Question

A survey of 1,673 randomly selected adults showed that 545 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 36 ​% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.05 significance level to complete parts​ (a) through​ (e).

Requried:
a. Is the test two-tailed, left-tailed, or right-tailed?
b. What is the test statistic?
c. What is the P-value?
d. What is the null hypothesis and what do you conchide about it?
e. Identify the null hypothesis.

Answer 1

Explanation:

We would set up the hypothesis test.

For the null hypothesis,

p = 0.36

For the alternative hypothesis,

P < 0.36

a) This is a left tailed test due to the inequality sign in the alternative hypothesis.

b)Considering the population proportion, probability of success, p = 0.36

q = probability of failure = 1 - p

q = 1 - 0.36 = 0.64

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 545

n = number of samples = 1673

P = 545/1673 = 0.33

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.33 - 0.36)/√(0.36 × 0.64)/1673 = - 2.56

c)We would determine the P-value from the normal distribution table. From the normal distribution table, the area below the test z score in the left tail 0.00523

P value = 0.00523

d) Since alpha, 0.05 > than the p value, 0.00523, then we would reject the null hypothesis.

e) The null hypothesis is the claim that 36​% of adults have heard of the new electronic reader.