Hello, does anyone have the manual calculations for Chapter 7, Problem 55 - Intr

Question

Hello, does anyone have the manual calculations for Chapter 7, Problem 55 - Introduction to the Practice of Statistics (8th Edition). Thanks

72 Comparing Two Means 447 7.52 Insulation study. A manufacturer of electric motors tests insulation at a high temperature (250°C) and records the number of hours until the insulation fails.18 The data for 5 specimens are 446 326 372 377 310 The small sample size makes judgment from the data difficult, but engineering experience suggests that the logarithm of the failure time will have a Normal distribution. Take the logarithms of the 5 observations, and user procedures to give a 90% confidence interval for the mean of the log failure time for insulation of this type. INSULAT 7.53 Power of the comparison of DXA machine operators. Suppose that the bone researchers in Exercise 7.43 wanted to be able to detect an alternative mean dif ference of 0.002. Find the power for this alternative for a sample size of 15. Use the standard deviation that you found in Exercise 7.43 for these calculations. 4. 7.54 Sample size calculations. You are designing a study to test the null hypothesis that -0 versus the alternative that is positive. Assume that is 15. Suppose that it would be important to be able to detect the alter- native -2. Perform power calculations for a variety of sample sizes and determine how large a sample you would need to detect this alternative with power of at least 0.80. 7.55 Determining the sample size. Consider Example 7.9 (page 435). What is the minimum sample size needed for the power to be greater than 80% when 0.75?

Explanation / Answer

Here alternative true mean = 0.75

and let say the sample size is n.

so standard error of sample mean = 1.5/ n

so here sample mean below which we cannot reject the null hypothesis is

x = 0 + t0.05,n-1 (1.5/ n ) = t0.05,n-1   1.5/ n

so for Power shall be greater than 0.8 we know that type II error < 1 - 0.8 = 0.2

so Pr(Type II error) = 0.2

Pr(x < t0.05,n-1 *1.5/ n ) = NORM(x <    t0.05,n-1 *1.5/ n ; 0.75 ; 1.5/ n) = 0.2

so Z - value for p - value = 0.2 is

Z = -0.842

-0.842 > (t0.05,n-1 *1.5/ n - 0.75)/ (1.5/ n)

0.75 > ( t0.05,n-1 *1.5+ 1.263)/ n

n < ( t0.05,n-1 *1.5+ 1.263)/ 0.75

so will will go for trial and error method.

for n = 25

t0.05,24 = 1.7109

( t0.05,n-1 *1.5+ 1.263)/ 0.75 = (1.7109 * 1.5 + 1.263)/0.75 = 5.1058

not good.

for n = 26,

t0.05,24 = 1.7081

( t0.05,n-1 *1.5+ 1.263)/ 0.75 = (1.7081 * 1.5 + 1.263)/0.75 = 5.10

sqrt(n) = sqrt(26) = 5.099

so yes both values are nearly equal. so n > 26

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