Question

The standard deviation of SCHOLTEST scores for students at a particular college is 285 points. Two highly skilled data science students, Wai and Yemane, want to estimate the average SCHOLTEST score of students at this college. They determine their margin of error should be no larger than 20 points.

(i) Yemane wants to use a 90 percent confidence interval. What minimum count value should her sample size equal?

(ii) Wai wants to use a 99 percent confidence interval. Should the size of his sample be greater or less than the size of Yemane’s sample? Why?

(iii) Calculate, for Wai’s 99 percent CI, the sample size he would need.

Among the answer choices that follow, CHOOSE THE BEST COMBINATIONS OF THE OPTIONS GIVEN IN (i) AND (ii) AND (iii).

Group of answer choices

a.

(i)

555

(ii)

Less than, because the tail probabilities are lower in Wai’s case, given the 99 percent CI he seeks.

(iii)

550

b.

(i)

555

(ii)

Greater than, because the higher confidence level entails a larger Z-statistic, which would cause the MOE to exceed 20 if the sample size were not increased.

(iii)

674

c.

(i)

550

(ii)

Greater than, because the higher confidence level entails a smaller p-value, which would cause the MOE to increase by 20 if the sample size were not increased.

(iii)

1348

d.

(i)

550

(ii)

Greater than, because the higher confidence level entails a larger Z-statistic, which would cause the MOE to exceed 20 if the sample size were not increased.

(iii)

1348

e.

None of the above calculations/reasoning is correct.

Answer 1

Solution:

Given:

Standard Deviation =

o= 285

E = Margin of Error = 20

(i) Yemane wants to use a 90 percent confidence interval. What minimum count value should her sample size equal?

c = 90% confidence level.

Z_c is z critical value for c = 90% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 = 0.9500

Use Excel command to get z value:

=NORM.S.INV( probability)

=NORM.S.INV(0.9500)

=1.645

Now use sample size formula:

$n= \left ( \frac{Z_{c}\times \sigma }{E} \right )^{2}$

(1.645 x 285 20

n = (23.44125)

n = 549.492202

n = 550

( Sample size is always rounded up)

(ii) Wai wants to use a 99 percent confidence interval. Should the size of his sample be greater or less than the size of Yemane’s sample? Why?

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Use Excel command to get z value:

=NORM.S.INV( probability)

=NORM.S.INV(0.9950)

=2.576

Thus Z_c= 2.576

Greater than, because the higher confidence level entails a larger Z-statistic, which would cause the MOE to exceed 20 if the sample size were not increased.

(iii) Calculate, for Wai’s 99 percent CI, the sample size he would need.

$n= \left ( \frac{Z_{c}\times \sigma }{E} \right )^{2}$

2.576 x 285 n=1 20

n = (36.708)

n = 1347.47726

n = 1348

Thus correct option is d)

d.

(i) 550

(ii) Greater than, because the higher confidence level entails a larger Z-statistic, which would cause the MOE to exceed 20 if the sample size were not increased.

(iii) 1348