Question

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of stress." Calculate and interpret a 95% confidence interval for the proportion of U.S. adults who considered themselves happy at that time. 1 How many successes and failures are there in the sample? Are the criteria for approximate normality satisfied for a confidence interval?
A What is the sample proportion?
B compute the margin of error for a 95% confidence interval.
C Interpret the margin of error you calculated in Question 1
C. Give the lower and upper limits of the 95% confidence interval for the population proportion (p), of U.S. adults who considered themselves happy in April, 2013.
D Give an interpretation of this interval.
E. Based on this interval, is it reasonably likely that a majority of U.S. adults were happy at that time?
H If someone claimed that only about 1/3 of U.S. adults were happy, would our result support this?

Answer 1

Answer:

number of successes

                 $k = 235$

number of failure

                 $y = 265$

The   criteria are met    

A

    The sample proportion is  $\r p = 0.47$

B

    $E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from  the true population proportion will not  more than 4.4%

Ci  

   $r = 0.514 = 51.4 \%$

 $v = 0.426 = 42.6 \%$

D

   This 95% confidence interval  mean that the the chance of the true    population proportion of those that are happy to be exist within the upper   and the lower limit  is  95%

E

  Given that 50% of the population proportion  lie with the 95% confidence interval  the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

 Yes our result would support the claim because

            $\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

     The sample size is  $n = 500$

     The sample proportion is  $\r p = 0.47$

 

Generally the number of successes is mathematical represented as

             $k = n * \r p$

substituting values

             $k = 500 * 0.47$

            $k = 235$

Generally the number of failure  is mathematical represented as

           $y = n * (1 -\r p )$

substituting values

           $y = 500 * (1 - 0.47 )$

           $y = 265$

for approximate normality for a confidence interval  criteria to be satisfied

          $np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

  Given that the confidence level is  95%  then the level of confidence 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

                 $E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

                 $E = 0.044$

=>               $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from  the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

          $\r p - E < p < \r p + E$

substituting values

        $0.47 - 0.044 < p < 0.47 + 0.044$

         $0.426 < p < 0.514$

The upper limit of the 95% confidence interval is  $r = 0.514 = 51.4 \%$

The lower limit of the   95% confidence interval is  $v = 0.426 = 42.6 \%$

This 95% confidence interval  mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit  is  95%

Given that 50% of the population proportion  lie with the 95% confidence interval  the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

            $\frac{1}{3 } < \frac{1}{2} (50\%)$