Question

The physician-recommended dosage of a new medication is 14 mg. Actual administered doses vary slightly from dose to dose and are Normally distributed with mean u. A representative of a medical review board wishes to see if there is any evidence that the mean dosage is more than recommended and so intends to test the hypotheses given below

(a) (5 pts.) The physician-recommended dosage of a new medication is 14 mg. Actual administered doses vary slightly from dose to dose and are Normally distributed with mean u. A representative of a medical review board wishes to see if there is any evidence that the mean dosage is more than recommended and so intends to test the hypotheses given below. H: = 14, H.: > 14 To do this, he selects 16 doses at random and determines the weight of each. He finds the sample mean to be x = 14.13 mg and the sample standard deviation to be s = 0.24 mg. Based on these data, test the hypotheses using significance level 0.05. (b) (2 pts.) In the context of this example, describe in words the meaning of type I and type II errors. (c) (2 pts. Determine a 99% confidence interval for the population mean u.

Answer 1

a)

b)

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho = 14 against alternative hypothesis H1 : > 14 Letc denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, Ž= 14.130, s2 = 0.057600,s= 0.24, n = 16 The test statistic can be written as : vna - 14.00) which under Ho follows at distribution with n - 1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > t0.05.n-1 = 1.753050 density 0.0- - 4 2 0 2 Now, Value of the test statistic: The value of the test statistic tobs = V16 (14.1300 – 14). = 2.166667 0.2400 2.17 associated degrees of freedom = 15 The critical value = 1.753050 1.75 P-value = P(t15 > 2.166667) =1- P(t15 <2.166667) =1-0.976614 = 0.023386 – 0.0234 Conclusion : Since p-value < 0.05 and tobs > tcritical = 1.753050, so we reject Ho at 0.05 level of significance Hence, we can conclude that population mean is significantly greater than 14

Type I error : Reject H0 when H0 is actually true , in this case supporting the claim that mean dosage is more than recommended while it is actually not true

Type II error : Aceept H0 when H0 is not actually true , in this case rejecting the claim that mean dosage is more than recommended while it is actually not true

c)

Here, 0.240 w = 14.130, s2 = 0.057600, s = 0.240, n = 16, standard error = = 0.060000 ~ 0.060 V16 alue is tcritical = t 1-0.99.15 = to.005,15 = 2.946713 ~ 2.947 At 99 percent confidence level, The margin of error is: 0.240 ME= 0.240 - * 2.947 = 0.176880 - * t 1-0.99 15 = * t0.005.15 = 0.177 V16 a 99 percent confidence interval of population mean u is S =īt 'n ** 10 15 0.240 = 14.130 + * t0.005,15 V16 = 14.130 + 0.240 * 2.947 V16 = 14.130 – 0.17688 = (13.953120, 14.306880) ~ (13.95, 14.31)