29 APRIL 2013 LECTURE

STATISTICS - HYPOTHESIS TESTINGS

WE all know last week the standard formula for ‘z’ score from big sample is

1 mean test

Z =

Based from that sacred formula:

2 mean test from 2 different samples: Independent Sampling

Z = () where

ONE-TAILED TEST

Rejection region : z < -

OR

[ z > when Ha: ( ]

TWO TAILED TEST

Rejection region: |z| >

Example

A large sample test for 2 different means

A dietician has developed a diet that is low in fats, carbohydrates and cholesterol. Although the diet was initially intended to be used by people with heart problems, the dietician wishes to examine the effect this diet has on the weights of obese people. Two random samples of 100 obese people each are selected , and one group of 100 is placed on the low fat diet. The other 100 are placed on a diet that contains approximately the same quantity of food but is not as low in fats, carbohydrates and cholesterols. For each person, the amount of weight lost in a 3 week period is recorded.

The data is shown below. [ from a book, photo stated Monday 29 th April, 2013]

Form a 95% confidence interval for the difference between the population mean weights losses for the 2 different diets. Interpret the result.

10 marks

Another way to compare the mean weight loss for 2 different diets is to conduct a hypothesis.

Step 1

Either use OR OR

Thus the elements of test are as follows

Ho is NULL HYPOTHESIS and H1 is ALTERNATIVE HYPOTHESIS

Ho: ()=0 -à and D0=0 for this hypothesis test

H1: () 0 -à ( the same way by saying it as )

Now the next step is

Step 2: Pasang Confidence Interval 95% --à + - 1.96

Step 3:

Compute the test statistic z = = ?

Step 4:look at your drawing. Put 95% confidence interval and put your z values clearly on your drawing

Rejection region: z< -= -1.96 or z > = 1.96 [ by looking at your own diagram]

Now substituting all the data from the above table dietician: mean from the first sample is 9.31 and mean from the second sample is 7.40. The standard deviations are also given from the book as s1=4.67; s2=4.04; n1=100; n2=100.

Z = = = 1.91/ 0.617 = 3.09

Step 4:

Now check your drawing

If 3.09 > z ; the computed ‘z’ value intrudes the critical region right?

So reject the Null Hypothesis

Else you accept Ho.

Step 5:

Conclude

Since you rejected Ho, you could write something like this

“..The computed z falls in the critical region right?

Therefore the sample provide sufficient evidence at = 0.05 , for the dietician to conclude that the mean weight losses for the 2 diets really are different types. i.e.,

Exercise 100:

Q1. Study the table below

Sample 1 / Sample 2
N1= 17 / N2= 12
S1= 3.4 / S2= 4.8

a)  Conduct the test Ho:( against H1: (.

b)  Find and interpret the 95% confidence interval for ( ).

**

What is a p-value?

The observed significance level or p-value , for a specific statistical test is the probability (assuming Ho is true) of observing a value of the test statistic that is at least as contradictory to the null hypothesis , and supportive of the alternative hypothesis, as the actual one computed from the sample data.

Drawing is given here/ copy from the whiteboard.

Pg 383- 385

Exercise 200:

Q1. Given mean first sample is 4.50; and mean from the second sample is 5.50; standard deviation first sample 2.34 and standard deviation from second sample is 2.43.

Construct the hypothesis testing as similar to the above example ‘dietician’. What did you get?

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NEW SUB TOPIC

*** Large Sample test of hypothesis about a population proportion.

Proportion is ‘kadar’ In Bahasa Malaysia. Example given 100 pengundi in PRU13. Out of that number 87 are not known whether they are true melayu / cino or India Malaysia. In technical term they are called ‘pengundi hantu’ created by SPR and only SPR knows that through and through. The proportion of hantu here is Proportion of pengundi tulen is

NOW FOCUS LET US SAY:

H0: p = 0.2

H1: p =/= 0.2

Test statistic for proportion

Z =

=

Rejection region is

Z< -z = -1.96 OR

Z > = 1.96

Computing for standard deviation =

Imagine we do breast screening from 140 ladies . It was found only 12 are positives

= 0.086

Computed

Z = = =

Summary of large sample test of hypothesis about ‘p’

One tailed test

Ho : p=po

H1: p < p0 atau p>p0

Test statistic is z =

IF TWO TAILED TEST how?

Ho: p = p0

H1: p =/= p0

Test statistic z =

Rejection region z < - OR z >

EXERCISE 300:

Q3. A manufacturer of alkaline batteries may want to be reasonably certain that fewer than 5% of its batteries are defective. Suppose 300 bateries are randomly selected from a very large shipment: each is tested and 10 defective batteries are found. Does this provide sufficient evidence for the manufacturer to conclude that the fraction defective in the entire shipment is less than 0.05? use

Solution:

Step 1

Ho : p = 0.05

H1: p < - z0.01= -2.33

Step 2

Draw

Step 3

Compute z = = [(10/300) - 0.05] / = 0.03-0.05 / = -1.35

Step 4

Refer to your drawing again plz.

Since our computed z is not in the critical region; we have to accept H0 then

Step 5

Conclude - there is insufficient evidence at the 0.01 level of significance to indicate that the shipment contains less than 5% defective batteries.

THE END.

Good Luck in studying towards 2BIT; 2BIC; 2BIP; 2BIM; 2BIS; 2BIE; ….