Answer:
99% confidence interval for the population average examination score is between a lower limit of 69.2872 and an upper limit of 82.7128.
Step-by-step explanation:
Confidence interval = mean +/- margin of error (E)
mean = 76
variance = 144
sd = sqrt(variance) = sqrt(144) = 12
n = 25
degree of freedom (df) = n - 1 = 25 - 1 = 24
confidence level (C) = 99% = 0.99
significance level = 1 - C = 1 - 0.99 = 0.01 = 1%
t-value corresponding to 24 df and 1% significance level is 2.797
E = t×sd/√n = 2.797×12/√25 = 6.7128
Lower limit = mean - E = 76 - 6.7128 = 69.2872
Upper limit = mean + E = 76 + 6.7128 = 82.7128
99% confidence interval is (69.2872, 82.7128)
using proportional law 5.6
Answer:
Your answer to the problem is D
Answer:
Copy all that then paste it to Google and there is a site that I found the answer on but I can't copy it it's to long.
