Question

In a simple random sample of 76 Alzheimer's patients, 12 of them were allergic to a drug. When testing the claim that more than 15% of all Alzheimer's patients are allergic to this drug: a) Specify the null hypothesis and the alternate hypothesis and identify the claim b) State the conditions and show how all conditions for inference have been satisfied c) Specify the type of test statistic, show its formula, show the correct numbers plugged in and give its value d) State the conclusion at 5% level of significance e) Describe a type II error

Answer 1

Sample size = n = 76

Sample proportion of Alzheimer patients who are allergic to the drug = p = 12/76 = 0.16

a) Null hypothesis : Population proportion of Alzheimer patients who are allergic to the drug (P) is 15%

P = 0.15

Alternative hypothesis : More than 15% of Alzheimer patients are allergic to the drug.

P > 0.15

(that is one tailed test)

b) Since, n*Po = 76*0.15 = 11.4 > 10

n*(1-Po) = 76*(1-0.15) = 64.6 > 10

Hence, the population can be assumed to be approximately normal.

c) Test statistic is given by -

$z = \frac{p - P_0}{\sqrt{\frac{P_0*(1-P_0)}{n}}}$

where, Po is the specified value of population proportion under the null hypothesis = 0.15

  $= \frac{0.16 - 0.15}{\sqrt{\frac{0.15*(1-0.15)}{76}}}$

  $= \frac{0.01}{0.041}$

= 0.244

d) At 5% level of significance, the critical value of z for one tailed test is 1.65

(It can be obtained from the z table by finding the z for which area is approximately equal to 0.05)

Since, the value of the test statistic (0.244) > Level of significance (0.05), we may reject the null hypothesis.

Hence, we conclude that more than 15% of Alzheimer patients are allergic to the drug.

e) Type II errror is the error of accepting the null hypothesis when null hypothesis is false or the alternative hypothesis is true.