Question

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?

Answer 1

Answer:

$z=\frac{0.787 -0.8}{\sqrt{\frac{0.8(1-0.8)}{47}}}=-0.223$  

$p_v =2*P(z$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ 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 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

$\hat p=\frac{37}{47}=0.787$ estimated proportion of operations successful

$p_o=0.8$ 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 different from 0.8 or 80%:  

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:  

$z=\frac{0.787 -0.8}{\sqrt{\frac{0.8(1-0.8)}{47}}}=-0.223$  

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:  

$p_v =2*P(z$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ 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 5% of significance the proportion of hospital's success rate is not significanlty different from 0.8 or 80%.