One of the reasons health care costs have been rising rapidly in recent years is the increasing cost of malpractice insurance for physicians. Also, fear of being sued causes doctors to run more precautionary tests (possibly unnecessary) just to make sure they are not guilty of missing something. These precautionary tests also add to health care costs. Data in the Excel Online file provided below are consistent with findings in the Reader's Digest article and can be used to estimate the proportion of physicians over the age of 55 who have been sued at least once. Construct a spreadsheet to answer the following questions.
150 participants
92 answered yes
58 answered no
hypothesized population portion 0.50
level of significance 0.05
a. Use the info above to compute the sample proportion of physicians over the age of 55 who have been sued at least once.
b.Calculate the value of the test statistic (4 decimal)
c. what is the p-value (to 4 decimal places)? Do not round intermediate calculations.

So the p value obtained was a very low 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of of physicians over the age of 55 who have been sued at least once is significantly higher than 0.5 .

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=92 represent the adults that answer yes

Part a

[tex]\hat p = \frac{92}{150}=0.613[/tex] estimated proportion of adults that adults that answer yes

[tex]p_o=0.5[/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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5.:

Null hypothesis:[tex]p\leq 0.5[/tex]

Alternative hypothesis:[tex]p > 0.5[/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].

Part b: Calculate the statistic

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

[tex]z=\frac{0.613 -0.5}{\sqrt{\frac{0.5(1-0.5)}{150}}}=2.7679[/tex]

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.

Part c: P value

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

[tex]p_v =P(z>2.7679)=0.0028[/tex]

And we can find this using the following excel code:

"=1-NORM.DIST(2.7679,0,1,TRUE)"

So the p value obtained was a very low 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of of physicians over the age of 55 who have been sued at least once is significantly higher than 0.5 .