Question

In clinical trials of the allergy medicine Clarinex (5 mg), it was reported that 50 out of 1655 individuals in the Clarinex group and 31 out of 1652 individuals in the placebo group experienced dry mouth as a side effect of their respective treatments. Test the hypothesis that a greater proportion of individuals in the experimental group experienced dry mouth compared to the individuals in the control group at the α=0.01 level of significance.

Null Hypothesis:

Alternate Hypothesis:

P-value:

Conclusion:

Interpretation:

Construct a 95% confidence interval for the difference between the two population proportions Experimental – Control. Explain what it means.

Answer 1

Solution:

Here, we have to use z test for the difference in two population proportions.

Null hypothesis: H₀: There is no any significant difference in the proportion of individuals in the experimental group and control group experienced dry mouth.

Alternative hypothesis: H_a: A greater proportion of individuals in the experimental group experienced dry mouth compared to the individuals in the control group.

H₀: p₁ = p₂ versus H_a: p₁ > p₂

This is an upper tailed or right tailed test.

We are given

Level of significance = α = 0.01

Test statistic for z test for two population proportions is given as below:

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

Where,

X1 =50

X2 =31

N1 =1655

N2 =1652

P = (X1+X2)/(N1+N2) = (50 + 31)/(1655+1652) = 0.0245

P1 = X1/N1 = 50/1655 = 0.03021148

P2 = X2/N2 = 31/1652 = 0.018765133

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

Z = (0.03021148 – 0.018765133) / sqrt(0.0245*(1 – 0.0245)*((1/1655) + (1/1652)))

Z = 2.1292

P-value = 0.0166

(by using z-table)

P-value > α = 0.01

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that a greater proportion of individuals in the experimental group experienced dry mouth compared to the individuals in the control group.

Now, we have to find the 95% confidence interval for difference between two population proportions.

Confidence interval for difference between two population proportions:

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

Confidence level = 95%

Z = 1.96

(by using z-table)

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

Confidence interval = (0.03021148 – 0.018765133) ± 1.96*sqrt(0.0245*(1 – 0.0245)*((1/1655) + (1/1652)))

Confidence interval = 0.011446347 ± 1.96*0.0054

Confidence interval = 0.011446347 ± 0.0105

Lower limit = 0.011446347 - 0.0105 = 0.0009

Upper limit = 0.011446347 + 0.0105 = 0.0220

Confidence interval = (0.0009, 0.0220)

We are 95% confident that the difference between population proportions of all individuals in the experimental group and control group will lies within 0.0009 and 0.0220.