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7nadin3 [17]
3 years ago
13

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

ess." 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?
Mathematics
1 answer:
Brilliant_brown [7]3 years ago
3 0

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\%)

 

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