SOLUTION: A survey showed that 73% of the random sample of 1506 people interviewed
favored drug tests for professional athletes. 68% said that professional athletics
using drug for the fir
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-> SOLUTION: A survey showed that 73% of the random sample of 1506 people interviewed
favored drug tests for professional athletes. 68% said that professional athletics
using drug for the fir
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Question 1201047: A survey showed that 73% of the random sample of 1506 people interviewed
favored drug tests for professional athletes. 68% said that professional athletics
using drug for the first time should be banned or suspended from the professional
sports. Find 95% confidence interval for the proportion of the public who favored
drug tests for professional athletes. Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
We focus on the portion that says
"A survey showed that 73% of the random sample of 1506 people interviewed favored drug tests for professional athletes."
and ignore the part that says "68% said that professional athletics
using drug for the first time should be banned or suspended from the professional
sports".
At 95% confidence, the z critical value is roughly z = 1.96
Use a table like this https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value. Look at the bottom row labeled "Z" and above the 95% confidence level.
A stats calculator can also compute this value.
Let's compute the margin of error for the proportion.
E = margin of error
E = z*sqrt(phat*(1-phat)/n)
E = 1.96*sqrt(0.73*(1-0.73)/1506)
E = 0.02242264791327
E = 0.022
This value is approximate.
Now we can compute the boundaries.
L = lower boundary of the confidence interval
L = phat - E
L = 0.73 - 0.022
L = 0.708
and
U = upper boundary of the confidence interval
U = phat + E
U = 0.73 + 0.022
U = 0.752
These values are approximate.
The 95% confidence interval in the format (L, U) is approximately (0.708, 0.752)
The 95% confidence interval in the format L < p < U is approximately 0.708 < p < 0.752
This second format is a bit more descriptive in terms of which population parameter we're trying to measure.
We are 95% confident the population proportion p is somewhere between 0.708 and 0.752
Meaning we are 95% confident the true percentage of people in favor of drug tests for professional athletes is somewhere between 70.8% and 75.2%
Side note:
An alternative confidence interval format is which in this case is roughly
(