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fredd [130]
3 years ago
10

in a program designed to help patients stop smoking 232 patients were given sustained care and 84.9% of them were no longer smok

ing after one month use a 0.05 significance level to test the claim that 80% of the patients. Smoking when given sustained care​
Mathematics
1 answer:
grandymaker [24]3 years ago
6 0

Answer:

z=\frac{0.849 -0.8}{\sqrt{\frac{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% .  

Step-by-step 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 \neq 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{\frac{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{\frac{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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dlinn [17]

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Step-by-step explanation:

Isolate t to get the unknown number:

(-4.5t)(49.5t) = 40

      Multiply (-4.5t)(49.5t):

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