What percentage of the population live in their state of birth? According to the 2010 U. S. census, 14% of people born in Nevada, 26% of people born in Florida, 35% of people born in Virginia, 42% of people born in California, and 50% of people born in Arkansas live in that state. Suppose you wanted to see if these percentages still hold true today, so you randomly selected 40 people born in Virginia and 17 said they live in that state. Based on your findings, can you conclude the proportion of all people born in Virginia that live in that state is now lower at α=.1? For the hypothesis stated above, what is the decision rule?

The p-value obtained is greater than the significance level at which the test was performed, hence, we fail to reject the null hypothesis & say that there is enough evidence to conclude that the proportion of all people born in Virginia that live in that state is now more or equal to 0.35.

That is, we reject the alternative hypothesis and say that the proportion of all people born in Virginia that live in that state is NOT now lower than 0.35.

Step-by-step explanation:

- The census claim is that 35% of people that were born in Virginia still live there.

- Randomly selected 40 people born in Virginia and 17 said they live in that state

- Can you conclude the proportion of all people born in Virginia that live in that state is now lower at α = 0.1

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 proportion of all people born in Virginia that live in that state is now more or equal to 0.35

The alternative hypothesis will now be that the proportion of all people born in Virginia that live in that state is now lower than 0.35

Mathematically,

The null hypothesis is represented as

H₀: p ≥ 0.35

The alternative hypothesis is represented as

Hₐ: p < 0.35

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ)/σₓ

x = sample proportion of the 40 people born in Virginia that still live there = (17/40) = 0.425

μ = p₀ = The census proportion for people born in Virginia that still live there = 0.35

σₓ = standard error = √[p(1-p)/n]

where n = Sample size = 40

σₓ = √[0.425×0.575/40] = 0.078162491 = 0.07816

t = (0.425 - 0.35) ÷ 0.07816

t = 0.9595701126 = 0.96

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

Degree of freedom = df = n - 1 = 40 - 1 = 39

Significance level = 0.10

The hypothesis test uses a one-tailed condition because we're testing only in one direction.

p-value (for t = 0.96, at 0.10 significance level, df = 39, with a one tailed condition) = 0.171485

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 = 0.10

p-value = 0.171485

0.171485 > 0.10

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & say that there is enough evidence to conclude that the proportion of all people born in Virginia that live in that state is now more or equal to 0.35.

That is, we reject the alternative hypothesis and say that the proportion of all people born in Virginia that live in that state is NOT now lower than 0.35.

Hope this Helps!!!

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