1). A class survey in a large class for first-year college students asked, "About how many hours do you study in a typical week?". The mean response of the 432 students was x¯¯¯ = 15 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation 8 hours in the population of all first-year students at this university.

What is the 99% confidence interval (±0.001) for the population mean?

Confidence interval is from _ to _ hours.

2). We have the survey data on the body mass index (BMI) of 650 young women. The mean BMI in the sample was x¯¯¯=26.5. We treated these data as an SRS from a Normally distributed population with standard deviation σ=7.

Find the margins of error for 99% confidence based on SRSs of N young women.

N margins of error (±0.0001)

106 ?

380 ?

1597 ?

Step-by-step explanation:

The confidence interval is:

CI = x ± ME, where x is the mean and ME is the margin of error.

The margin of error is:

ME = CV × SE, where CV is the critical value and SE is the standard error.  For a normal distribution, the critical value is the z-score at the given confidence interval.

The standard error is:

SE = σ / √n, where σ is the population standard deviation and n is the sample size.

1) At 99% confidence, the z-score is 2.576.  The standard error is:

SE = 8 / √432

SE = 0.385

The margin of error is:

ME = 2.576 × 0.385

ME = 0.992

The confidence interval is:

CI = 15 ± 0.992

CI = (14.008, 15.992)

2) At 99% confidence, the z-score is 2.576.

The margin of error is:

ME = 2.576 × 7 / √n

ME = 18.032 / √n

When n = 106, ME = 1.7514.

When n = 380, ME = 0.7943.

When n = 1597, ME = 0.3875.

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6

40 i don’t know lololol