Support the Monkey! Tell All your Friends and Teachers
 Home MonkeyNotes Printable Notes Digital Library Study Guides Study Smart Parents Tips College Planning Test Prep Fun Zone Help / FAQ How to Cite Request a New Title

For n = 10 - 9 = 9, t 0.05 = 2.57. Therefore , 0.71 < 2.57.

The null hypothesis is accepted i.e. the testing is reliable.

Example Boys from a certain class were given a test in statistics. They were given a six month’s further extra coaching and a second test of an equal level was held. Do the marks give evidence that the students have benefitted by the extra coaching ?

Boys                   1     2   3    4    5    6   7    8   9    10   11

Marks (1st test) 23  20  19  21  18  20  18  18  23  16  19

Marks (2nd test)24  19  22  18  20  22  20  20  23  20  17

 Your browser does not support the IFRAME tag.

Solution: The null hypothesis Ho : m1 - m2 or Ho : m1 - m2 = 0 i.e. D = 0

alternative hypothesis Ha : m1 ¹ m2 or Ha : m1 - m2 ¹ 0

The number of degree of freedom is 11 - 1 = 10

t0.05, n = 10 = 2.228.

Thus, 1.482 < 2.228 i.e. the calculated value of t - it considerably less than the table value. Hence we accept null hypothesis Ho, that the marks do not give any evidence that the boys have been benefitted by the extra coaching.

Example A dietitian decided to try out a new type of diet program on 10 boys. Before applying this program, he recorded their weights. Three months later, he recorded their weights again. The weight would have grown by an average of 6 kgs. during this period even without this special diet program. Did this special diet program help ? Use 0.05 level of significance.

Boys               1     2     3     4     5     6     7     8     9     10

Before weight 36  32   31   36   23    28    25   26   35   28

After weight    45  37   39   45   31   34    29   36    42   35

Solution: The null hypothesis Ho : m1 ¹ m2 or Ho : m1 - m2 ¹ 0

or more clearly Ho : m1 - m2 i.e. D £ 6

alternative hypothesis Ha : D > 6

The degree of freedom n (df) = 10 - 1 = 9. The test is one-tailed because we are asked only whether the new diet program increases the weight and not whether it reduces it.

Now t0.05, n = 9 = 2.26 ( i,e. t0.025, n = 9 = 2.25 for one tailed test) Because 2.87 > 2.26. The null hypothesis can be rejected. Thus the test has provided evidence that the new diet program caused the weight increase in boys. Though the amount of actual increase was not large ( 1.6 kgs over normal weight) but is was statistically significant.

Index

8.1 Population
8.2 Sample
8.3 Parameters and Statistic
8.4 Sampling Distribution
8.5 Sampling Error
8.6 Central Limit Theorem
8.7 Critical Region
8.8 Testing of Hypothesis
8.9 Errors in Tesitng of Hypothesis
8.10 Power o a Hypothesis Test
8.11 Sampling of Variables
8.12 Sampling of Attributes
8.13 Estimation
8.14 Testing the Difference Between Means
8.15 Test for Difference Between Proportions
8.16 Two Tailed and one Tailed Tests
8.17 Test of Significance for Small Samples
8.18 Students t-distribution
8.19 Distribution of 't' for Comparison of Two Samples Means Independent Samples
8.20 Testing Difference Between Mens of Two Samples Dependent Samples or Matched Paired Observations
8.21 Chi-Square
8.22 Sampling Theory of Correlation
8.23 Sampling Theory of Regression

Chapter 1