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in a program designed to help patients stop smoking 232 patients were given sustained care and 84.9% of them were no longer smoking after one month use a 0.05 significance level to test the claim that 80% of the patients. Smoking when given sustained care​

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Answer:


z=\frac{0.849 -0.8}{\sqrt{(0.8(1-0.8))/(232)}}=1.869


p_v =2*P(Z>1.869)=0.0616

If we compare the p value obtained and the significance level given
\alpha=0.05 we see that
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 5% of significance the proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .

Explanation:

1) Data given and notation

n=232 represent the random sample taken

X represent the adults were no longer smoking after one month


\hat p=0.849 estimated proportion of adults were no longer smoking after one month


p_o=0.80 is the value that we want to test


\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)


p_v represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.8.:

Null hypothesis:
p=0.8

Alternative hypothesis:
p \\eq 0.8

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.

3) Calculate the statistic

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


z=\frac{0.849 -0.8}{\sqrt{(0.8(1-0.8))/(232)}}=1.869

4) 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=0.05. The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:


p_v =2*P(Z>1.869)=0.0616

If we compare the p value obtained and the significance level given
\alpha=0.05 we see that
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 5% of significance the proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .

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