Download sample test

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(1) Example 1 : A sample of size 9 from a normal population gave
. Find 99% interval of population mean.
(M.U.
2009)
Sol. : Step 1 : For 99% confidence level and for
freedom the column under
gives the critical value
degrees of
Step 2 : Since the population standard deviation is not known and
sample is small
Step 3 : the confidence interval is
(2) Example 2 : Nine items of a sample had the following values
45, 47, 50, 52, 48, 47, 49, 53, 51
Does the mean of 9 items differ significantly from the assumed population mean
(M.U. 2002, 10)
Sol. : We first calculate sample mean and sample standard deviation (by assumed
mean method).
Calculation of and
X
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45
47
50
52
48
47
49
-3
-1
2
4
0
-1
1
9
1
4
16
0
1
1
53
51
Sum
5
3
10
25
9
66
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(i)
Awab Sir-89 76 104646
The null hypothesis
Alternative hypothesis
(ii)
Calculation of test statistic : Since the sample size is small, we use
t-distribution.
(iii)
(iv)
Level of significance :
Critical value : The value of at 5% level of significance for
degress of freedom is
Decision : Since the calculated value of
is less than the table
value
the null hypothesis is accepted.
(v)
The mean of nine items does not differ significantly from assumed
population mean
.
(3) Example 3 : ten individuals are chosen at random from a population and their heights are
found to be 63, 63, 64, 65, 66, 69, 69, 70, 70, 71 inches. Discuss the suggestion that the
mean height of universe is 65 inches.
(M.U. 2003)
Sol. : We first calculate sample mean
and sample standard deviation
Calculation of
X
(i)
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.
and
63
63
64
65
66
69
69
70
70
71
Sum
-3
-3
-2
-1
0
3
3
4
4
5
10
9
9
4
1
0
9
9
16
16
25
98
The null hypothesis
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Awab Sir-89 76 104646
Alternative hypothesis
(ii)
Calculation of test statistic :
(iii)
Level of significance :
(iv)
Critical value : The value of
at 5% level of significance for
degress of freedom is
(v)
Decision : Since the calculated value of
value
is less than the table
the null hypothesis is accepted.
The mean height of the universe may be 65 inches.
(4) Example 4 : Tests made on breaking strength of 10 pieces of a metal wire gave the
following results. 578, 572, 570, 568, 572, 570, 570, 572, 596, and 584 in kgs.
Test if the breaking strength of the metal wire can be assumed to be 577 kg. ?
(M.U. 2002, 04, 05, 06)
Sol. : We first calculate the mean and sample standard deviation of the sample.
Calculation of and
X
(i)
578
572
570
568
572
570
570
572
596 584
-2
-8
-10
-12
-8
-10
-10
-8
6
4
-48
4
64
100
144
64
100
100
64
256
16
864
Sum
The null hypothesis
Alternative hypothesis
(ii)
Calculation of test statistic :
(iii)
Level of significance :
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(iv)
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Critical value : The value of
at 5% level of significance for
degress of freedom is
(v)
Decision : Since the calculated value of
value
is less than the table
the null hypothesis is accepted.
The mean is 577.
(5) Example 5 : The means of two random samples of size 9 and 7 are
and
respectively. The sum of the squares of the deviation from the means are
and
repectively. Can the samples be considered to have been drawn from
the same population ?
(M.U. 2004)
Sol. :
(i)
Null Hypothesis :
Alternative Hypothesis
(ii)
Calculation of test statistic : unbiased estimate of common population
standard deviation is
Standard error of the difference between the means
(iii)
Level of Significance :
(iv)
Critical value : The table value of at
degrees of freedom is
(v)
Decision : Since the computed value of
table value
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is greater than the
, the null hypothesis is rejected.
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The sample cannot considered to have been drawn from the same population.
(6) Example 6 : Samples of two types of electric bulbs were tested for length of life and the
following were obtained,
Type I
No. of samples
Type II
8
7
Mean of the samples (in hours)
1134
1024
Standard deviation (in hours)
35
40
Test at 5% level of significance whether the difference in the sample means is significant.
(Table value of t for 13 d.f. is
, for 14 d.f. is
and for 15 d.f. is
).
(M.U. 2004, 06)
Sol. : We have
(i)
Null Hypothesis :
Alternative Hypothesis
(ii)
Calculation of test statistic : Since the sizes of the sample are small we
use t-distribution.
The Standard error of the difference between the two means is given by
(iii)
Level of Significance :
(iv)
Critical value : The table value of at
degrees of freedom is
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(v)
Awab Sir-89 76 104646
Decision : Since the computed value of
table value
is greater than the
, the null hypothesis is rejected.
The difference is significant..
(7) Example 7 : The heights of six randomly chosen sailors are in inches : 63, 65, 68, 69, 71,
and 72. The heights of ten randomly chosen soldiers are : 61, 62, 65, 66, 69, 69, 70, 71,
72, and 73.
Discuss in the light that these data throw on the suggestion that the soldiers on an
average are taller than sailors.
(M.U. 1997, 2002)
Sol. : We first calculate the mean and standard deviation of the heights of both sailors and
soldiers.
Sailors
Soldiers
Heights
63
65
68
69
71
72
Heights
-5
-3
0
1
3
4
25
9
0
1
9
16
61
62
65
66
69
69
70
71
72
73
0
0
Now,
The unbiased estimate of the common population
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(i)
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Null Hypothesis :
Alternative Hypothesis
(ii)
Calculation of test statistic :
Now,
(iii)
Level of Significance :
(iv)
Critical value : The table value of at
degrees of freedom is
(v)
Decision : Since the computed value of
table value
is smaller than the
, the hypothesis is accepted.
The means are equal i.e. the suggestion that the soldiers on the average
are taller than sailors cannot be accepted.
(8) Example 8 : A certain injection administered to 12 patients resulted in the following
changes of blood pressure :
5, 2, 8, -1, 3, 0, 6, -2, 1, 5, 0, 4
Can it be concluded that the injection will be in general accompanied by an increase in
blood pressure ?
(M.U. 1998, 2004, 11)
Sol. : We first calculate
.
Calculation of and
X
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5
2
8
-1
3
0
6
-2
1
5
0
4
3
0
6
-3
1
-2
4
-4
-1
3
-2
2
9
0
36
9
1
4
16
16
1
9
4
4
Page 7
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(i)
Awab Sir-89 76 104646
The null hypothesis
Alternative hypothesis
(ii)
Calculation of test statistic : Since the sample sizes is small, we use
students t-distribution.
(The positive sign of t denotes the increase in the level i.e. in blood pressured.)
(iii)
Level of significance :
(iv)
Critical value : The value of
at 5% level of significance for
degress of freedom for one tailed test
at 10%
LOS for two tailed test
(v)
Decision : Since the calculated value of
critical value
is greater than the
the hypothesis is rejected.
The is rise in B.P.
(9) Example 9 : In a certain experiment to compare two types of pig-tools A and B, the
following result of increasing weight were obtained.
Pig Number
: 1 2 3 4 5 6 7 8
Increase in weight X kg by A
: 49 53 51 52 47 50 52 53
Increase in weight Y kg by B
: 52 55 52 53 50 54 54 53
(i) Assuming that two sample of pigs are independent, can we conclude that food B is
better than food A.
(ii) Examine the case if the same set of pigs were used in both the cases. (M.U. 2004,06)
Sol. : (a) We first calculate
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.
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Calculation of
etc.
Food A
49
53
51
52
47
50
52
53
-2
2
0
1
-3
-1
1
2
-1
Food B
4
4
0
1
16
1
1
4
31
52
55
52
43
50
54
54
53
-1
2
-1
0
-3
1
1
0
-1
1
4
1
0
9
1
1
0
17
And
(i)
Null Hypothesis :
Alternative Hypothesis
(ii)
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Calculation of test statistic :
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(The negative value indicates the decrease in the level i.e. in the increase in
weight.)
(iii)
Level of Significance :
(iv)
Critical value : The table value of at
degrees of freedom is
(v)
Decision : Since the computed value
is less than the table
value
, the hypothesis is rejected at 5% level of significance
(One tailed test.)
Food B is superior to A.
(b) If the same set of pigs were used in the two tests : We first calculate the difference
between the weights in the two tests and from these we calculate
.
Calculation of and
X
(i)
49
53
51
52
47
50
52
53
Total
52
55
52
53
50
54
54
53
-3
-2
-1
-1
-3
-4
-2
0
-5
-4
-3
-3
-5
-6
-3
-2
-32
25
16
9
9
25
36
16
4
140
The null hypothesis
Alternative hypothesis
(ii)
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Calculation of test statistic :
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Awab Sir-89 76 104646
(iii)
Level of significance :
(iv)
Critical value : The value of
at 5% level of significance for
degress of freedom is
(v)
(one tailed test).
Decision : Since the calculated value of
critical value
is less than the
the hypothesis is rejected.
Food B is superior to food A.
(10) Example 10 : Investigate the association between the darkness of eye colour in father
and son from the following data.
(M.U. 2010)
Colour of father’s eyes
Colour of son’s eyes
Dark
Not dark
Total
Dark
48
80
128
Not dark
90
782
872
Total
138
862
1000
Sol. :
(i)
Null Hypothesis :
There is no association between the darkness of
eye colour in father and son.
Alternative Hypothesis
(ii)
There is an association.
Calculation of test statistic : On the basis of this hypothesis the
excepted frequency of dark eyed sons with dark eyed father
Where, A= number of dark eyed fathers (total of first column)
B= number of dark eyed sons (total of first row)
N= total number of observation
Expected frequency
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(This is because if there is no association, Since the ratio of dark eyed father to
the total is 128 / 1000 out of 138 dark eyed sons there will be
dark eyed sons.)
Having obtained the expected frequency in the first cell, Since the totals remain
the same, the figures in other cells can be easily obtained as
.
We thus get the following table.
Colour of father’s eyes
Colour of son’s eyes
Dark
Not dark
Total
Dark
18
110
128
Not dark
120
752
872
Total
138
862
1000
Calculation of
O
E
48
18
900
80
110
900
90
120
900
782
752
900
Total
(iii)
Level of Significance :
Degrees of freedom
(iv)
.
Critical value : For 1 d.f. at5% level of significance the table value of
is
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(v)
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Decision : Since the calculated value of
the table value of
is much greater than
, the hypothesis is rejected.
There is an association between darkness of colour of fathers and sons.
(11) Example 11 : Two batches of 12 animals each are give test inoculation. One batch was
inoculated and the other was not. The number of dead and surviving animals are given in
the following table for both cases. Can the inoculation be regarded as effective against
the disease at 5% level of significance. (Make Yates correction)
Dead
2
8
10
Inoculated
Not – Inoculated
Total
Surviving
10
4
14
Total
12
12
24 (M.U. 2004, 11)
Sol. :
(i)
Null Hypothesis :
There is no association between inoculation and
death.
Alternative Hypothesis
There is association between inoculation an
death.
(ii)
Calculation of test statistic : On the basis of this hypothesis the number
in the first cell
Where, A= Total in the first column,
B= total in the first row,
N= total number of observation
The observation in the first cell
The remaining cell frequencies may be calculated in the same manner or
may be obtained by subtracting this frequency from the row total and column
total. We then apply Yates correction and prepare the table as shown in next page.
(iii)
Level of Significance :
Degrees of freedom
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.
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(iv)
O
Awab Sir-89 76 104646
Critical value : For 1 degree of freedom at5% level of significance the
table value of
is
Calculation of
E
2
10
8
4
(v) Decision : Since the calculated value of
is greater than the
table value of
, the hypothesis is rejected.
There as association between inoculation and death i.e. inoculation is
effective against the disease.
(12) Example 12 : A die was thrown 132 time and the following frequencies were observed.
No. obtained
: 1,
2,
3,
Frequency
: 15, 20, 25,
Test the hypothesis that die is unbiased
Sol. :
(i)
Null Hypothesis :
5,
29,
6.
28.
Total
132
(M.U. 2010)
The die is unbiased.
Alternative Hypothesis
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4,
15,
The die is not unbiased.
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(ii)
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Calculation of test statistic : On the hypothesis that the die in unbiased
we should expect the frequency of each number to be 132 / 6 =22.
Calculation of
No.
O
E
1
15
22
49
2
20
22
4
3
25
22
9
4
15
22
49
5
29
22
49
6
28
22
36
Total
(iii)
196
Level of Significance :
Number of Degrees of freedom
(iv)
Critical value : For 5 degree of freedom at5% level of significance the
table value of
(v)
.
is
Decision : Since the calculated value of
value of
is less than the table
, the hypothesis is accepted.
The die is unbiased.
(13) Example 13 : The number of car accidents in a metropolitan city found 20, 17, 12, 6, 7,
15, 8, 5, 16 and 14 per month respectively. Use - test to check whether these
frequencies are in agreement with the belief that occurrence of accidents was the same
during 10 months period. Test at 5% level of significance. (table value of
)
(M.U.2001)
Sol. :
(i)
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Null Hypothesis :
Accidents occur equally on all months.
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Alternative Hypothesis
Awab Sir-89 76 104646
Accidents do not occur equally on all
months.
(ii)
Calculation of test statistic : On the basis of this hypothesis, the number
of accidents per month = (total) / 10 = (20+17+12+6+7+15+8+5+16+14) /
10 = 120 / 10 = 12.
(iii)
Level of Significance :
Number of Degrees of freedom
(iv)
Critical value : For 9 degree of freedom at5% level of significance the
table value of
(v)
.
is
Decision : Since the calculated value of
table value of
is greater than the
, the hypothesis is rejected.
Accidents do not occur equally on all months.
(14) Example 14 : Theory predicts that the production of beans in the four groups A, B, C, D
should be 9 : 3 : 3 : 1. In an experiment among 1600 beans the number in the four groups
were 882, 313, 287 and 118. Does the experimental result support the theory ?
(M.U. 2001, 06)
Sol. :
(i)
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Null Hypothesis :
The production of the beans in the four groups A,
B, C, D, is the given proportion 9 : 3 : 3 : 1.
Alternative Hypothesis
The proportion is not as given above.
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(ii)
Calculation of test statistic : On the basis of the above hypothesis, Sinec
the sum is 9+3+3+1 = 16, the number of beans in the four groups will be.
(iii)
Level of Significance :
Degrees of freedom
(iv)
Critical value : For 3 degree of freedom at5% level of significance the
table value of
(v)
.
is
Decision : Since the calculated value of
value of
is less than the table
, the null hypothesis is accepted.
The proportion 9 : 3 : 3 : 1 is correct.
(15) Example 15 : The figures given below are (a) the observed frequencies of distribution.
(b) the frequencies of the normal distribution, having the same mean, standard deviation
and the total frequency as in (a).
(a) 1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1.
(b) 2, 15, 66, 210, 484, 799, 943, 799, 484, 210, 66, 15, 2.
Apply
test of goodness of fit.
(M.U. 2004)
Sol. : Since the frequencies at the beginning and end are less than 10, we group them and
then apply the
test.
Calculation of
O
1,
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12, 66,
220,
495,
792,
924,
792,
495,
220, 66, 12, 1
Page 17
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,
E
(i)
66,
210,
484,
Null Hypothesis :
Awab Sir-89 76 104646
799,
943,
799,
484,
210, 66,
The fit is good.
Alternative Hypothesis
The fit is not good.
(ii)
Calculation of test statistic :
(iii)
Level of Significance :
.
Number Degrees of freedom : There are originally 13 classes. Since they
are reduced to 11 by grouping twice, the degrees of freedom is reduced by 2.
Further, since the mean, the standard deviation and the total frequency of original
data are used, three constraint are introduced, reducing the degree of freedom by
3. Fro calculating the mean and the standard deviation three sums
are required. Hence the degree of freedom is reduced by 3.
Thus, the
(iv)
.
Critical value : For 8 degree of freedom at5% level of significance, the
table value of
(v)
Decision : Since the calculated value of
value of
is less than the table
, the hypothesis is accepted.
The fit is good.
(16) Example 16 : the number of defects in printed circuit board is hypothesized to follow
Poisson distribution. A random sample of 60 printed boards showed the following data.
Number of defects
:
0
1
2
3
Observed frequency
:
32
15
9
4
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Does the hypothesis of Poisson distribution seen appropriate.
(M.U. 2004)
Sol. :
(i)
Null Hypothesis :
The defects follow Poisson distribution.
Alternative Hypothesis
The defects do not follow Poisson
distribution.
(ii)
Calculation of test statistic : The expected frequencies of Poisson
distribution are given by
Expected frequencies
Where,
m = mean of the distribution,
x = random variable,
N = number of observations.
Here,
Exp. Freq.
Of zero defects
Of one defects
Of two defects
Of three defects
Calculation of
No. of defects
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O
E
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0
32
1
15
2
9
3
4
Awab Sir-89 76 104646
Total
(iii)
Level of Significance :
Number of Degrees of freedom
.
(The number of degrees of freedom for each class is one. There are originally 4
classes. Hence, the degrees of freedom originally is 4. But we reduced the classes by one, thus,
reducing the degree by one. Further, while calculating the parameter m, we used two sums
, reducing the degree of by two.)
(iv)
Critical value : For 1 degree of freedom at 5% level of significance the
table value of
(v)
is
Decision : Since the calculated value of
value of
is less than the table
, the hypothesis is accepted.
The defects follow Poisson’s distribution.
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