Question

Most cases of cervical cancer are linked to a few strains of the human papilloma virus (HPV). The pharmaceutical company Merck developed a vaccine ("Gardisil") against these HPV strains. Worldwide clinical trials followed young women after vaccination or administration of placebo for two to four years for signs of HPV-caused cervical cancer. Based on a sample of 100 women in the Gardisil group, 32% developed cervical cancer and based on a sample of 100 women in the placebo group, 42% developed cervical cancer. Test the claim that the proportion of women who developed cervical cancer in the Gardisil group is less than the proportion of women who developed cervical cancer in the placebo group at α=0.01.α=0.01.

Round your answers to three decimal places.

(a). Select the correct hypotheses where  p1p1   denotes the proportion of women who developed cervical cancer in the Gardisil group and p2p2 denotes the proportion of women who developed cervical cancer in the placebo group.

• H0:ˆp1=ˆp2H0:p^1=p^2, Ha:ˆp1>ˆp2Ha:p^1>p^2
• H0:p1=p2H0:p1=p2, Ha:p1≠p2Ha:p1≠p2
• H0:p1=p2H0:p1=p2, Ha:p1>p2Ha:p1>p2
• H0:p1=p2H0:p1=p2, Ha:p1<p2Ha:p1<p2

(b). Without rounding any interim calculations, compute the test statistic:     

(c). Compute the critical value:   

(d). Using your answer from part (b), compute the p-value:

Answer 1

(a) Here hypothesis are

H₀ : p₁ = p₂

H_a : p₁ < p₂

(b) Here pooled proportion = p = (0.32 * 100 + 0.42 * 100)/(100 + 100) = 0.37

standard error of proportion = sqrt [p₁(1-p₁)/n₁ + p₂(1-p₂)/n₂] = sqrt [0.32 * (1-0.32)/100 + 0.42 * (1 - 0.42)/100] = 0.0679

Test statistic

Z = (p^{^}₁ - p^{^}₂)/se = (0.32 - 0.42)/0.0679 = -1.4725

(c) Here alpha = 0.01

so here critical value = Z_critical = NORMSINV(0.01) = -2.326

(d) Here p - value = P(Z < -1.4725) = 0.0704

Here p - value > 0.01 so we would fail to reject the null hypothesis