![An automatic bottling machine fills coke into two liter (2000 cc) bottles. A consumer advocate wants to test the
null hypothesis that the average amount filled by the machine into a bottle is at least 2000 cc. A random sample of
40 bottles coming out of the machine was selected and the exact content of the selected bottles are recorded. The
sample mean was 1999.6 cc. The population standard deviation is known from past experience to be 1.30 cc.
Test the null hypothesis at the 5% significance level.
H0: 2000
H1: 2000
n = 40
For = 0.05, the critical value
of z is 1.645
The test statistic is:
Accept H0 if: [z≤ 1.645]
Reject H0 if: z >]
0
n
x
z
0
H
Reject
95
.
1
|
1.95
|
=
=
0
|
|
1.3
=
1999.6
=
x
40
=
n
40
1.3
2000
-
1999.6
n
x
z
EXAMPLE](https://image.slidesharecdn.com/examplesintest-250123032424-9dc1746b/85/examples-in-test-pptx-H-V-VM-CCFFFFFFFXF-5-320.jpg)
examples in test.pptx H V VM CCFFFFFFFXF
- 1. EXAMPLES-TEST OF SIGNIFICANCE Dr. Karuna.M Dept. of Public Health Science KIUniversity
- 2. The general formula for z is: The Z is a standard normal distribution If Z=1.96, then area under right and left side (two sided) of it will be 5% (0.05) Testing of single sample Mean Proportion
- 3. COMPARISON OF MEAN VALUES Single group study -To compare with established result Two groups -Dependent sample (single group Pre- post design) -Independent samples Multiple group
- 4. SINGLE SAMPLE TEST H0: Sample mean=Population mean Large sample Small sample ~
- 5. An automatic bottling machine fills coke into two liter (2000 cc) bottles. A consumer advocate wants to test the null hypothesis that the average amount filled by the machine into a bottle is at least 2000 cc. A random sample of 40 bottles coming out of the machine was selected and the exact content of the selected bottles are recorded. The sample mean was 1999.6 cc. The population standard deviation is known from past experience to be 1.30 cc. Test the null hypothesis at the 5% significance level. H0: 2000 H1: 2000 n = 40 For = 0.05, the critical value of z is 1.645 The test statistic is: Accept H0 if: [z≤ 1.645] Reject H0 if: z >] 0 n x z 0 H Reject 95 . 1 | 1.95 | = = 0 | | 1.3 = 1999.6 = x 40 = n 40 1.3 2000 - 1999.6 n x z EXAMPLE
- 6. Example Study conducted to measure the psychological distress among 25 street children in a city. The mean GHQ-28 was 19.57 ± 5.76 with range (8-28). It is known that from previous study the GHQ score of these children is 18.8. Is there any difference in mean GHQ score of the present study with earlier one? H0: µ=18.8 Ha:µ≠18.8 ~ P>0.05 Not significant (i.e. No difference)
- 7. COMPARISON OF TWO GROUPS(MEANS) It is natural to use of Z statistic to measure the difference between the two means: But for small sample
- 8. EXAMPLE: Whether any significant difference in HAM-D score between two group of patients with depression Unpaired Data – Randomly select Group1 : 10 depression patients Group 2: 10 depression patients with anxiety Observe the magnitude of the HAM-D score Using two-sample t-test
- 9. EXAMPLE… Drug 1: 17.9 ± 4.3 Drug 2: 20.2 ± 5.9 Substituting the values in the test statistic =0.9962 Table t value 18 d.f = 2.101, hence P=0.332 (i.e. P>0.05) Inference: There is no significant difference between group 1 and group 2
- 10. ASSUMPTIONS FOR T TEST Estimation of one variable in two groups Groups are independent i.e. nothing in group helped to determine who was in the second group (unrelated samples) If groups are related, then to compare differences, a test called a paired t test can be used Variables must be measured at the interval or ratio level
- 11. T-TEST Assumptions cont… Homogeneity of variance- variable is equally distributed in the over all population Normal distribution of observations Independence of observations Population variance is not known but are assumed to be equal Random sampling has been used
- 12. COMPARISON OF MEAN (PAIRED SAMPLE T-TEST) Sometimes, the data are collected in pairs. In that case, it may be more appropriate to use paired t- test to test the change from before to after.
- 13. PAIRED DATA One set of individuals or objects are taken Two observations made on each individual (Before & after) In this data the post observations depends on pre observations They always paired, hence the no. of observation on pre and post are equal.
- 14. PAIRED T-TEST Candidates IQ before(X) IQ after(Y) X-Y (X-Y)2 1 110 120 -10 100 2 120 118 2 4 3 123 125 -2 4 4 132 136 -4 16 5 125 121 4 16 Assume that the data consists of n subjects measured pre and post as pairs (X1, Y1), (X2, Y2), …,(Xn, Yn) •An IQ test was administered to find persons before and after they trained. The result are given below: Test whether there is any change in IQ after the training programme. Under Ho mean change
- 15. Let Di=Xi -Yi , then mean difference Testing Test statistic: Substituting the values Level of significance: α=0.01 Table value=4.6 at 4df We concluded that there is no change in the IQ after training programme.
- 16. THANK YOU