Answer:
33.293 ± 0.01= 33.303 and 33.383,
Step-by-step explanation:
We first need to fit a normal distribution , but neither the mean nor the standard deviation is given . We therefore estimate the sample mean and sample standard deviation <em>s</em>. Using the data we find ∑fx=<u>378 </u> and
<u>∑fx²=1344 </u> so that mean x` = 2.885 or 2.9 and standard deviation s =1.360
x f fx x² fx²
1 27 27 1 27
2 30 60 4 120
3 29 87 9 261
4 21 84 16 336
<u>5 24 120 25 600 </u>
<u> ∑f=131 ∑fx=378 ∑fx²=1344 </u>
Mean = x`=<u> </u> ∑fx/ <u> </u>∑f= 2.9
Standard Deviation = s= √∑fx²/∑f-(∑fx/∑f)²
s= √1344/131 - (378/131)²
s= √10.26-(2.9)²
s= √10.26- 8.41
s= √1.85= 1.360
Next we need to compute the expected frequencies for all classes and the value of chi square. The necessary calculations for expected frequencies , ei`s ( ei= npi`) where pi` is the estimate of pi together with the value of chi square are shown below.
Categories zi` P(Z<z) pi` Expected Observed
frequency ei Frequency Oi
1 -1.39 0.0823 0.0823 10.78 27
2 -0.66 0.2546 0.1723 22.57 30
3 0.07 0.5279 0.2733 35.80 29
4 0.808 0.7881 0.2602 34.08 21
5 1.54 0.937 0.1489 19.51 24
Next we need to compute the expected frequencies for all classes and the value of chi square. The necessary calculations for expected frequencies , ei`s ( ei= npi`) where pi` is the estimate of pi together with the value of chi square are shown below.
Categories Expected Observed (oi-ei)²/ei
frequency ei Frequency Oi OBSERVED VALUE
1 10.78 27 24.41
2 22.57 30 1.54
3 35.80 29 1.29
4 34.08 21 5.02
5 19.51 24 1.033
<u>Total 131 33.293</u>
There are five categories , we have used the sample mean and sample standard deviation , so the number of degrees of freedom is 5-1-2= 2
The critical region is chi square ≥ chi square (0.001)(2) =9.21
<u>CONCLUSION:</u>
Since the calculated value of chi square =9.21 does not fall in the critical region we are unable to reject our null hypothesis and conclude normal distribution provides a good fit for the given frequency distribution.
