A study was conducted with people who have retinitis pigmentosa (an eye disease
ID: 3362540 • Letter: A
Question
A study was conducted with people who have retinitis pigmentosa (an eye disease that causes tunnel vision) to assess the health effects of high dose vitamins A and E. One of the purposes of the study was to discover the influence of vitamin A on serum retinol (a fat-soluble vitamin important in vision). In conducting tests of statistical significance investigators wanted to have confidence that serum retinol levels were approximately normally distributed. Their data are shown below. Test the null hypothesis that the data point to a population where the serum retinol values are normally distributed with a mean of 1.89 mol/L and a standard deviation of 0.36 mol/L . Provide a brief answer to this question: Does it look as though the variable can be treated as being normally distributed? (one or two sentences will do)Explanation / Answer
Back-up Theory
Let X ~ N(µ, 2).
To test if the given sample observations on X conform to N(µ, 2), the process involved is given below.
Let Li and Ui be respectively the lower and upper boundaries of the ith class (class interval) of the frequency distribution of given observations.
Define the corresponding Z-values as follows:
Z(Li) = (Li – Xbar)/s and Z(Ui) = (Ui – Xbar)/s where Xbar and s are respectively the sample mean and sample standard deviation.
Using Excel Function or Tables of Standard Normal Distribution, evaluate the following:
(Li) = P[Z Z(Li)] and (Ui) = P[Z Z(Ui)].
[(Ui) can also be directly found out using Excel Function for the given mean and standard deviation]
Then,
Expected probability of ith class = Pi = (Ui) - (Li) or Pi = (Ui) - (Ui-1) and
Expected frequency of ith class = Pi x n, where n = total frequency of sample observations.
Test Statistic = 2 = [1,k]{(Oi - Ei)2/Ei}, where Oi and Ei are respectively the observed and expected frequencies of ith class and k = number of classes.
The null hypothesis, H0, that the given sample observations on X conform to N(µ, 2) is rejected at % level of significance if the calculated value 2cal > 2crit where 2crit is the upper % point of Chi-square distribution with (k - 1) degrees of freedom.
Calculations
Expected Class Probabilities and Frequncies, Pi, EI
µ =
1.89
=
0.36
n =
73
Class (i)
Ui
(Ui)
Pi
Ei
1
1.405
0.088954
0.08895434
6.493667
2
1.755
0.35383
0.2648759
19.33594
3
2.105
0.72482
0.37099025
27.08229
4
2.455
0.941728
0.2169073
15.83423
5
> 2.455
0.05827221
4.253871
Total
1
73
Chi-square Calculation
k =
5
Class (i)
Oi
Ei
(Oi-Ei)^2/Ei
1
6
6.493667
0.0375299
2
22
19.33594
0.36704774
3
22
27.08229
0.95374721
4
20
15.83423
1.09595535
5
3
4.253871
0.36959108
Total
73
2.82387128
=
0.05
2cal =
2.82387128
DF =
4
2crit =
9.48772904
p-value =
0.58771918
Since 2cal < 2crit, null hypothesis is accepted. Also confirmed by p-value > .
So, the given observations can be assumed to have come from a Normal Population with mean 1.89 and standard deviation 0.36 ANSWER
Expected Class Probabilities and Frequncies, Pi, EI
µ =
1.89
=
0.36
n =
73
Class (i)
Ui
(Ui)
Pi
Ei
1
1.405
0.088954
0.08895434
6.493667
2
1.755
0.35383
0.2648759
19.33594
3
2.105
0.72482
0.37099025
27.08229
4
2.455
0.941728
0.2169073
15.83423
5
> 2.455
0.05827221
4.253871
Total
1
73
Chi-square Calculation
k =
5
Class (i)
Oi
Ei
(Oi-Ei)^2/Ei
1
6
6.493667
0.0375299
2
22
19.33594
0.36704774
3
22
27.08229
0.95374721
4
20
15.83423
1.09595535
5
3
4.253871
0.36959108
Total
73
2.82387128
=
0.05
2cal =
2.82387128
DF =
4
2crit =
9.48772904
p-value =
0.58771918
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