Question

An SRS of 350 high school seniors gained an average of x¯=21 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ=52 .(a) Find a 99% confidence interval for the mean change in score μ in the population of all high school seniors. (Enter your answers rounded to two decimal places.)lower bound of confidence interval: ???upper bound of confidence interval: ???(b) What is the margin of error for 99% ? (Enter your answer rounded to two decimal places.)margin of error: ????(c) Suppose we had an SRS of just 100 high school seniors. What would be the margin of error for 99% confidence? (Enter your answer rounded to three decimal places.)margin of error: ????(d) How does decreasing the sample size change the margin of error of a confidence interval when the confidence level and population standard deviation remain the same?A. Decreasing the sample size keeps the margin of error the same, provided the confidence level and population standard deviation remain the same.B. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation alsodecrease.C. Decreasing the sample size decreases the margin of error, provided the confidence level and population standard deviation remain the same.D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

Answer 1

Answer:

a) $21-2.58\frac{52}{\sqrt{350}}=13.83$  

$21+2.58\frac{52}{\sqrt{350}}=28.17$  

So on this case the 99% confidence interval would be given by (13.83;28.17)  

b) $ME=2.58\frac{52}{\sqrt{350}}=7.17$

c) $ME=2.58\frac{52}{\sqrt{100}}=13.42$

d) D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

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".  

Part a

$\bar X=21$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$\sigma=52$ represent the population standard deviation  

n=350 represent the sample size  

99% confidence interval  

The confidence interval for the mean is given by the following formula:  

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)  

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$  

Now we have everything in order to replace into formula (1):  

$21-2.58\frac{52}{\sqrt{350}}=13.83$  

$21+2.58\frac{52}{\sqrt{350}}=28.17$  

So on this case the 99% confidence interval would be given by (13.83;28.17)  

Part b

The margin of error is given by:

$ME=2.58\frac{52}{\sqrt{350}}=7.17$

Part c

The margin of error is given by:

$ME=2.58\frac{52}{\sqrt{100}}=13.42$

Part d

As we can see when we reduce the sample size we increase the margin of error so the best option for this case is:

D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.