Question
For the data provided test the goodness of fit (a = 0.05)that the frequency distribution of battery lives may beapproximated by a normal distribution with mean m = 3.0 andstandard deviation s = 0.7. Note: you should have atleast 5 observations per cell (so, you’ll likely collapse thefirst three into one cell and the last two intervals into one).
Class
Interval
Frequency
1.5-1.9
1
2.0-2.4
2
2.5-2.9
4
3.0-3.4
15
3.5-3.9
10
4.0-4.4
5
4.5-4.9
4
Note to potential helper: I just need help for finding theexp't values for each of the four (after compression) ClassIntervals. After that I know how to do the rest. Thanks.
Class
Interval
Frequency
1.5-1.9
1
2.0-2.4
2
2.5-2.9
4
3.0-3.4
15
3.5-3.9
10
4.0-4.4
5
4.5-4.9
4
Explanation / Answer
If we wish to run a goodness of fit to a normal distibution,with mean 3 and std. dev. .7, we need to calculate the expectedprobability of obtaining these values, then multiply it by thetotal number of observations (41). Use z-scores and calculate areaunder normal curve for each class interval. a)1.5-1.9=(1.5-3)/.7 to (1.9-3)/.7= z-score range from-2.14286 to -1.5714.(normalcdf(z1,z2)=normalcdf(-2.14,-1.57)=.04198times 41=1.72 for Class Interval 1.5-1.9 . . b)2.0-2.4=(2-3)/.7 to (2.4-3)/.7=z-score range from -1.4286 to-.85714.(normalcdf(z1,z2)=normalcdf(-1.4286,-.85714)=.11912times 41=4.88 for Class Interval2.0-2.4 . . c)2.5-2.9=(2.5-3)/.7 to (2.9-3)/.7=z-score range from -.714286to -.14286(normalcdf(z1,z2)=normalcdf(-.714286,-.14286)=.20568times 41=8.4327 for Class Interval 2.5 to2.9 *Note for yourcompression method you can just add these 3 up to get the expectedfor Interval 1.5 to 2.9 as 15.03. Or, ifyou find the probability of 1.5 to 2.9 (z-scores -2.143 to -.14286)you get 17.513, which might be the oneyou are looking for; I can't tell from the context of theproblem. *Note for yourcompression method you can just add these 3 up to get the expectedfor Interval 1.5 to 2.9 as 15.03. Or, ifyou find the probability of 1.5 to 2.9 (z-scores -2.143 to -.14286)you get 17.513, which might be the oneyou are looking for; I can't tell from the context of theproblem. d)3.0-3.4=(3-3)/.7 to (3.4-3)/.7=z-score range from 0 to.5714286(normalcdf(z1,z2)=normalcdf(0,.5714286)=.216145times 41=8.862 for Class Interval 3.0 to3.4. . . e)3.5-3.9=(3.5-3)/.7 to (3.9-3)/.7= z-score range from .7143to 1.2857(normalcdf(z1,z2)=normalcdf(.7143,1.2857)=.1382537times 41=5.668 for Class Interval 3.5 to3.9 . . f)4.0-4.4=(4-3)/.7 to (4.4-3)/.7=z-score range from 1.42857 to2.(normalcdf(z1,z2)=normalcdf(1.42857,2)=.05381times 41=2.206 for Class Interval 4.0 to4.4. . . g)4.5-4.9=(4.5-3)/.7 to (4.9-3)/.7=z-score range from 2.142857to 2.7142857.(normalcdf(z1,z2)=normalcdf(2.14286,2.714286)=.01274times 41=.52239 for Class Interval 4.4 to4.9 *Note you can add upthese last two for your compression method to get the expected for4.0 to 4.9 as 2.728. Ifyou just find the probability of getting 4.0 to 4.9 (z-scores from1.42857 to 2.7143) you get about 3 as the expected value, but I'mnot sure which one you would use (results should be verysimilar). Hope this helps! (It took awhile...) *Note you can add upthese last two for your compression method to get the expected for4.0 to 4.9 as 2.728. Ifyou just find the probability of getting 4.0 to 4.9 (z-scores from1.42857 to 2.7143) you get about 3 as the expected value, but I'mnot sure which one you would use (results should be verysimilar). Hope this helps! (It took awhile...)