In a survey of 1,000 American adults conducted in April 2012, 43% reported having gone through an entire week without paying for anything in cash. Test to see if this sample provides evidence that the proportion of all American adults going a week without paying cash is greater than 40%. Use the fact that a randomization distribution is approximately normally distributed with a standard error of 0.016. Show all details of the test and use a 5% significance level.
1. State the null and alternative hypotheses.
2. What is the test statistic? Round your answer to two decimal places.
3. What is the p-value? Round your answer to two decimal places.
4. What is the conclusion?
A. Do not reject H_0 and find evidence that the proportion is greater than 40%.
B. Do not reject H_0 and do not find evidence that the proportion is greater than 40%
C. Reject H_0 and find evidence that the proportion is not greater than 40%.
D. Reject H_0 and find evidence that the proportion is greater than 40%.

1) The null hypothesis is represented as

H₀: p ≤ 0.40

The alternative hypothesis is represented as

Hₐ: p > 0.40

2) Test statistic = 1.91

3) p-value = 0.028067

4) Conclusion is Option D.

Reject H_0 and find evidence that the proportion is greater than 40%.

This is because the p-value < significance level for this question.

Step-by-step explanation:

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, the null hypothesis is that the sample provides no significant evidence that the proportion of all American adults going a week without paying cash is greater than 40%. That is, the proportion of all American adults going a week without paying cash is lesser than or equal to 40%.

The alternative hypothesis will now be that the sample provides evidence that the proportion of all American adults going a week without paying cash is greater than 40%.

Mathematically,

The null hypothesis is represented as

H₀: p ≤ 0.40

The alternative hypothesis is represented as

Hₐ: p > 0.40

To do this test, we will use the z-distribution because, the degree of freedom is large enough, it checks out.

So, we compute the z-test statistic

z = (x - μ)/σₓ

x = sample proportion = 0.43

μ = p₀ = standard to be tested against = 0.40

σₓ = standard error of the poll proportion = √[p(1-p)/n]

where n = Sample size = 1000

σₓ = √[0.43×0.57/1000] = 0.0156556699 = 0.0157

z = (0.43 - 0.40) ÷ 0.0157

z = 1.911 = 1.91

checking the tables for the p-value of this z-statistic

p-value (for z = 1.91, at 0.05 significance level, with a one tailed condition) = 0.028067

Note that it is one tailed because we're checking if the proportion is greater than a value; moving in only one direction.

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 5% = 0.05

p-value = 0.028067

0.028067 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis and accept the alternative hypothesis that the sample provides evidence that the proportion of all American adults going a week without paying cash is greater than 40%.

Hope this Helps!!

actually your concept is wrong

step-by-step explanation:

150% of x =300

150x=300×100

x=(300×100)÷150

x=200