Question

A random sample is selected from a normally distributed population. The following sample statistics are obtained: n = 20, = 30, and s = 10. Based on this information, and using a 95% confidence level, which of the following statements is true? a. The margin of error is approximately 4.68. b. The margin of error is approximately 3.24. c. The standard deviation of the sampling distribution of is .50. d. The critical value is 1.96. e. The critical value is 1.7921.

Answer 1

Answer:

b) The margin of error is approximately 3.24

e) The critical value is 1.7921.

Step-by-step explanation:

Step(i):-

Given sample size 'n' =20

Given sample standard deviation 's' = 10

Margin of error

The margin of error is determined by

$M.E = \frac{t_{\alpha } S.D }{\sqrt{n} }$

The level of significance ∝ =0.95

The degrees of freedom = n-1 = 20-1=19

t₀.₉₅ = 1.729

$M.E = \frac{1.729 X10 }{\sqrt{20} }$

Margin of error = 3.866

Step(ii):-

Margin of error

The margin of error is determined by

$M.E = \frac{t_{\alpha } S.D }{\sqrt{n} }$

Given another sample size n =30

The level of significance ∝ =0.95

The degrees of freedom = n-1 = 30-1=29

t₀.₉₅ = 1.70

$M.E = \frac{1.7021 X10 }{\sqrt{30} } =3.24$

Margin of error = 3.24