(90,100] I would appreciate if you tell me the steps on how to go about creating

Question

(90,100]

I would appreciate if you tell me the steps on how to go about creating the graph.

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. lines and no markers) on the same axes, as follows: -Predicted -Observed ..slh. (0-10] 10-20] (20-30 30-40] (40-50] (50-60] (60-70] Score Range

Explanation / Answer

Ans:

H0:grades follows normal distribution.

H1:grades do not follow normal distribution.

Use NORMDIST function in excel to calculate expected or perdicted count.

e.g. P(0<X<10)=NORMDIST(10,49.67,14.43,TRUE)-NORMDIST(0,49.67,14.43,TRUE)=0.0027

Now,multiply it by 75,0.0027*75=0.20

Test statistic:

Calculated chi square score=8.029

Critical chi square(significance level=0.05 and df=9)=16.92

As,calculated chi square score<critical chi square score,We fail to reject null hypothesis,H0.

We have sufficient evidence to conclude that grades follow normal distribution.

Range Observed(O) Probability Expected(E) (O-E)^2/E 0-10 0 0.0027 0.20 0.202 10-20 1 0.0169 1.27 0.056 20-30 5 0.0665 4.99 0.000 30-40 11 0.1650 12.37 0.152 40-50 26 0.2577 19.33 2.301 50-60 12 0.2538 19.04 2.602 60-70 11 0.1576 11.82 0.057 70-80 8 0.0617 4.62 2.465 80-90 1 0.0152 1.14 0.017 90-100 0 0.0024 0.18 0.176 Total 75 75.0 8.029

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