Question

A random sample of 200 adult residents in the U.S reveals that the mean minutes of sleep they received the night prior to the interview was 528 minutes with a standard deviation of 141 minutes. Using a computer, a random sample of 200 observations was taken (with replacement) from these data and an average was computed and recorded. This was repeated 10,000 times and a visualization was made of this sampling distribution. The mean of the sampling distribution was 528 minutes and the standard deviation was 10 minutes. The sampling distribution was symmetric and unimodal. Which of the following is an approximate 95% confidence interval for the mean amount of sleep of all US residents?

a. Lower bound: 528-141, upper bound: 528+141; (387, 669)
b. Lower bound: 528-10, Upper bound: 528+10; (518,538)
c. Lower bound: 528-20141, upper bound: 528 * 2*141; 246,810)
d. lower bound: 528-2-10, upper bound: 528+2*10; (508, 548)

Answer 1

d. lower bound: 528-2-10, upper bound: 528+2*10; (508, 548)

Explanation:

The computation of the approximate 95% confidence interval is shown below:

Given that

mean = 528,

standard deviation = 10

`Mean \mp z_{(0.05)/(2)}* standard\ deviation \\\\=528 \mp 2 * 10\\\\=(508,548)`

hence, the correct option is d.

Answer 2

The confidence interval gives a range of estimate for the true population mean at a given confidence level. The confidence interval for the mean is lower bound: 528-2-10, upper bound: 528+2*10; (508, 548)

The confidence interval for the mean is obtained using the relation :

• μ ± Z*σ
• Z* = 95% critical level = ±1.96
• μ = sample mean = 528
• σ = standard deviation = 10 minutes

Plugging in the values :

528 ± (1.96 × (10))

Approximating Z* (1.96) = 2.00

Lower boundary = 528 - (2 × 10) = 528 - 20 = 508

Upper boundary = 528 + (2 × 10) = 528 + 20 = 548

Therefore, the confidence interval is lower bound: 528-2-10, upper bound: 528+2*10; (508, 548)