Complete Question
A researcher is funded to obtain an estimate for the population proportion of smokers who have tried using e-cigarettes. She plans to interview 100 smokers. Previous studies have estimated that 20% of smokers have tried e-cigarettes. She finds that 23% of smokers have tried e-cigarettes.
Which of the following is correct?
A
0.23 is the population proportion
B
The margin of error for a 95% confidence interval is 8%
C
The standard error of the sample proportion is 0.177
D
This is a biased estimate because it is based on smokers none of the above
Answer:
The correct option is B
Step-by-step explanation:
From the question we are told that
The sample size is n = 100
The population proportion is p = 0.20
The sample proportion is [tex]\r p = 0.23[/tex]
Generally the standard error is mathematically represented as
[tex]SEp = \sqrt{\frac{\r p(1 - \r p)}{n} }[/tex]
=> [tex]SEp = \sqrt{\frac{0.23 (1 - 0.23)}{100} }[/tex]
=> [tex]SEp = 0.04208[/tex]
Generally for a 95% confidence level is level of significance is
[tex]\alpha = (100 - 95)\%[/tex]
[tex]\alpha = 5\%[/tex]
[tex]\alpha = 0.05[/tex]
Now the critical value of [tex]\frac{\alpha }{2}[/tex] obtained from the normal distribution table is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the percentage margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * SE_p * 100[/tex]
[tex]E = 1.96* 0.04208 * 100[/tex]
[tex]E = 8\%[/tex]
