A group of professors investigated first-year college students' knowledge of astronomy. One concept of interest was the Big Bang theory of the creation of the universe. In a sample of 149149 freshmen students, 3232 believed that the Big Bang theory accurately described the creation of planetary systems. Based on this information, is it correct at the alphaαequals=0.100.10 level of significance to state that more than 20% of all freshmen college students believe the Big Bang theory describes the creation of planetary systems? State the null and alternative hypotheses. Choos

Step-by-step explanation:

The question is incomplete. The complete question is:

A group of professors investigated first-year college students' knowledge of astronomy. One concept of interest was the Big Bang theory of the creation of the universe. In a sample of 149 freshmen students, 32 believed that the Big Bang theory accurately described the creation of planetary systems. Based on this information, is it correct at the alpha = 0.01 level of significance to state that more than 20% of all freshmen college students believe the Big Bang theory describes the creation of planetary systems? State the null and alternative hypotheses. Choose the correct answer below. H_0: p = 0.20 H_a: p not equal to 0.20 H_0: p not equal to 0.20 H_a: p = 0.20 H_0: p = 0.20 H_a: p 0.20 If alpha = 0.05, find the rejection region for the test. Choose the correct answer below. z > 1.645 z > 1.96 z

Solution:

We would set up the null and alternative hypothesis. The correct options are

For null hypothesis,

p ≥ 0.2

For alternative hypothesis,

p < 0.2

This is a left tailed test.

Considering the population proportion, probability of success, p = 0.2

q = probability of failure = 1 - p

q = 1 - 0.2 = 0.8

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 32

n = number of samples = 149

P = 32/149 = 0.21

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.21 - 0.2)/√(0.2 × 0.8)/149 = 0.31

The calculated test statistic is 0.31 for the right tail and - 0.31 for the left tail

Since α = 0.05, the critical value is determined from the normal distribution table.

For the left, α/2 = 0.05/2 = 0.025

The z score for an area to the left of 0.025 is - 1.96

For the right, α/2 = 1 - 0.025 = 0.975

The z score for an area to the right of 0.975 is 1.96

In order to reject the null hypothesis, the test statistic must be smaller than - 1.96 or greater than 1.96

Therefore, the rejection region is z > 1.96

y=mx+b is the final answer!   : )

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