Question

Life rating in Greece. Greece faced a severe economic crisis since the end of 2009. Suppose a Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 27% of them said they would rate their lives poorly enough to be considered "suffering." Round all answers to four decimal places. 1. What is the population parameter of interest? A. The 25% of Greeks in the sample who believe they are suffering. B. The actual proportion of Greeks who believe they are suffering. C. The score on the survey that corresponds to suffering. D. All the people who live in Greece. 2. What is the value of the point estimate of this parameter? 3. Construct a 95% confidence interval for the proportion of Greeks who are "suffering." ( , ) 4. If we decided to use a higher confidence level, the confidence interval would be: A. wider B. narrower C. stay the same 5. If we used the same confidence level with a larger sample, the confidence interval would be: A. wider B. narrower C. stay the same

Answer 1

Solution :

Given that,

1)  B. The actual proportion of Greeks who believe they are suffering.

n = 1000

2) Point estimate = sample proportion = = 0.27

1 - = 1 - 0.27 = 0.73

3) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.960}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.27 * 0.73) / 1000)

= 0.028

A 95% confidence interval for population proportion p is ,

± E  

= 0.27 ± 0.028

= ( 0.242, 0.298 )

4)  A. wider, because confidence level increases, margin of error increases

5) B. narrower because sample size increases, margin of error decreases