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Suppose 229 subjects are treated with a drug that is used to treat pain and 52 of them develop nausea use a 0.01 significance level to test the claim that more than 20% of users develop nausea identify the Noel and alternative hypothesis for this test

User Bsarrazin
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Answer:

The null hypothesis is
H_o: p \leq 0.2

The alternate hypothesis is
H_a: p > 0.2

The pvalue of the test is 0.1515 > 0.01, which means that we do not reject the null hypothesis that 20% of less users develop nausea, that is, we have no sufficient evidence that this proportion is larger than 20%.

Explanation:

Test the claim that more than 20% of users develop nausea

This means that the null hypothesis is that 20% or less of the users develop nausea, that is:


H_o: p \leq 0.2

And the alternate hypothesis is that more than 20% develop, so:


H_a: p > 0.2

The test statistic is:


z = (X - \mu)/((\sigma)/(√(n)))

In which X is the sample mean,
\mu is the value tested at the null hypothesis,
\sigma is the standard deviation and n is the size of the sample.

0.2 is tested at the null hypothesis:

This means that
\mu = 0.2, \sigma = √(0.2*0.8)

Suppose 229 subjects are treated with a drug that is used to treat pain and 52 of them develop nausea.

This means that
n = 229, X = (52)/(229) = 0.2271

Value of the test statistic:


z = (X - \mu)/((\sigma)/(√(n)))


z = (0.2271 - 0.20)/((√(0.2*0.8))/(√(229)))


z = 1.03

Pvalue of the test an decision:

Probability of finding a proportion above 0.2271, which is 1 subtracted by the pvalue of z = 1.03

Looking at the z-table, z = 1.03 has a pvalue of 0.8485

1 - 0.8485 = 0.1515

The pvalue of the test is 0.1515 > 0.01, which means that we do not reject the null hypothesis that 20% of less users develop nausea, that is, we have no sufficient evidence that this proportion is larger than 20%.

User Benomite
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