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.

So the p value obtained was a very high value and using the significance level given we have 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.

Step-by-step explanation:

Data given and notation  

n=244 represent the random sample taken

X=51 represent the subjects with nausea

estimated proportion of subjects with nausea

is the value that we want to test

represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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:  

Alternative hypothesis:  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

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 . The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

So the p value obtained was a very high value and using the significance level given we have 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.

The null hypothesis, should not be rejected

Step-by-step explanation:

Out of 244 subjects, 51 subjects develop nausea:

Sample proportion,

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

i.e whether or not

i.e whether p > 0.2 or not

Null hypothesis, H₀ : p ≤ 0.2

Alternative hypothesis,

The test statistic is given by the formula:

n = 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, should not be rejected

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

over the argument its not good to fight.

gmt is 4 hrs ahead of edt so just add four hours to 1: 14: 20 (minutes and seconds stay the same)

gmt is 5: 14: 20

hope this !

~just a girl in love with shawn mendes