Question

A sample of 52 Elementary Statistics students includes 13 women. Assuming the sample is random:

(a) Estimate the percentage of women taking Elementary Statistics by computing an appropriate confidence interval with 98% confidence.

(b) At 10% significance, test whether less than 40% of the enrollment in all Elementary Statistics classes consists of women.

Answer 1

Solution :

Given that,

n = 52

x = 13

= x / n = 13 / 52 = 0.25

1 -  = 1 - 0.25 = 0.75

(a)

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.326 * (((0.0.250 * 0.750) / 52)

= 0.140

A 95% confidence interval for population proportion p is ,

- E < P <  + E

0.250 - 0.140 < p < 0.250 + 0.140

0.110 < p < 0.390

(11.0% , 39.0%)

(b)

This is the left tailed test .

The null and alternative hypothesis is

H₀ : p = 0.40

H_a : p <  0.40

P₀ = 0.40

1 - P₀ = 1 - 0.40

z =  - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.25 - 0.40 / [(0.250 * 0.750) / 52]

= -2.208

This is the left tailed test .

P(z < -2.208) = 0.0136

P-value = 0.0136

= 10% = 0.10

P-value <

Reject the null hypothesis .