Mathematics, 13.07.2019 08:40 arionaking59p71cfc

In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor's degree in four years or less. find the sample size needed to estimate that percentage. use a 0.03 margin of error and use a confidence level of 95%. complete parts (a) through (c) below. a.) assume that nothing is known about the percentage to be estimated. n=b.)assume prior studies have shown that about 55%of full-time students earn bachelor's degrees in four years or less. n=c. does the added knowledge in part (b) have much of an effect on the sample size?

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The sample size needed to estimate the population proportion of a data within a margin of error, M, to a confidence level of 1 - α, with a previous population proportion estimate, p, is given by:

$n=p(1-p)\left( \frac{z_{\alpha/2}}{M} \right)^2$

Part A:

Given a margin of error of 0.03 a confidence level of 95%. Since we have know prior knowledge of the proportion estimate, we make use of the consevartive proportion estimate of 50% or 0.05.

$z_{\alpha/2}$ for 95% confidence is 1.96.

Therefore, the required sample size is given by

$n=0.5(1-0.5)\left( \frac{1.96}{0.03} \right)^2 \\ \\ =0.5(0.5)(65.33)^2=0.25(4,268) \\ \\ =1,067.11$

Therefore, the required sample size is 1,068.

Part B:

Given that prior studies have shown that about 55% of full-time students earn bachelor's degrees in four years or less.

Thus, p becomes 55% = 0.55 and the required sample size is given by

$n=0.55(1-0.55)\left( \frac{1.96}{0.03} \right)^2 \\ \\ =0.55(0.45)(65.33)^2=0.2475(4,268) \\ \\ =1,056.44$

Therefore, the required sample size is 1,057.

Part C:

When we dont know the additional infromation (i.e. in part A), we obtained that the required sample size is 1,068 but after we know the information our sample size becomes 1,057.

Though there is a difference between both sample sizes but the difference is not much and hence, the knowledge of the additional information does not have much of an effect on the sample size.

Answer from: Janznznz1121

The required sample size is 1,068

A) Significance Level, $\alpha = 0.05$ , Margin of Error, E=0.03

The provided estimate of proportion p is, p=0.5

The critical value for significance level, $\alpha = 0.05$ is 1.96 .

The following formula is used to compute the minimum sample size required to estimate the population proportion p within the required margin of error:

$n=p(1-p)\dfrac{Z_c}{E^2}^{2}$

Put Values , we get

$n=0.5(1-0.5)\dfrac{1.96}{0.03^2}^{2}\\\\n=0.5(0.5)(65.33)^2\\\\n=0.25(4,268)$

n=1067.11

Therefore, the sample size needed to satisfy the condition n=1067.11 and it must be an integer number, we conclude that the minimum required sample size is n=1068

Sample size, 1068

B) Significance Level, $\alpha = 0.05$ , Margin of Error, E=0.03

The provided estimate of proportion p is, p=0.55

The critical value for significance level, $\alpha = 0.05$ is 1.96 .

The following formula is used to compute the minimum sample size required to estimate the population proportion p within the required margin of error:

$n=p(1-p)\dfrac{Z_c}{E^2}^{2}$

Put Values , we get

$n=0.55(1-0.55)\dfrac{1.96}{0.03^2}^{2}\\\\n=0.55(0.45)(65.33)^2\\\\n=0.2475(4,268)$

n=1056.44

Therefore, the sample size needed to satisfy the condition n=1056.44 and it must be an integer number, we conclude that the minimum required sample size is n=1057

Sample size, 1057

C)The difference between both the part a and b is very less so added knowledge of the extra information does not have much effect on sample size.

Learn more about z score here;

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Answer from: Quest

$answer \:$ paul's with a y-intercept of $ 2856

Answer from: Quest

p = 6s

step-by-step explanation:

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