We have the survey data on the body mass index ( BMI) of 664 young women. The mean BMI in the sample was . We treated these data as an SRS from a Normally distributed population with standard deviation 7.4. Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence. Interval (0.01)? margins of error (0.0001)?

Step-by-step explanation:

Margin of error = critical value × standard error

The standard error is:

s = σ / √n

s = 7.4 / √664

s = 0.287

The population is normally distributed, so the critical value can be found from a z-score table.

For 90% confidence, z = 1.645.

For 95% confidence, z = 1.960.

For 99% confidence, z = 2.576.

The margins of error are:

90%: 1.645 × 0.287 = 0.4724

95%: 1.960 × 0.287 = 0.5629

99%: 2.576 × 0.287 = 0.7398

The mean is 26.6, so the confidence intervals are:

90%: (26.6 − 0.47, 26.6 + 0.47) = (26.13, 27.07)

95%: (26.6 − 0.56, 26.6 + 0.56) = (26.04, 27.16)

99%: (26.6 − 0.74, 26.6 + 0.74) = (25.86, 27.34)

step-by-step explanation:

if u want da anser kill your teeser ifu hav wun

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;