Question

You may need to use the appropriate appendix table or technology to answer this question.

A random sample of 49 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 210. There are 48 degrees of freedom. Find the margin of error at 95% confidence.

A. 2.011

B. 8.617

C. 50.317

D. 60.319

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 1400

sample standard deviation = s = 210

sample size = n = 49

Degrees of freedom = df = 48

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,48 = 2.01063

Margin of error = E = t/2,df * (s /n)

  E = 2.01063 * (210 / 49)

E = 60.3189

E = 60.319

Margin of error is 60.319

Correct option :- D. 60.319