Question

Problem

In 2012, at the beginning of California's most recent drought, 14% of front yards in Los Angeles County were drought tolerant. Note: A "drought tolerant" front yard is a yard consisting of artificial turf, rocks, or water efficient plants. A hydrologist believes that, due to landscaping rebate programs, the population proportion of drought-tolerant front yards in Los Angeles County has increased since 2012.

The hydrologist observes a random sample of 800 front yards from across Los Angeles County and finds that 140 out of the 800 front yards are drought tolerant.

Is there sufficient evidence to conclude that the population proportion of drought-tolerant front yards in Los Angeles County has increased since 2012? Use an α = 0.01 level of significance.

Question 1

State the null and alternative hypotheses (using symbols).

Question 2

Explain what it would mean for the hydrologist's study to make a Type I Error.

Question 3 (Optional)

Verify the requirements for this hypothesis test. Briefly show that each requirement is satisfied.

Question 4

Test the hypothesis using either the classical (test statistic) approach or the P-value approach. Show all steps to your approach. Then state whether you reject H 0 or do not reject H 0.

Question 5

Is there sufficient evidence to conclude that the population proportion of drought-tolerant front yards in Los Angeles County has increased since 2012? State your conclusion in one or two sentences.

Answer 1

The following information is provided:

The sample size is N = 800,

the number of favorable cases is X = 140,

and the sample proportion is

and the significance level is α = 0.01

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p=0.14 , the population proportion of drought-tolerant front yards in Los Angeles County has 14%  since 2012

Ha: p>0.14 , the population proportion of drought-tolerant front yards in Los Angeles County has increased more than 14% since 2012

This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used.

type 1 error

Hypothesis testing is the art of testing if variation between two sample distributions can just be explained through random chance or not.

If we have to conclude that two distributions vary in a meaningful way, we must take enough precaution to see that the differences are not just through random chance.

Traditionally we try to set Type I error as .05 or .01 - as in there is only a 5 or 1 in 100 chance that the variation that we are seeing is due to chance.

This is called the 'level of significance'

for hydrologist level of significance at 1% would mean 1 in 100 chance that front randomly selected yard in los angels has drought tolerance just by chance.

Requirement for assumption of normal population -

n*p > 10 and n*(1-p)>10

now n = 800 and p = 0.175

hence n*p = 800 * 0.175 = 140 >10

and n * (1 - 0.175) = 800(0.825) = 660 > 10

both conditions met.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the critical value for a right-tailed test is

z_c = 2.33

The rejection region for this right-tailed test is

R={z:z>2.33}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that

Z = 2.853 > z c​ = 2.33,

it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p = 0.0022,

and since p=0.0022<.01,

it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected.

Therefore, there is enough evidence to claim that the population proportion p is greater than 14%​, at the α=0.01 significance level.

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