In: Statistics and Probability
Assuming the population has an approximate normal distribution, if a sample size n = 24 has a sample mean ¯ x = 49 with a sample standard deviation s = 2 , find the margin of error at a 99% confidence level. THEN YOU MUST ROUND TO TWO DECIMALS..
Solution :
Given that,
Point estimate = sample mean = = 49
sample standard deviation = s = 2
sample size = n = 24
Degrees of freedom = df = n - 1 = 24 - 1 = 23
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t /2,df = t0.005,23 = 2.807
Margin of error = E = t/2,df * (s /n)
= 2.807 * (2 / 24)
Margin of error = E = 1.15