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Suppose 244 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. Use a 0.01 significance level to test the claim that more than 20​% of users develop nausea.

User Apohl
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Answer:

The null hypothesis,
H_(0) should not be rejected

Explanation:

Out of 244 subjects, 51 subjects develop nausea:

Sample proportion,
\hat{p} = (51)/(244) \\


\hat{p} = 0.209

We want to test whether Population proportion, p > 20% or not

i.e whether
p > (20)/(100) or not

i.e whether p > 0.2 or not

Null hypothesis, H₀ : p ≤ 0.2

Alternative hypothesis,
H_(a) : p > 0.2

The test statistic is given by the formula:


z = \frac{\hat{p} -p_(0) }{\sqrt{(p_(0)(1-p_(0) ) )/(n) } }

n = 244


p_(0) = 0.2


1 - p_(0) = 1-0.2 = 0.8


\hat{p}-p_(0) = 0.209 - 0.2 = 0.009


z = \frac{0.009 }{\sqrt{(0.2(0.8 ) )/(244) } }

z = 0.3515

p(z>0.3515) = 1 - p(z≤0.3515) = 1 - 0.637

p(z>0.3515) = 0.363

p value = 0.363

α = 0.01

Since 0.363 > 0.01

The null hypothesis,
H_(0) should not be rejected

There is no evidence that supports the claim that the null hypothesis should be rejected at 0.01 significance level

User Ujifgc
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3 votes

Answer:


z=\frac{0.209 -0.2}{\sqrt{(0.2(1-0.2))/(244)}}=0.351


p_v =P(z>0.351)=0.363

So the p value obtained was a very high 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 FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of subjects with nauseas is not higher than 0.2 or 20% at 1% of significance.

Explanation:

Data given and notation

n=244 represent the random sample taken

X=51 represent the subjects with nausea


\hat p=(51)/(244)=0.209 estimated proportion of subjects with nausea


p_o=0.2 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

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.2.:

Null hypothesis:
p \leq 0.2

Alternative hypothesis:
p > 0.2

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{0.209 -0.2}{\sqrt{(0.2(1-0.2))/(244)}}=0.351

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 right tailed test the p value would be:


p_v =P(z>0.351)=0.363

So the p value obtained was a very high 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 FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of subjects with nauseas is not higher than 0.2 or 20% at 1% of significance.

User Jason Leach
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