b) A confidence interval was used to estimate the proportion of statistics stude
ID: 3225482 • Letter: B
Question
b) A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Using the information above, what total size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence? Explain/show how you obtain your answer
b) A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Using the information above, what total size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence? Explain/show how you obtain your answer
b) A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Using the information above, what total size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence? Explain/show how you obtain your answer
Explanation / Answer
answer
CHOOSING THE SAMPLE SIZE, POPULATION PROPORTION
= POPULATION PROPORTION 54 (54%)
B = SPECIFIED BOUND ON THE ERROR 0.08
Level of Confidence 95
'z critical value' from Look-up Table for 95% 1.96
Using Excel NORMSINV(probability) = NORMSINV ( 0.5*( 1 - 95 / 100 )))
Returns 'z critical value' from the inverse of the standard normal cumulative distribution. The distribution has a mean of zero and a standard deviation of one.
significant digits 3
B = 'z critical value' * SQRT [ * (1 - )/n]
Algebraically solve for SAMPLE SIZE n :
n = * (1 - ) * [ 'z critical value' / B ]^2 = 0.54 * (1 - 0.54) * [ 1.96 / 0.08 ]^2
SAMPLE SIZE = 150
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