Answer:
1.646%
Step-by-step explanation:
Sample size (n) = 2000
Sample Proportion p = 17% = 0.17
α = 0.95
Margin of Error :
Z(1-α)/2 * √((p*(1 - p)) /n)
Z(1 - 0.95/2) * √((0.17 * (1 - 0.17) / 2000)
Zscore = 1.96
1.96 * √(0.1411/2000)
1.96 * √0.00007055
1.96 * 0.0083994
Hence,
1.96 * 0.0083994
= 0.0164628
= 0.0164628 * 100%
= 1.646%
The margin of error for the random sample 2,000 driver's license applications is 1.646%.
The probability or the chances of error while choosing or calculating a sample in a survey is called the margin of error.
It can be found out using the following formula.
[tex]MOE=Z_{\frac{\alpha}{2}}\times\left(\sqrt{p\times\dfrac{1-p}{n}}\right)[/tex]
Here, α is the level of significance, n is the sample size and p is the sample proportion.
A random sample of 2,000 driver's license applications found that 17% reported their eye color as blue.
Let level of significance is 0.05. At this level of significance the critical value of Z is 1.96. Put the values in the above formula as,
[tex]MOE=(1.96)\times\left(\sqrt{0.17\times\dfrac{1-0.17}{2000}}\right)\\MOE=0.01646\\MOE=1.646\%[/tex]
Thus, the margin of error for the random sample 2,000 driver's license applications is 1.646%.
Learn more about the margin of error here;
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