LAB 1 DATE: 2080/08/21
TITLE: Implement the Sampling Distribution and Estimation Using SPSS
PROJECT 1.1 (FOR UNGROUPED DATA):
Enter the following vales in SPSS and calculate mean, standard deviation, range, mode and
median:
Weights: 25, 35, 45, 55, 65, 75
WORKING EXPRESSION:
Mean ( x ̅ ) =
∑x
N
√
2
( )
Standard Deviation (σ) = Σ x−μ
N
Range = Maximum- Minimum
Mode = Data with the highest frequency
n+1 th item
2
Median =
PROCEDURE:
1. Select Analyze Descriptive Statistics Frequencies.
2. Click Frequencies Move Weights into Variable(s).
3. Click Statistics. Select Mean, Median, Mode, Range and standard deviation.
4. Click Continue OK.
CALCULATION (FROM SPSS):
Weights
Valid 6
N
Missing 0
Mean 50.0000
Median 50.0000
Mode 25.00a
Std. Deviation 18.70829
Range 50.00
CONCLUSION:
The average weight is found to be 50 units and the median is also same as that of the average
weight.
Since multiple mode exist, the smallest weight is taken i.e., 25 units.
The provided weights were deviated by 18.70829.
The difference between maximum and minimum is 50 units.
PROJECT 1.2 (FOR UNGROUPED DATA):
Enter the following values in SPSS and calculate mean, standard deviation, range, mode and
median:
Weight (x) Mid-value (m) Frequency (f)
20-30 25 4
30-40 35 6
40-50 45 7
50-60 55 21
60-70 65 23
70-80 75 2
WORKING EXPRESSION:
Mean ( x ̅ ) =
∑x
N
√
2
( )
Standard Deviation (σ) = Σ x−μ
N
Range = Mid-Value of the Highest-Class Interval – Mid-Value of the Lowest Class
Interval
Mode = L + ( f 1−fm
)
2 f 1−fm−f 2
×h
Median =
( n+12 )
PROCEDURE:
1. Enter the Data Editor Window.
2. Select Data Weight Cases.
3. Move Frequency into Frequency Variable.
4. Click Ok. Select Analyze Descriptive Statistics Frequencies
5. Click the frequencies Move MidValue into Variable(s)
6. Click the Statistics. Select Mean, Standard Deviation, Range, Mode and Median.
7. Click Continue. Click Ok.
CALCULATION (FROM SPSS):
MidValue
Cumulative
Frequency Percent Valid Percent Percent
Valid 25.00 4 6.3 6.3 6.3
35.00 6 9.5 9.5 15.9
45.00 7 11.1 11.1 27.0
55.00 21 33.3 33.3 60.3
65.00 23 36.5 36.5 96.8
75.00 2 3.2 3.2 100.0
Total 63 100.0 100.0
MidValue
N Valid 63 63
Missing 0 0
Mean 54.3651
Median 55.0000
Mode 65.00
Std. Deviation 12.55607
Range 50.00
CONCLUSION:
The average weight is found to be 54.3651 units and the median is 55 units.
The weight with highest frequency is 65 units.
The provided weights were deviated by 12.55607.
The difference between maximum and minimum is 50 units.
PROJECT 1.3 (CONFIDENCE INTERVAL FOR POPULATION MEAN (µ) ):
Enter the following values in SPSS and create a confidence interval assuming normal
distribution:
Length: 125, 120, 121, 123, 122, 130, 124, 122, 120, 122, 118, 119, 123, 124, 122, 124, 121,
122, 138, 149, 123, 128, 122, 130, 120, 122, 124, 134, 137, 128, 122, 121, 125, 120, 13 2,
130, 122, 124
WORKING EXPRESSION:
Confidence interval for population mean µ is given by,
x ± z α S.E. ( x )
Where,
( x ) : Sample Mean
z α : Z- value for α level of significance
S.E. ( x ): Standard error for mean
PROCEDURE:
1. Enter the data
2. Select Analyze Compare Means One Sample T Test
3. Click Options Type 95% Confidence Interval
4. Click Continue OK.
CALCULATION FROM SPSS:
One-Sample Statistics
N Mean Std. Deviation Std. Error Mean
Length 40 125.2750 6.14770 .97204
One-Sample Test
Test Value = 0
95% Confidence Interval of
Mean the Difference
t df Sig. (2-tailed) Difference Lower Upper
Length 128.879 39 .000 125.27500 123.3089 127.2411
CONCLUSION:
Hence, using SPSS it was found that the confidence interval of mean of the population is
123.3089 to 127.2411.
LAB 2 DATE: 2080/03/19
TITLE: Perform testing of hypothesis
PROJECT 2.1 (HYPOTHESIS TESTING OF SINGLE MEAN OF LARGE SAMPLE):
The following values are the lengths of 40 steel rods selected for lab test from a factory:
Length: 125, 120, 121, 123, 122, 130, 124, 122, 120, 122, 118, 119, 123, 124, 122, 124, 121,
122, 138, 149, 123, 128, 122, 130, 120, 122, 124, 134, 137, 128, 122, 121, 125, 120, 13 2,
130, 122, 124
Test whether this sample of size 40 has come from a population whose mean length is 125
cm.
PROCEDURE:
1. Enter the data in the data editor.
2. Select Analyze Compare Means One Sample T test. Type in Test Value box.
3. Click Options Type 95 in confidence interval percentage box.
4. Click on Continue and then Ok.
SOLUTION:
STEP I:
Null hypothesis: µ = 125 cm
i.e., there is no significance difference between the sample mean and the population mean.
STEP II:
Alternative Hypothesis: µ ≠ 125 cm
i.e., there is significance difference between the sample mean and the population mean.
STEP III:
Test Statistics is given by:
= σ
x−μ
√n
CALCULATION OF TEST STATISTICS FROM SPSS:
One-Sample Statistics
N Mean Std. Deviation Std. Error Mean
X 40 125.2750 6.14770 .97204
One-Sample Test
Test Value = 125
95% Confidence Interval of
Mean the Difference
t df Sig. (2-tailed) Difference Lower Upper
X .283 39 .779 .27500 -1.6911 2.2411
So, P value = 0.779
STEP IV:
Given level of significance,
ɑ = 5% = 0.05
STEP V:
Decision:
Since P > ɑ, accept and reject
CONCLUSION:
Hence, there is no significance difference between the sample mean and the population mean.
PROJECT 2.2 (HYPOTHESIS TESTING BETWEEN TWO POPULATION MEANS
FOR MATCHED PAIRED):
The sales of a product of a company after and before advertisement are as follows:
Month 1 2 3 4 5 6
Before X 120 140 160 140 180 190
After X 200 210 150 200 220 240
Is advertisement effective at 5%?
PROCEDURE:
1. Enter the data into data editor.
2. Select Analyze Compare Means Paired-Sample T test.
3. Click Options Select Desired Options Continue Ok
SOLUTION:
STEP I:
Null hypothesis: µ1 = µ2
i.e., there is no significance difference between the mean before and after advertisement.
STEP II:
Alternative Hypothesis: µ1 ≠ µ2
i.e., there is significance difference between the mean before and after advertisement.
STEP III:
Test Statistics is given by:
x 1−x 2
√
2 2
= σ1 σ
+¿ 2 ¿
n1 n2
CALCULATION OF TEST STATISTICS FROM SPSS:
Paired Samples Statistics
Mean N Std. Deviation Std. Error Mean
Pair 1 Before 155.0000 6 26.64583 10.87811
After 203.3333 6 30.11091 12.29273
Paired Samples Correlations
N Correlation Sig.
Pair 1 Before_X & After_X 6 .374 .465
Paired Samples Test
Paired
Differences
95%
Confidence
Interval of the
Std. Std. Difference
t
Mean Deviatio Error Sig. (2-
Mean Lower Upper tailed)
n
Pai Before_X - 31.8852 13.0170 - - - .014
r1 - After_X 48.3333 1 8 81.7948 14.8718 3.713
3 1 6
So, P value = 0.014
STEP IV:
Given level of significance,
ɑ = 5% = 0.05
STEP V:
Decision:
Since P < ɑ, accept and reject
CONCLUSION:
Since there is significance difference between the means before and after the advertisement,
the advertisement is proved to be effective.
PROJECT 2.3 (HYPOTHESIS TESTING WHEN RAW DATA FOR INDPENDENT
SAMPLE IS GIVEN):
The monthly advertising cost of a company for two products X and Y were as follows during
6-month period:
Month 1 2 3 4 5 6 7
Cost I (X) 220 240 160 240 280 290 -
Cost II (Y) 100 110 150 100 120 140 145
Is there sufficient evidence to conclude that average cost on advertising on product Y is more
than on product X?
PROCEDURE:
1. Enter the data into Data editor.
2. Select Analyze Compare Means Independent Samples T Test.
3. Move value into Test Variable(s) and type into grouping variable.
4. Click define groups and type 1 and 2 into group 2.
5. Click Options Continue Ok.
SOLUTION:
STEP I:
Null hypothesis: µ = µ
i.e., there is no significance difference between the average cost of advertising on products X
and Y.
STEP II:
Alternative Hypothesis: µ < µ
i.e., the average cost of advertising on product Y is more than that on product X.
STEP III:
Test Statistics is given by:
x 1−x 2
√
2 2
= s1 s2
+
n1 n 1
CALCULATION OF TEST STATISTICS FROM SPSS:
Group Statistics
Type N Mean Std. Deviation Std. Error Mean
Cost Cost_I(X) 6 238.3333 46.65476 19.04673
Cost_II(Y) 7 123.5714 21.35304 8.07069
Independent Samples Test
Leven’s Test t-test for Equality of Means
for Equality
of Variances
F Sig. t df Sig. Mean Std. Error 95% confidence Interval
(2- Difference Difference
of the Difference
tailed)
Lower Upper
Equal 1.357 .269 5.862 11 .000 114.76190 19.57600 71.67541 157.84840
Variances
Cost
assumed
Equal 5.548 6.775 .001 114.76190 20.68608 65.51535 164.00846
Variances
not
assumed
So, P value= 0.269
STEP IV:
Given level of significance,
ɑ = 5% = 0.05
STEP V:
Decision:
Since P > ɑ, accept and reject
CONCLUSION:
There is no significance difference between the average cost of advertising on product X and
product Y.
