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# Testing Hypotheses - PowerPoint PPT Presentation

Testing Hypotheses. Hypothesis about a mean. Example

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Example

A drug company develops a new cholesterol lowering drug that they think is more effective than Loopitor. Suppose it is known that a 10 mg dose of Loopitor lowers cholesterol levels by an average of 43 ppm, with a standard deviation of 16 ppm. The company plans to conduct a test to determine if its new drug is more effective than Loopitor at lowering cholesterol levels. The company will assume that the standard deviation will be the same in the population of their patients as it is for the users of Loopitor.

Null Hypothesis – mean value for the new drug is the same as for the old --  = 43 ppm

Null Hypothesis is true

Null Hypothesis is false

Null hypothesis is accepted

Correct Decision

Type II error

Null hypothesis is rejected

Type I error

Correct Decision

The burden of proof is placed upon the new drug demonstrating that it has a high probability of being superior to the existing product.

Let the probability of making a type I error (rejecting the null hypothesis when it is true) be  where

 is typically taken to be 0.05 or 0.01

The alternate hypothesis (H1) is  > 0 = 43 -- the new drug is more effective than Loopitor

Sampling distribution – assuming  = 0 and x’ = /n

The one-tail test

H0: new drug mean same as Loopitor mean --  = 0 = 43

H1: new drug mean > 0

Choose  = 0.05 – probability of Type I error(rejecting H0 when true)

Assume standard deviation of new population same as for Loopitor users

Region of rejection

Region of acceptance

 = 0.05

E = z/2(/n)

In the experiment that we have designed, we want to put the burden of proof on the new drug to demonstrate with high probability that it is superior to Loopitor.

We make rejection of the null hypothesis to occur with very low probability, if in fact the null hypothesis (no difference) is true.

On the other hand, the new drug will be a significant money maker if a 10 mg dose of it lowers cholesterol by an average of 53 ppm or more (10 ppm more than Loopitor), and we want the probability of making a type II error – not rejecting the null hypothesis when the true population mean is greater than or equal to 53 – to also be low.

x’

0 = 43

1 = 53

Probability of type I error

Probability of type II error

The One-tail Test

We superimpose the sampling distributions with population means of 0 = 43 and 1 = 53 with the same standard deviation /n below

Region of acceptance of H0

Region of rejection of H0

E0

E1

x’

0 = 43

1 = 53

Probability of type II error

The Power of the Test

The power of the test is defined to be 1 - , where  is the probability of making a type II error when the population mean is greater than or equal to 53 for this example.

We can increase the power of the test, 1 - , by increasing the sample size n, thus decreasing the standard deviation of the sampling distribution /n

Region of acceptance of H0

Let’s assume we select , the probability of rejecting the null hypothesis H0 when it is true to equal 0.05, and  the probability of accepting H0 when the population mean is greater than or equal to 53 to be 0.10. The power of the test will then be 1 -  = 0.90.

We then have from the previous slide

E0 = z/n and E1 = z/n

and E0 + E1 = 53 – 43 = 10

Where z = 1.645 and z = 1.28 for the above values of  and 

E0 + E1 = z/n + z/n = 10 solving for n we get

n = (1.645 + 1.28)2(16)2/100 = 21.90  sample size of 22 or more

One-tail test with  unknown

If the population standard deviation is unknown, use the sample statistic s to estimate  and use the t-distribution and proceed as before

If in the previous example, the population standard deviation were unknown, and s = 16 was an estimate of the population parameter, with n = 30 we would have

E0 = t(29 df)s/n = 4.96

Where t = 1.699 and the region of acceptance is x’ < 47.96

 = 0.05

x’R

One-tail test with  unknown

With the population standard deviation unknown, we use the sample standard deviation to estimate the population parameter, and use the t-tables to find t with 29 degrees of freedom to find a value such that  = 0.05 lies to the right of t in a standardized t-distribution.

Region of acceptance

x’

E = t(29 df)s/n = 4.96 and x’R = 43 + 4.96 = 47.96

Accept H0 if the sample mean, x’, is less than 47.96

Inferences about 2 means -- 1 and 2 known

Suppose we want to test the effectiveness of a drug. We would select a random sample from some population and then randomly assign different individuals from this initial sample to either receive the drug or a placebo. Effectively this gives us two separate randomly selected samples of size n1 and n2.

To test the effectiveness of the drug we will determine the means from the two samples and use the difference in these two means to test the effectiveness of the drug.

H0: 1 = 2 the new drug is no better than the placebo

H1: 1 < 2 the new drug reduces some effect

Inferences about 2 means -- 1 and 2 known

To test the null hypothesis we will use the following statistic:

z = ((x’1 – x’2) – (1 - 2))/(12 /n1 + 22 /n2)

This statistic has standard normal distribution with mean 0 (assuming 1 - 2 = 0) and standard deviation 1. To test at the 5% significance level, we get from the tables or from invNorm(0.05) = -1.645 the critical value.

We will accept H0 if the sample data gives a value of z > -1.645

And accept the alternative that the drug is effective only if the z value falls below this critical value.

Inferences about 2 means -- 1 and 2 unknown

Case 1: 1 and 2 unknown and assumed different

• In this case the experimenter must collect from each population

• The two sample sizes n1 and n2

• The two sample means x’1 and x’2

• The two sample standard deviations s1 and s2

The statistic used to test the null hypothesis will now have a t-distribution with number of degrees of freedom 1 less than the smaller of n1 and n2

t = ((x’1 – x’2) – (1 - 2))/(s12 /n1 + s22 /n2)

Inferences about 2 means -- 1 and 2 unknown

Case 2: 1 and 2 unknown but assumed equal

If we can assume that the standard deviations from the two populations from which the samples were drawn are the same, we can pool the two sample standard deviations to get a better estimate with a larger number of degrees of freedom.

Sp2 = ((n1 – 1)s12 + (n2 – 1)s22 )/((n1 – 1) + (n2 – 1))

With df = n1 + n2 – 2

The test statistic is now

t = ((x’1 – x’2) – (1 - 2))/(sp2 /n1 + sp2 /n2)

• We have examined 3 different cases:

• 1 and 2 known

• 1 and 2 unknown and assumed different

• 1 and 2 unknown but assumed the same

In the first case the test statistic is

z = ((x’1 – x’2) – (1 - 2))/(12 /n1 + 22 /n2)

And it has standard normal distribution

In the second and third cases the test statistic has t-distribution with the population standard deviations in the formula above replaced by the sample standard deviations s12 and s22

In the third case the standard deviations from the two samples are pooled to get a larger number of degrees of freedom and hence a tighter bound on the region of acceptance.

You must first read the problem to determine which of the three cases applies, then extract the important information:

n1 and n2, x’1 and x’2, and s1 and s2 or 1 and 2

Next formulate the hypotheses that you want to test

H0:1 = 2

H1: 1 < 2 //one tail test (one value is “better” than the second)

or

H1: 1  2 //two tail test (the two means are not equal)

Next step—determine the significance level for your test

Significance level  is the probability of making a type I error

• Next find the critical value – for accepting H0

• For case 1 zc = invNorm() for 1 < 2, invNorm(1-) for 1 > 2, invNorm(1 - /2) for 1  2

For cases 2 and 3 the critical value for the t-statistic is similarly found in the t-tables

Lastly plug the data from the two samples into the test statistic formula for the appropriate case, and determine if the value obtained falls within the region of acceptance for the critical value that you obtained from the tables above.