Answer to Question #287763 in Statistics and Probability for Tey

Question

 The Pharmacy Department of Serdang Hospital reported that the average annual spending on medicines was different from RM838 per patient. A random sample of 60 patients was selected and found that the average annual spending on medicines was RM745 per patient with a standard deviation of RM300. (i) At the 5% level of significance, can the department report be rejected? (ii) Explain the Type I and Type II error in this test. 

Answer 1

Solution:

Sample size n = 60

Sample mean x bar = 745

Sample standard deviation s = 300

a.

Ho: population mean mu = 838

H1: population mean mu is not equal to 838

Since the population standard deviation is unknown and sample size is large (>30) ,we use a Z-test

Test statistic: $z=\dfrac{\bar x-\mu}{s/\sqrt n}$

Z = -93/38.729

Z = -2.40125

p value = 0.00820

Using z-score table shown below:

Since p value is less than 0.05, we reject the null hypothesis at 5% level of significance

There is sufficient evidence to say population mean is different from 838

b.

Type 1 error occurs when we reject true Null hypothesis

Type 2 error occurs when we accept false Null hypothesis

Type 1 error is also called as Producers risk

Type 2 error is also called as Consumers risk

Here,

Type 1 error : Pharmacy department reports that average annual spending is not RM838, when originally it is RM838

Type 2 error : Pharmacy department reports that average annual spending is RM838, when originally it is not RM838