An experiment was conducted to test the effect of a new drug on a viral infectio

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, p1 and p2, in the two groups were found to be 0.36 and 0.60, respectively.

(a) Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Use = 0.05. (Round your answers to two decimal places.)

1-2. Null and alternative hypotheses:

H0: (p1 p2) = 0 versus Ha: (p1 p2) 0

H0: (p1 p2) = 0 versus Ha: (p1 p2) < 0

H0: (p1 p2) < 0 versus Ha: (p1 p2) > 0

H0: (p1 p2) 0 versus Ha: (p1 p2) = 0

H0: (p1 p2) = 0 versus Ha: (p1 p2) > 0

3. Test statistic: z =

4. Rejection region: If the test is one-tailed, enter NONE for the unused region.

z >

z <

5. Conclusion:

H0 is rejected. There is insufficient evidence to indicate that the drug is effective in treating the viral infection.

H0 is not rejected. There is sufficient evidence to indicate that the drug is effective in treating the viral infection.

H0 is rejected. There is sufficient evidence to indicate that the drug is effective in treating the viral infection.

H0 is not rejected. There is insufficient evidence to indicate that the drug is effective in treating the viral infection.

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

Explanation / Answer

H0: (p1 p2) = 0 versus Ha: (p1 p2) < 0

Two-Sample T-Test and CI

Sample    N Mean StDev SE Mean
1       111 71.1   20.7      2.0
2       222 73.4   19.3      1.3

Difference = (1) - (2)
Estimate for difference: -2.27
95% CI for difference: (-6.91, 2.38)
T-Test of difference = 0 (vs ): T-Value = -0.96 P-Value = 0.337 DF = 207

Two-Sample T-Test and CI

Sample    N Mean StDev SE Mean
1       111 71.1   20.7      2.0
2       222 73.4   19.3      1.3

Difference = (1) - (2)
Estimate for difference: -2.27
95% CI for difference: (-6.79, 2.26)
T-Test of difference = 0 (vs ): T-Value = -0.98 P-Value = 0.326 DF = 331
Both use Pooled StDev = 19.8059

Test and CI for Two Proportions

Sample   X   N Sample p
1       18 50 0.360000
2       30 50 0.600000

Difference = p (1) - p (2)
Estimate for difference: -0.24
95% upper bound for difference: -0.0804577
Test for difference = 0 (vs < 0): Z = -2.40 P-Value = 0.008

Fisher’s exact test: P-Value = 0.014

3)

z = -2.40

4)

z < -1.645

5)

p-value = 0.008 < 0.05

we reject the null hypothesis

H0 is rejected. There is sufficient evidence to indicate that the drug is effective in treating the viral infection.

b)

95% CI for difference: (-6.91, 2.38)

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