Respuesta :
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  s. Using the data we find ∑fx=378  and
∑fx²=1344  so that mean x` = 2.885 or 2.9  and standard deviation s =1.360
x     f         fx    x²     fx²
1 Â Â Â 27 Â Â Â Â Â Â 27 Â Â Â Â 1 Â Â Â Â Â 27
2 Â Â Â 30 Â Â Â Â Â 60 Â Â Â Â 4 Â Â Â Â Â 120
3 Â Â Â 29 Â Â Â Â Â 87 Â Â Â Â 9 Â Â Â Â Â 261
4 Â Â Â 21 Â Â Â Â Â Â 84 Â Â Â Â 16 Â Â Â Â 336
5 Â Â Â 24 Â Â Â Â Â 120 Â Â Â 25 Â Â Â Â 600 Â Â Â Â Â
   ∑f=131    ∑fx=378       ∑fx²=1344 Â
Mean = x`= ∑fx/  ∑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 Â Â Â Â Â Â Â Â Â Â -1.39 Â Â Â 0.0823 Â Â 0.0823 Â Â Â 10.78 Â Â Â Â Â Â Â Â Â 27
2 Â Â Â Â Â Â Â Â Â -0.66 Â Â Â 0.2546 Â Â 0.1723 Â Â Â Â 22.57 Â Â Â Â Â Â Â Â 30
3 Â Â Â Â Â Â Â Â Â Â 0.07 Â Â Â Â 0.5279 Â Â Â 0.2733 Â Â Â 35.80 Â Â Â Â Â Â Â Â 29
4 Â Â Â Â Â Â Â Â Â 0.808 Â Â Â 0.7881 Â Â Â 0.2602 Â Â Â Â 34.08 Â Â Â Â Â Â Â Â 21
5 Â Â Â Â Â Â Â Â Â Â 1.54 Â Â Â Â 0.937 Â Â Â Â Â 0.1489 Â Â Â Â 19.51 Â Â Â Â Â Â Â Â 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 Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10.78 Â Â Â Â Â Â Â Â Â 27 Â Â Â Â Â Â Â Â Â Â 24.41
2 Â Â Â Â Â Â Â Â Â Â Â Â Â 22.57 Â Â Â Â Â Â Â Â 30 Â Â Â Â Â Â Â Â Â Â Â Â 1.54
3 Â Â Â Â Â Â Â Â Â Â Â Â Â 35.80 Â Â Â Â Â Â Â Â 29 Â Â Â Â Â Â Â Â Â Â Â Â 1.29
4 Â Â Â Â Â Â Â Â Â Â Â Â 34.08 Â Â Â Â Â Â Â Â 21 Â Â Â Â Â Â Â Â Â Â Â Â Â 5.02
5 Â Â Â Â Â Â Â Â Â Â Â Â Â 19.51 Â Â Â Â Â Â Â Â 24 Â Â Â Â Â Â Â Â Â Â Â Â 1.033
Total                        131          33.293
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
CONCLUSION:
Since the calculated value of chi square =9.21 Â 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.
Using the Chisquare test statistic relation, the observed value of the Chisquare statistic is 2.09
The observed value of Chisquare can be calculated thus:
- χ² = [tex] \frac{(observed - Expected)^{2}}{Expected} [/tex]
Since the samples are uniformly distributed ;
- Expected value = (27+30+29+21+24)/5 = 131/5 = 26.2
Substituting the Parameters into the equation :
χ² =[tex] \frac{(27 - 26.2)^{2}}{26.2} + \frac{(30 - 26.2)^{2}}{26.2} + \frac{(29 - 26.2)^{2}}{26.2} + \frac{(21 - 26.2)^{2}}{26.2} + \frac{(24 - 26.2)^{2}}{26.2} [/tex]
χ² =[tex] 0.0244 + 0.5511 + 0.2992 + 1.0321 + 0.1847 [/tex]
χ² =[tex] 2.0915 [/tex]
Hence, the Chisquare value is χ² [tex] = 2.09 [/tex]
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