Mathematics, 05.08.2020 03:01 honeytolentino08

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?

Answers: 2

Show answers

Answers

Answer from: iamabeast51

number of successes

k = 235

number of failure

y = 265

The criteria are met

The sample proportion is $\r p = 0.47$

$E =4.4 \%$

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%

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

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

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

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

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