Assume that you have 12 well plates available for culturing cells. At the 24 hou
ID: 3201337 • Letter: A
Question
Assume that you have 12 well plates available for culturing cells. At the 24 hour mark, you collect an image from 7 of the wells, count the number of subregions in the image that are populated with cells and determine you have a mean of 122.25 subregions and a standard deviation of 41.86 subregions. There are problems with the image archival that prevent collection from all wells. At the 48 hour mark, you repeat the procedure and determine you have a mean of 217.25 subregions and a standard deviation of 47.74 subregions from 8 of the wells.
What is the standard normal deviate for the difference in the populations from day 1 to day 2?
What is the lower bound on the 95% confidence interval?
What is the upper bound on the 95% confidence interval?
Explanation / Answer
x1 = 122.25 , x2 =217.25 , s1 =41.86, s2 =47.74 , n1 = 7, n2 =8
Pooled standard Error = Sp * sqrt (1/n1 + 1/n2)
Sp = sqrt [( n1 -1 ) *s1^2 + (n2 -1) * s2^2 / n1 + n2 -2]
= sqrt [ ( 7 -1 ) 41.86^2 + ( 8-1) * 47.74^2 / 7 + 8 -2]
= 45.12
Standard error = Sp * sqrt(1/n1 + 1/n2)
= 45.12 * sqrt ( 1/7 + 1/8)
= 23.35
lower bound on the 95% confidence interval?
t value at 95% CI with 14 df = +/- 2.145
Lower bound = (x1 - x2) - t * standard error
= -95 - 2.145 * 23.35
= -145.08
Upper Bound = (x1 - x2) + t * standard error
= -95 + 2.145 *23.35
= -44.91