Question

10.2.17- Assigned Media Question Help In a clinical trial, 22 out of 864 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 2.2% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 2.2% of this drug's users experience flulike symptoms as a side effect at the a = 0.01 level of significance? less than 5% of the population size, and the sample can be reasonably assumed to be random, the requirements for Because np. (1-P) = 18.6 10, the sample size is testing the hypothesis are satisfied (Round to one decimal place as needed.) What are the null and alternative hypotheses? Hop 0.022 versus H. p> 0.022 (Type integers or decimals. Do not round.) Find the test statistic, Zo 20=(Round to two decimal places as needed.)

Answer 1

Solution :-

10.2)

  Because np₀ (1-P₀) = 18.6 > 10. The sample size is cannot be more than 5% of the population size ,and the sample cannot be random the requirements for testing the hypothesis is not satisfied.

This is the more tailed test .

The null and alternative hypothesis is

H₀ : p = 0.022

H_a : p > 0.022

$\hat p$ = x / n = 22/864 = 0.0255

P₀ = 0.022

1 - P₀ = 0.978

Test statistic = z

= $\hat p$ - P₀ / [ $\sqrt$ P_{0 *} (1 - P₀ ) / n]

=0.0255-0.022 / [ $\sqrt$ 0.022*(0.978) /864 ]

= 0.69

P(z > 0.69) = 1 - P(z < 0.69) = 0.2451

P-value = 0.2451

$\alpha$ = 0.01

0.2451 >0.01

Do not reject the null hypothesis .

There is in.sufficient evidence to suggest that  

Because np₀ (1-P₀) = 268.0 > 10. The sample size is cannot be more than 5% of the population size ,and the sample cannot be random the requirements for testing the hypothesis is not satisfied.

This is the left tailed test .

The null and alternative hypothesis is

H₀ : p = 0.41

H_a : p <0.41

$\hat p$ = x / n = 1108/435 =0.3946

P₀ = 0.41

1 - P₀ = 1-0.41 = 0.59

Test statistic = z

= $\hat p$ - P₀ / [ $\sqrt$ P_{0 *} (1 - P₀ ) / n]

=0.3946-0.41 / [ $\sqrt$ 0.41*(0.59) /1108 ]

= -1.18

P(z > -1.18) = 1 - P(z <-1.18 ) = 0.1190

P-value = 0.1190

$\alpha$ = 0.01

0.1190> 0.01

Do not reject the null hypothesis .

There is insufficient evidence to suggest that