Your professor wishes to estimate the proportion of ALL high school students enrolled in college-level courses each school year. A sample of 1500 students revealed that 18.3% were enrolled in college-level courses. Find the margin of error for a 90% confidence interval for a proportion.
Solution :
Given that,
n = 1500
Point estimate = sample proportion = =18.3%=0.183
1 - = 1-0.183=0.817
At 90% confidence level
= 1 - 90%
= 1 - 0.90 =0.10
/2
= 0.05
Z/2
= Z0.05 = 1.645 ( Using z table )
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.645 (((0.183*0.817) / 1500)
= 0.016
Your professor wishes to estimate the proportion of ALL high school students enrolled in college-level courses...
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