Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute indicate that the percentage is now 46% (the Wall Street Journal, October 5, 2012)a. Develop appropriate hypotheses such that rejection of H0 will support the conclusion that a smaller proportion of American families own stocks or stock funds in 2012 than 10 years ago.
b. Assume the Investment Company Institute sampled 300 American families to estimate that the percent owning stocks or stock funds was 46% in 2012. What is the p-value for your hypothesis test?
c. At α = .01, what is your conclusion?

Answers

Answer 1

Answer:

Using the z-distribution, as we are working with a proportion, it is found that:

a) $H_0: p = 0.53$ , $H_1: p < 0.53$

b) The p-value is of 0.0075.

c) Since the p-value of the test is of 0.0075 < 0.01 for the left-tailed test, it is found that there is enough evidence to reject the null hypothesis and conclude that a smaller proportion of American families own stocks or stock funds in 2012 than 10 years ago.

What are the hypothesis tested?

At the null hypothesis, it is tested if the proportion is still of 53%, that is:

$H_0: p = 0.53$

At the alternative hypothesis, it is tested if the proportion is now smaller, that is:

$H_1: p < 0.53$

Item a:

The hypothesis are:

$H_0: p = 0.53$

$H_1: p < 0.53$

Item b:

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{(p(1-p))/(n)}}$

In which:

$\overline{p}$ is the sample proportion.

p is the proportion tested at the null hypothesis.

n is the sample size.

In this problem, the parameters are:

$\overline{p} = 0.46, p = 0.53, n = 300$ .

Hence, the value of the test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{(p(1-p))/(n)}}$

$z = \frac{0.46 - 0.53}{\sqrt{(0.53(0.47))/(300)}}$

$z = -2.43$

Using a z-distribution calculator, considering a left-tailed test, as we are testing if the proportion is less than a value, with z = -2.43, it is found that the p-value is of 0.0075.

Item c:

Since the p-value of the test is of 0.0075 < 0.01 for the left-tailed test, it is found that there is enough evidence to reject the null hypothesis and conclude that a smaller proportion of American families own stocks or stock funds in 2012 than 10 years ago.

More can be learned about the z-distribution at brainly.com/question/26454209

Answer 2

Answer:

Answer:

a) Null hypothesis: $p\geq 0.53$

Alternative hypothesis: $p < 0.53$

b) $z=\frac{0.46 -0.53}{\sqrt{(0.53(1-0.53))/(300)}}=-2.429$

$p_v =P(Z<-2.429)=0.0076$

c) So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

$\hat p=0.46$ estimated proportion of American families owning stocks or stock funds

$p_o=0.53$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

Part a

We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:

Null hypothesis: $p\geq 0.53$

Alternative hypothesis: $p < 0.53$

Part b

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.46 -0.53}{\sqrt{(0.53(1-0.53))/(300)}}=-2.429$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(Z<-2.429)=0.0076$

Part c

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .

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Answers

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Answer:

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Answers

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Answers

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Answers

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