A study of 40 bowlers showed that their average score was 186. The standard devi
ID: 3200252 • Letter: A
Question
A study of 40 bowlers showed that their average score was 186. The standard deviation of the population is 6. Construct a 95% confidence interval for the mean score of all bowlers. A health care professional wishes to estimate the birth weights of infants. How large a sample must she select if she desires to be 90% confident that the true mean is within 6 ounces of the sample mean? The standard deviation is estimated to be 8 ounces. A random sample of 78 students were interviewed and 59 said they would re-elect the current president as student government president. Let p represent the proportion of all students who would re-elect the current president. Find a 90% confidence interval for p.Explanation / Answer
6.) µ = 186; = 6; n = 40; 1 - = 0.05
z/2 = 1.96
Standard Error (SE) = / n = 6/40 = 0.9486
CI = [µ ± z*E] = [184.1406, 187.8594]
7.) Standard Error (SE) = / n
Margin of Error (E) = z/2* / n
E = 6
= 8
n = [z/2* / E]2
1 - = 0.9
z/2 = 1.645
n = [1.645*8 / 6]2
= 4.81 ~ 5
8.) p = 59/78 = 0.7564
1 - p = 0.2435
SE = p*(1-p) / n
= 0.0486
= 0.1
z/2 = 1.645
CI = [p ± z/2*SE] = [0.6764, 0.8363]
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