Question

An experiment was conducted to test the effect of a new drug on a viral infection. After the infection was induced in 100 mice, the mice were randomly split into two groups of 50. The first group, the control group, received no treatment for the infection, and the second group received the drug. After a 30-day period, the proportions of survivors, p̂1 and p̂2, in the two groups were found to be 0.36 and 0.64, respectively.

(a) Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Use α = 0.05.

(b) Find the test statistic and rejection region. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)

(c) Use a 95% confidence interval to estimate the actual difference (p₁ − p₂) in the survival rates for the treated versus the control groups. (Round your answers to two decimal places.)

Answer 1

Answer:

a &b)

Null hypothesis:

H₀ : p₁ = p₂

Alternative hypothesis:

H_a : p₁ < p₂

Here it is a one –tailed / left-tailed / lower tailed test.

Now consider the test statistic formula

i.e.,

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Where,

X1 = 50*0.36  = 18

X2 = 50*0.64 = 32

N1 = 50

N2 = 50

Now,

P = (X1+X2)/(N1+N2)

substitute the given values

= (18 + 32) / (50 + 50)

= 0.5

P1 = X1/N1 = 18/50

= 0.36

P2 = X2/N2 = 32/50

= 0.64

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Z = (0.36 – 0.64) / sqrt(0.5*(1 – 0.5)*((1/50) + (1/50)))

Z = - 2.8

P-value = 0.0026

(by using z-table)

Here p-value < α = 0.05 So that we reject the null hypothesis

Hence we conclude that there is sufficient evidence to indicate that the drug is effective in treating the viral infection.

c)

To give the 95% interval

Consider,

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Now,

P1 = X1/N1 = 18/50

= 0.36

P2 = X2/N2 = 32/50

= 0.64

N1 = 50

N2 = 50

Here for the Confidence level = 95% the Critical Z value = 1.96

Now consider,

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

substitute the values

Confidence interval = (0.36 – 0.64) ± 1.96*sqrt[(0.36 *(1 – 0.36)/ 50) + (0.64*(1 – 0.64)/ 50)]

= - 0.28 ± 0.1882

Lower limit = - 0.28 - 0.1882

= - 0.4682

Upper limit = - 0.28 + 0.1882

= - 0.0918

Confidence interval = (- 0.47, - 0.09)