Consider the observed frequency distribution for a set of grouped random variabl
ID: 2949655 • Letter: C
Question
Consider the observed frequency distribution for a set of grouped random variables available below. Perform a chi-square test using a a 0.05 to determine if the observed frequencies follow the normal probability distribution with ? 99 and ?=19. EB Click the icon to view the observed frequency distribution State the appropriate null and altemative hypotheses. What is the null hypothesis? A. Ho: The mean number of the random variable is not equal to 99. O B. Ho: The random variable follows the normal probability distribution Oc. ??: The mean numberofthe random varable is equal to 99. 0 D. ??-The random variable does rot 10lowite normal probablity darbunon. Click to select your answer and then click Check Answer Clear All 8Explanation / Answer
Solution:
The null and alternative hypotheses for the given test are given as below:
Null hypothesis: H0: The random variable follows the normal probability distribution.
Alternative hypothesis: Ha: The random variable does not follow the normal probability distribution.
We are given
Level of significance = ? = 0.05
Mean of normal distribution = µ = 99
SD of normal distribution = ? = 19
Test statistic is given as below:
Chi square = ?[(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
Observed frequencies, Expected frequencies, and calculations are provided in the following table:
X
O
P
E
(O - E)^2
(O - E)^2/E
Less than 80
10
0.158655254
7.932762697
4.273470069
0.538711447
80 to 99
14
0.341344746
17.0672373
9.407944676
0.551228328
99 to under 118
18
0.341344746
17.0672373
0.870046248
0.050977568
118 and more
8
0.158655254
7.932762697
0.004520855
0.000569897
Total
50
1
50
1.141487239
Probabilities of normal distribution are calculated by using z-table or excel.
[Excel command: =normdist() in case particular score or normsdist() in case z-score]
Chi square = ?[(O – E)^2/E] = 1.141487239
Test statistic = 1.141487239
We are given
N = 4, df = 4 – 1 = 3, ? = 0.05
Critical value = 7.814727764 (by using Chi square table or excel command =chiinv(?,df))
P-value = 0.767069502 (by using Chi square table or excel command =chidist(x,df))
Here, P-value > ? = 0.05
So, we do not reject the null hypothesis that the random variable follows the normal probability distribution.
There is sufficient evidence to conclude that the random variable follows the normal probability distribution.
X
O
P
E
(O - E)^2
(O - E)^2/E
Less than 80
10
0.158655254
7.932762697
4.273470069
0.538711447
80 to 99
14
0.341344746
17.0672373
9.407944676
0.551228328
99 to under 118
18
0.341344746
17.0672373
0.870046248
0.050977568
118 and more
8
0.158655254
7.932762697
0.004520855
0.000569897
Total
50
1
50
1.141487239
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