Question 1196734: In a study of 805 randomly selected medical malpractice lawsuits, it was found that 495 of them were dropped or dismissed. Use a 0.01 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed.
Found 2 solutions by math_tutor2020, ewatrrr:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
p = population proportion of lawsuits that are dropped or dismissed
The phrasing "most are dropped or dismissed" implies p > 0.5, which is where the claim resides. This is in the alternative hypothesis
Null: p ≤ 0.5
Alternative: p > 0.5
We have a right-tailed test due to the inequality sign in the alternative hypothesis.
x = 495 = number of lawsuits dropped or dismissed
n = 805 = sample size
phat = sample proportion of lawsuits dropped or dismissed
phat = x/n
phat = 495/805
phat = 0.614907 approximately
According to the sample taken, about 61.4% of the lawsuits were dropped or dismissed. This is over 50%, so does this confirm the claim? Not necessarily since this might have happened purely by chance. Perhaps we could have taken another random sample and got a phat smaller than 0.50. This is where the hypothesis test is needed.
Compute the standard error (SE)
SE = sqrt(p*(1-p)/n)
SE = sqrt(0.5*(1-0.5)/805)
SE = 0.017623 approximately
Then compute the test statistic
z = (phat - p)/SE
z = (0.614907 - 0.5)/0.017623
z = 6.52028598989957
z = 6.52
Now we'll need the critical value. We're going to use an inverse normal calculator such as this one
https://statisticshelper.com/inverse-normal-distribution-calculator/
Use 0 for the mean and 1 for the standard deviation.
Type 0.01 for the area, which is the alpha level.
The result it should show is -2.33
We'll look at the positive version
So P(Z > 2.33) = 0.01 approximately
You can use something like this to confirm
https://davidmlane.com/normal.html
and you can also use that same calculator to find the critical values. The nice thing about that second calculator is that it shows what the diagram looks like when shading under the normal curve.
The key takeaway is that z = 2.33 is the approximate critical z value here. Anything larger will be in the rejection region. This applies to our test statistic of z = 6.52; therefore, we reject the null.
If you were to use the p-value approach, then you should find the p-value is so small it's practically zero.
Since the p-value is smaller than alpha = 0.01, it's another way to see why we reject the null.
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Summary:
Hypotheses
Null: p ≤ 0.5
Alternative: p > 0.5
Claim is in the alternative
This is a right-tailed test
Test statistic: z = 6.52
Critical value: z = 2.33 (stuff to the right of this is in the rejection region)
p-value = so very small it's practically 0
Decision: Reject the null
Interpretation: The majority (ie more than 50%) of the lawsuits were indeed dropped or dismissed.
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website!
Hi
In a study of 805 randomly selected medical malpractice lawsuits, it was found that 495 of them were dropped or dismissed.
x̄ = 495/805 = .614907
Use a 0.01 significance level. Note: critical value = | invNorm(.01)| = 2.326
test the claim that "most"( > 50% ) medical malpractice lawsuits are dropped or dismissed
Ho: u ≤ .50 Note: Ho MUST contain equality:: = or ≤ or ≥
Ha: u > .50 claim
 = 6.520
If the absolute value of the t-value is greater than the critical value, you reject the null hypothesis.
6.520 > 2.326 REJECT Ho
that is: "most"( > 50% ) medical malpractice lawsuits ARE dropped or dismissed
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