inferences - hypothesis testing (earthquake insurance)

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fastfred

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May 25, 2006
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I don't have any idea how to do this, and I've been stuck on this question for ages.

Q1. A newspaper article revealed that only 133 of 337 randomly selected residences in Los Angeles County were protected by earthquake insurance.

a) What are the appropriate null and alternate hypotheses to test the research hypothesis that fewer than 40% of residents of Los Angeles County were protected by earthquake insurance?

This is as far as I got:
. . .Ha: mean < 0.4
. . .H0: mean = 0.4

b) Does the data provide sufficient evidence to support the research hypothesis? Use the 10% level of significance. Make sure you check if the test is appropriate in this situation.

c) Calculate and interpret the P-value for the test.
 
The claim is <40% have earthquake insurance. A random sample

revealed \(\displaystyle \frac{133}{337}\approx{.39466}\) or about 39.5%

had insurance.

\(\displaystyle H_{0}:p\geq{0.4}\;\ and\;\ H_{a}:<0.4\;\ (Claim)\)

Level of significance: \(\displaystyle {\alpha}=0.10\)

Left-tailed where \(\displaystyle z_{0}=-1.28\), rejection region is z<-1.28.

\(\displaystyle \L\\\frac{0.39466-0.40}{\sqrt{\frac{(0.39466)(0.60534)}{337}}}\approx{-0.2006}\)

z is not in the rejection region. Do not reject the null hypothesis.

There is not enough evidence to support the claim that less than 40%

have quake insurance.

Also, another way to tell:

p=0.42

Since \(\displaystyle p>{\alpha}\), you should fail to reject the null hypothesis.

How do you find the p-value, you ask?.

The p-vlaue in a left-tail test is the area in the left tail.

Look up -0.20 in the z-table. It corresponds to an area of 0.42.
 
wow thanks alot.

I'll try study this :arrow:


btw is this correct for

An advertisement for a certain brand of cigarettes claims that there is no more
than 18 mg of nicotine per cigarette on average.
A test of 12 cigarettes gave a sample mean of 19.1 mg of nicotine with a sample
standard deviation of 1.9 mg.

Do you think that the claim is true?

Conduct an appropriate test at the 5% significance level (i.e. α = 0.05) to provide
evidence to support your answer.
What can be said (if anything) about the P-value for this test?




n=12
sample mean=19.1
sample SD=1.9
alpha=0.05

Ha: mean > 18
H0: mean =18
df=11


rejection region is t>1.796
t=2.0055

so reject at alpha=0.05


so is this right and how would i do the p-value for the t-test?
 
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