population mean weekday sleep time, how many students should he include in the sample?
Find zα2=zα2=
Round your answer (n) UP to the next whole number.
The required
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