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?

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

[tex]r = 0.514 = 51.4 \%[/tex]

[tex]v = 0.426 = 42.6 \%[/tex]

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

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

Step-by-step explanation:

From the question we are told that

The sample size is [tex]n = 500[/tex]

The sample proportion is [tex]\r p = 0.47[/tex]

Generally the number of successes is mathematical represented as

[tex]k = n * \r p[/tex]

substituting values

[tex]k = 500 * 0.47[/tex]

[tex]k = 235[/tex]

Generally the number of failure is mathematical represented as

[tex]y = n * (1 -\r p )[/tex]

substituting values

[tex]y = 500 * (1 - 0.47 )[/tex]

[tex]y = 265[/tex]

for approximate normality for a confidence interval criteria to be satisfied

[tex]np > 5 \ and \ n(1- p ) \ >5[/tex]

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

[tex]\alpha = 100 - 95[/tex]

[tex]\alpha = 5 \%[/tex]

[tex]\alpha =0.05[/tex]

Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table, the value is

[tex]Z_{\frac{ \alpha }{2} } = 1.96[/tex]

Generally the margin of error is mathematically represented as

[tex]E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }[/tex]

substituting values

[tex]E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }[/tex]

[tex]E = 0.044[/tex]

=> [tex]E =4.4 \%[/tex]

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

[tex]\r p - E < p < \r p + E[/tex]

substituting values

[tex]0.47 - 0.044 < p < 0.47 + 0.044[/tex]

[tex]0.426 < p < 0.514[/tex]

The upper limit of the 95% confidence interval is [tex]r = 0.514 = 51.4 \%[/tex]

The lower limit of the 95% confidence interval is [tex]v = 0.426 = 42.6 \%[/tex]

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

[tex]\frac{1}{3 } < \frac{1}{2} (50\%)[/tex]