Question 1132264: An irate patient complained that the cost of doctor's visit was to high. She randomly surveyed 20 other patients and found that the mean amount of money they spent on each doctor's visit was $44.80. the standard deviation of the sample was 43.53.
a) Find a point estimate of the population mean.
b) Find the 95% confidence interval of the population mean. Assume the variable is normally distributed.
c) Test the hypothesis that the cost of doctor's visit was $35 at 5%
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! point estimate is $44.80, the sample mean
half-interval is t(df=19, 0.975)=2.093 multiplied by s/sqrt(n)
this is 2.093*43.53/sqt(20)=20.37
The interval is ($24.43, $65.17)
Because $35 is in the interval, one cannot reject the null hypothesis.
t=(35-44.80)/43.53/sqrt(20)
=-0.96, which would not reach significance. The doctor's visit certainly could cost $35.
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