Question

It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the same among the 25 to 30 age group. Question 1. How many of this younger age group must we survey in order to estimate the proportion of non-grads to within .10 with 90% confidence? Use the value of p from the over-50 age group. (Round up to the nearest integer.) n = Question 2. Suppose we still want 90% confidence but we want to cut the margin of error to .04. What is the necessary sample size? (Round up to the nearest integer.) n = Question 3. What sample size is needed to estimate the proportion of non-grads to within .04 with 95% confidence? (Round up to the nearest integer.) n =

Answer 1

Answer:

1)  n=48  

2) n=298

3) n=426

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$  

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.10$ and $\alpha/2 =0.05$. And the critical value would be given by:  

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$  

The margin of error for the proportion interval is given by this formula:  

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)  

And on this case we have that $ME =\pm 0.1$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:  

$n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63$  

And rounded up we have that n=48  

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7$  

And rounded up we have that n=298  

Part 3

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:  

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$  

The margin of error for the proportion interval is given by this formula:  

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)  

And on this case we have that $ME =\pm 0.04$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:  

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.96})^2}=425.22$  

And rounded up we have that n=426  

Answer 2

The probability shows that the number of younger age group that must will be surveyed will be 48.

How to compute the probability?

From the information given, the confidence level is 90%, margin or error is 0.1. The critical value from the z table is given as 1.645.

The number of samples will be:

= 0.23 × (1 - 0.23) × (1.645/0.1)²

= 0.23 × 0.77 × (1.645/0.1)²

= 48

The sample size when we are 90% confidence but we want to cut the margin of error to .04 will be computed thus:

= (1.645/.004)² × 0.23 × (1 - 0.23)

= (41.125)² × 0.1771

= 300

The sample size that is needed to estimate the proportion of non-grads to within .04 with 95% confidence will be:

= (1.96/0.04)² × 0.23 × (1 - 0.23)

= 48.99 × 0.1771

= 426

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