Doctors nationally believe that 80% of a certain type of operation are successful. In a particular hospital, 47 of these operations were observed and 37 of them were successful. At = .05, is this hospital's success rate different from the national average?

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] 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 hospital's success rate is not significanlty different from 0.8 or 80%.

Step-by-step explanation:

1) Data given and notation

n=47 represent the random sample taken

X=37 represent the operations successful

[tex]\hat p=\frac{37}{47}=0.787[/tex] estimated proportion of operations successful

[tex]p_o=0.8[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 different from 0.8 or 80%:

Null hypothesis:[tex]p=0.8[/tex]

Alternative hypothesis:[tex]p \neq 0.8[/tex]

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

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic

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

[tex]z=\frac{0.787 -0.8}{\sqrt{\frac{0.8(1-0.8)}{47}}}=-0.223[/tex]

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 [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

[tex]p_v =2*P(z<-0.223)=0.824[/tex]

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] 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 hospital's success rate is not significanlty different from 0.8 or 80%.