Question

GRE Test: The Graduate Record Examination (GRE) is a test required for admission to many US graduate schools. Student’s scores on the quantitative portion of the GRE follow a normal distribution with standard deviation of 8.8. Suppose a random sample of 10 students took the test, and their scores are given below: 152, 126, 146, 149, 152, 164, 139, 134, 145, 136 a) Find a point estimate of the population mean. b) Construct a 95% confidence interval for the true mean score for this population. c) How many students should be surveyed to estimate the mean score within 3 points with 95% confidence? d) How many students should be surveyed to estimate the mean score within 1 point with 95% confidence? e) How many students should be surveyed to estimate the mean score within 0.5 points with 95% confidence?

Answer 1

Answer:

a) $\bar X=144.3$

b) The 95% confidence interval would be given by (138.846;149.754)      

c) n=34

d) n=298

e) n=1190

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".

We can calculate the sample mean with this formula:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

$\bar X=144.3$ represent the sample mean  

$\mu$ population mean (variable of interest)

$\sigma=8.8$ represent the population standard deviation

n=10 represent the sample size  

a) Find a point estimate of the population mean

We can calculate the sample mean with this formula:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

$\bar X=144.3$

b) Construct a 95% confidence interval for the true mean score for this population.

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 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $t_{\alpha/2}=1.96$

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

$144.3-1.96\frac{8.8}{\sqrt{10}}=138.846$    

$144.3+1.96\frac{8.8}{\sqrt{10}}=149.754$

So on this case the 95% confidence interval would be given by (138.846;149.754)    

Part c

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

And on this case we have that ME =+20 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$, replacing into formula (5) we got:

$n=(\frac{1.960(8.8)}{3})^2 =33.05 \approx 34$

So the answer for this case would be n=34 rounded up to the nearest integer

Part d

$n=(\frac{1.960(8.8)}{1})^2 =297.493 \approx 298$

So the answer for this case would be n=298 rounded up to the nearest integer

Part e

$n=(\frac{1.960(8.8)}{0.5})^2 =1189.97 \approx 1190$

So the answer for this case would be n=1190 rounded up to the nearest integer