Bin Range Count N 75 (0,10] 0 Mean 49.67 (10,20] 1 Std. Dev. 14.43 (20,30] 5 (30
ID: 3293214 • Letter: B
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Bin Range Count N 75 (0,10] 0 Mean 49.67 (10,20] 1 Std. Dev. 14.43 (20,30] 5 (30,40] 11 (40,50] 26 (50,60] 12 (60,70] 11 (70,80] 8 (80,90] 1 (90,100] 0 Question 6 (20 points) The midterm exam statistics and histogram (with ten-point bins), are in the third tab of the attached Excel spreadsheet. You will be applying the chi-squared test to determine if the grades follow a Normal distribution, to a 5% level of significance. Considering the null hypothesis that the deviation from a Normal distribution is not statistically significant: a) Determine, the predicted number of scores in each bin nj assuming a Normal distribution. Plot the histogram values (as a bar chart) and the predicted values (as a scatter, with straight lines and no markers) on the same axes, as follows -Predicted 0 (0-10 (10-20] (20-30 (30-40] (40-50] (50-60] (60-70] Score Range Figure 3. Example combo chart for Question 6(a). b) Determine, for each interval, the value ofthe chi-squared parameter MAE 2381 Experimental Methods and Measurements Summer 2017 c) Determine d) Using the value from (c), determine from Table 4.6 in Figliola & Beasley (or some other 2 table), the level of significance corresponding to the data. (Note: it may be necessary to interpolate between columns. Do not simply take the closest value.) Determine 2 at the desired level of significance. Compare the 2 value from (e) to the value calculated in (c). Should we reject the null hypothesis, or not? Does this mean that the data is well-modeled by a Normal distribution, or not? Discuss. e) f) Note that, because the equations for the 2 distribution are a function oftwo statistical parameters (X and Sr), the number ofdegrees of feedom -N-2.Explanation / Answer
the chi- square table
Expected (E) can be calculated be normal distribution mean = 49.67 and standard deviation = 14.43
Expected value for bin range (0,10] = Normalcdf [10, , , True]
Expected value for bin range (10,20] = Normalcdf [20, , , True] - Normalcdf [10, , , True]
(c) Determine xi2 = [Oi - Ei ]2 /Ei
The values are given table
(d) Level of significance alpha = 0.05
(e) X2 = 8.051
NUmber of degrees of freedom dF = 10 -2 = 8
x2 critical = 15.507
so here X2 < x2 critical so we should reject the null hypothesis and conclude that the distribution is normal.
Bin Range Count (o) Expected (E) (O-E)^2/E = xi2 N 75 (0,10] 0 0.22 0.224 Mean 49.67 (10,20] 1 1.27 0.056 Std. Dev. 14.43 (20,30] 5 4.99 0.000 (30,40] 11 12.37 0.152 (40,50] 26 19.33 2.301 (50,60] 12 19.04 2.602 (60,70] 11 11.82 0.057 (70,80] 8 4.62 2.465 (80,90] 1 1.14 0.017 (90,100] 0 0.18 0.176 sum 8.051Related Questions
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