Question

A random sample of 75 students revealed 0.25 had acquired an internship for summer 2020 before February 2020. If an associated confidence interval started at 0.152 and ended at 0.348, what was the level of confidence?

Group of answer choices

a) 90

b)80

c)99

d)95

For a t distribution with n = 19, find the probability for the following region:

To the left of +2.5524

a)0.51

b)0.49

c)0.99

d)0.91

Answer 1

Given that, sample size (n) = 75 and sample proportion = 0.25

Confidence interval for population proportion is, (0.152, 0.348).

Margin of error (E) = (0.348 - 0.152)/2 = 0.196/2 = 0.098

we want to find, the confidence level (c),

Using z-table first we find, P(Z > 1.96)

P(Z > 1.96) = 1 - P(Z < 1.96) = 1 - 0.975 = 0.025

Here, 0.025 = alpha/2

alpha = 2 * 0.025 = 0.05

Therefore, confidence level (c) = 1 - alpha = 1 - 0.05 = 0.95

Answer : d) 0.95

Q.2) Given that, n = 19 and we want to find, the area to the left of +2.5524

Degrees of freedom = 19 - 1 = 18

Using Excel first we get, area to the right of +2.5524

Excel Command : =TDIST (2.5524, 18, 1) = 0.01

Therefore, the area to the left of +2.5524 is, 1 - 0.01 = 0.99

Answer : c) 0.99