Question

A) Assume that the significance level is α=0.05. Use the given information to find the​ P-value and the critical​ value(s). The test statistic of z = −1.09 is obtained when testing the claim that p<0.2.

B) Identify the type I error and the type II error that corresponds to the given hypothesis. The proportion of settled medical malpractice suits is 0.29.

C) A survey of 1,562 randomly selected adults showed that 579 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 33​% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.01 significance level to complete parts​ (a) through​ (e).

Sample​ proportion: 0.370679

Test​ statistic, ​z: 3.4191

Critical​ z: ± 2.5758

​P-Value: 0.0006

Answer 1

A)

Given that,

This is the left tailed test ,

z = -1.09 Reject the null hypothesis that the

= 0.05

P(z < -1.09) = 0.1379

P-value = 0.1379

Z_0.05 = -1.645

Critical value = -1.645

B)

Type I error occurs when rejecting the null when it is True .

Reject the null hypothesis that the proportion of settled medical malpractice suits is 0.29 when the

proportion is actually equal to 0.29

Type II error occurs when We fail to reject H0 when it is False

Failed to eject the null hypothesis that the proportion of settled medical malpractice suits is less than 0.29 when the

proportion is actually equal to 0.29 .

C)

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.33

H_a : p 0.33

n = 1562

x = 579

= x / n = 579 / 1562 = 0.3707

P₀ = 0.33

1 - P₀ = 1 - 0.33 = 0.67

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.3707 - 0.33 / [(0.33 * 0.67) / 1562]

= 3.419

Test statistic = 3.419

P(z > 3.419) = 1 - P(z < 3.419) = 1 - 0.9997 = 0.0003

P-value = 2 * 0.0003 = 0.0006

= 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Critical value = -2.576 and +2.576