A variable contains five categories. It is expected that data are uniformly distributed across these five categories. To test this, a sample of observed data is gathered on this variable resulting in frequencies of 27, 30, 29, 21, and 24. Using α = 0.01, the observed value of chi-square is:

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:

Since the samples are uniformly distributed ;

Substituting the Parameters into the equation :

χ² =

χ² =

χ² =

Hence, the Chisquare value is χ²

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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       27             27        1          27

2       30           60         4          120

3       29           87         9          261

4       21            84         16        336

5       24           120       25        600          

     ∑f=131      ∑fx=378             ∑fx²=1344  

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                     -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

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  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.

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation: