Question

The drug Lipitor is mean to reduce cholesterol and LDL cholesterol. I n clinical trials 19 out of 863 patients taking 10 mg of Lipitor daily complained of flulike symptoms. Suppose that it is known that 1.9% of patients taking competing drugs complain of flulike symptoms. Is there evidence to conclude that more than 1.9% of Liptor users experience flulike-smptoms as a side effect at α=0.05, level of significance ?

Which of the following is true In the hypothesis testing of the above problem?

1.The conclusion is do not reject

2.The p-value < α=0.01

3.The test is two tail test

4.The test statistic is z = 1.56

Answer 1

Solution:

The null and alternative hypothesis are

H₀ : p = 0.019 vs H_a : p > 0.019

n = 863

x = 19

Let  $url=https://latex.codecogs.com/gif.latex?%5Chat%20p&t=1596167392&ymreqid=f4764063-0d40-e2b0-1cde-f0008d012f00&sig=sTtj6fUQ557EVceE5O50Rw--~D$ be the sample proportion.

$url=https://latex.codecogs.com/gif.latex?%5Chat%20p&t=1596167392&ymreqid=f4764063-0d40-e2b0-1cde-f0008d012f00&sig=sTtj6fUQ557EVceE5O50Rw--~D$ = x/n = 19/863 = 0.0220

The test statistic z is

z =   $url=https://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Chat%20p-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D&t=1596167392&ymreqid=f4764063-0d40-e2b0-1cde-f0008d012f00&sig=3WpFpTtEmFdUVRm2kKWcfg--~D$

=  (0.0220 - 0.019)/ $\sqrt{}$ [0.019*(1 - 0.019)/863]

= 0.65

The Test Statistic is z = 0.65

> sign in Ha indicates that the test is "RIGHT TAILED"(One tailed right sided)

For right tailed z test ,

p value = P(Z > z) = P(Z > 0.65) = 1 - P(Z < 0.65) = 1 - 0.7422 = 0.2578

p value = 0.2578

p value > α = 0.01

The conclusion is do not reject.

Correct option is

The conclusion is do not reject