We would like to construct a 99% confidence interval for the mean, based on a sample of 50 observations. The sample standard deviation is 5.5 and the sample mean is 22. The population standard deviation is 3.6. Calculate the lower confidence limit. Round your answer to 3 decimals if needed

Lower confidence limit = 20.191

Step-by-step explanation:

Sample mean; x' = 22

Sample standard deviation; s = 5.5

Sample size; n = 50

Confidence interval = 99%

Since sample size > 30, we will use the formula;

CI = x' ± zs/√n

Where;

x' is sample mean

s is sample standard deviation

n is sample size

z is z-value of the confidence interval

From the table attached, z for 99% = 2.3263

Thus;

CI = 22 ± (2.3263 × 5.5)/√50

CI = 22 ± 1.809

CI = 23.809 or 20.191

Lower confidence limit = 20.191

4 & 3

step-by-step explanation:

factor of 108=2×2×3×3×3=2×2×3^3=the number is 4 because 2 only occurs 2 times and 2×2=4 and factor of 2187=3^3×3^3×3=the number is 3 because 3 onlyvoccur 1 time.

answer:   n ≤ -3 or n < -5

step-by-step explanation:

-5 + 2n ≥ -11

+5             +5

2n ≥ -6

2/2 ≤ -6/2

n ≤ -3

-6 - 10n > 44

+6           +6

-10n > 50

-10/-10 50/-10

n < -5