Question

Life rating in Greece. Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered "suffering".
a. Describe the population parameter of interest. What is the value of the point estimate of this parameter?

b. Check if the conditions required for constructing a confidence interval based on these data are met.

c. Construct a 95% confidence interval for the proportion of Greeks who are "suffering".

d. Without doing any calculations, describe what would happen to the confidence interval if we decided to use a higher confidence level.

e. Without doing any calculations, describe what would happen to the confidence interval if we used a larger sample.

Answer 1

Answer:

a

        The population parameter of interest is the true proportion of Greek who are suffering

    While the point estimate of this parameter is  proportion of those that would rate their lives poorly enough to be considered "suffering". which is 25%  

b

   The condition  is met

c

   The  95% confidence interval is   $0.223 < p < 0.277$

d

      If the confidence level is increased which will in turn reduce the level of significance but increase the critical value($Z_{\frac{\alpha }{2} }$) and this will increase the margin of error( deduced from  the formula for margin of error i.e  $E \ \alpha \ Z_{\frac{\alpha }{2} }$ ) which will make the confidence interval wider

e

  Looking at the formula for margin of error if the we see that if the  sample size is increased the margin of error will reduce making the  confidence level narrower

Step-by-step explanation:

From the question we are told that

    The sample size is  n  =  1000

     The  population proportion is  $\r p = 0.25$

     

Considering question a

   The population parameter of interest is the true proportion of Greek who are suffering

    While the point estimate of this parameter is  proportion of those that would rate their lives poorly enough to be considered "suffering". which is 25%  

Considering question b

The condition for constructing a confidence interval is

        $n * \r p > 5\ and \ n(1 - \r p ) >5$

So  

        $1000 * 0.25 > 5 \ and \ 1000 * (1-0.25 ) > 5$

         $250 > 5 \ and \ 750> 5$

Hence the condition  is met

Considering question c

    Given that the confidence level is  95%  then  the level of significance is mathematically evaluated as

          $\alpha = 100 - 95$    

          $\alpha = 5 \%$

          $\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is  

              $Z_{\frac{\alpha }{2} } = 1.96$        

Generally the margin of error is mathematically represented as

         $E = Z_\frac{ \alpha }{2} * \sqrt{ \frac{\r p (1 - \r p ) }{n} }$

substituting values

         $E = 1.96 * \sqrt{ \frac{ 0.25 (1 - 0.25 ) }{ 1000} }$

         $E = 0.027$

The  95% confidence interval is mathematically represented as

            $\r p - E < p < \r p + E$

substituting values  

           $0.25 - 0.027 < p < 0.25 + 0.027$

substituting values

           $0.223 < p < 0.277$

considering d

  If the confidence level is increased which will in turn reduce the level of significance but increase the critical value($Z_{\frac{\alpha }{2} }$) and this will increase the margin of error( deduced from  the formula for margin of error i.e  $E \ \alpha \ Z_{\frac{\alpha }{2} }$ ) which will make the confidence interval wider

considering e

     Looking at the formula for margin of error if the we see that if the  sample size is increased the margin of error will reduce making the  confidence level narrower