This question uses R. All relevant code would be greatly appreciated. 7. Suppose

Question

This question uses R. All relevant code would be greatly appreciated.

7. Suppose we are trying to find a faster drying glue. In a study done years age, Brand A and Brand B were tested 8 times each (on a total of 16 identical surfaces) and the drying times were Brand ABrand B 11.56 9.16 10.81 11.35 6.04 8.60 9.72 12.26 16.41 18.52 14.13 12.49 6.29 14.54 14.60 15.79 4 Now, suppose Brand B has now come out with a new and improved version and we want to collect a new sample of Brand A and the new Brand B drying times. We believe the drying time will be more similar and want to be able to compare mean drying times. To estimate the variability for our new study, we will use the pooled sample variance from the old study as an estimate for (a) What sample size do we need so that we can construct a 95% confidence interval estimating As-Ha that has a margin of error equal to 1 minute? (b) Consider a hypothesis test of Ho : As-Ha vs. Ha : As > A, using = 0:05. What sample size do we need in order to detect a difference of 1 minute with 90% power?

Explanation / Answer

a)

Two-Sample T-Test and CI: A, B

Two-sample T for A vs B

   N   Mean StDev SE Mean
A 8   9.94   2.01     0.71
B 8 15.35   1.82     0.64

Difference = (A) - (B)
Estimate for difference: -5.409
95% CI for difference: (-7.467, -3.351)
T-Test of difference = 0 (vs ): T-Value = -5.64 P-Value = 0.000 DF = 14
Both use Pooled StDev = 1.9189

1.9189

z * sqrt(sp^2 (1/n1 + 1/n2)) = e

n1 = n2 =n

n = (zs /e)^2 *sqrt(2)

here z = 1.96 , e = 1

= (1.96 /1)^2 * 2 * 1.9189^2

= 28.2909

hence n1 = n2 = 29

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