“Hypothesis Testing for Population Mean”

Arun Kumar, Ravindra Gokhale, and NagarajanKrishnamurthy

Quantitative Techniques-I, Term I, 2012Indian Institute of Management Indore

Case: Quality Wireless B

Daily hold time at the service call center has mean 79.50seconds and standard deviation 16.86 seconds.

Performance over ten day in Ray Jackson’s absence:Average hold time is 86.6 seconds.

Performance over ten day in Ray Jackson’s presence:Average hold time is 74.4 seconds.

Jackson’s Management Approach?

Not accounting for process variation.

Jackson’s Management Approach?

Not accounting for process variation.

How to make more informed decision?

To ascertain whether 74.4 and 86.6 seconds wait time, whichis different than known average wait time of 79.5, can beattributed to the natural process variation.

Statistical problem?

To ascertain whether wait time of 74.4 seconds (or 86.6seconds) is an outlier even when process variation is taken intoaccount.

Statistical problem?

To ascertain whether wait time of 74.4 seconds (or 86.6seconds) is an outlier even when process variation is taken intoaccount.

Framing a Hypothesis

Null Hypothesis

Stating what would be considered normal if no changehappened.H0 : µ = 79.5

Alternative Hypothesis

Stating what would happen when a change happens.HA : µ 6= 79.5

Framing a Hypothesis

Null Hypothesis

Stating what would be considered normal if no changehappened.H0 : µ = 79.5

Alternative Hypothesis

Stating what would happen when a change happens.HA : µ 6= 79.5

What conclusions we make?

We fail to reject the null hypothesis.

We reject the null hypothesis.

What conclusions we make?

We fail to reject the null hypothesis.

We reject the null hypothesis.

Two types of error

ConclusionNot Reject Reject

Truth Not Reject OK Type I Error (α)Reject Type II error (β) OK

Type I and Type II error

You can not control both simulatenously.

Type I and Type II error have inverse relationship for a fixedsample size.

Type I and Type II error

You can not control both simulatenously.

Type I and Type II error have inverse relationship for a fixedsample size.

Controlling Type II error

Effect of sample size on Type II error

Increasing the sample size reduces Type II error keeping Type Ierror constant. Of course, you can reduce both type I and typeII error to a desirable level by having a sufficiently large samplesize but cost will be a factor.

Power of a test

Reduction in β implies that 1− β increases. 1− β is alsoknown as the power of the test.

Be aware of tests that do not have much power.

Controlling Type II error

Effect of sample size on Type II error

Increasing the sample size reduces Type II error keeping Type Ierror constant. Of course, you can reduce both type I and typeII error to a desirable level by having a sufficiently large samplesize but cost will be a factor.

Power of a test

Reduction in β implies that 1− β increases. 1− β is alsoknown as the power of the test.

Be aware of tests that do not have much power.

Controlling Type II error

Effect of sample size on Type II error

Increasing the sample size reduces Type II error keeping Type Ierror constant. Of course, you can reduce both type I and typeII error to a desirable level by having a sufficiently large samplesize but cost will be a factor.

Power of a test

Reduction in β implies that 1− β increases. 1− β is alsoknown as the power of the test.

Be aware of tests that do not have much power.

Significance Level

Acceptable value of Type I error is also known as significancelevel.

The best practice is to set the significance level before data iscollected.

Significance Level

Acceptable value of Type I error is also known as significancelevel.The best practice is to set the significance level before data iscollected.

What significance level you want to set for Quality

Wireless Example?

5% and 1% are the commonly used levels of significance.

What significance level you want to set for Quality

Wireless Example?

5% and 1% are the commonly used levels of significance.

How to conduct the test?

Find out the test statistic.

Find out the critical value.

If the absolute value of the test statistic is greater thanthe absolute value of the critical value then we reject thehypothesis. Otherwise we conclude that the data doesnot provide sufficient evidence to reject the hypothesis.

How to conduct the test?

Find out the test statistic.

Find out the critical value.

If the absolute value of the test statistic is greater thanthe absolute value of the critical value then we reject thehypothesis. Otherwise we conclude that the data doesnot provide sufficient evidence to reject the hypothesis.

How to conduct the test?

Find out the test statistic.

Find out the critical value.

If the absolute value of the test statistic is greater thanthe absolute value of the critical value then we reject thehypothesis. Otherwise we conclude that the data doesnot provide sufficient evidence to reject the hypothesis.

How to conduct the test?

Find out the test statistic.

Find out the critical value.

If the absolute value of the test statistic is greater thanthe absolute value of the critical value then we reject thehypothesis. Otherwise we conclude that the data doesnot provide sufficient evidence to reject the hypothesis.

Test Statistic for µ

z =x̄ − µ0

σ/√n

µ0 is the hypothesized value for µ.

Test statistic in Quality Wireless case

z = (74.4−79.5)

(16.86/√

10)= −0.95

Critical value

Critical value is Zα/2, where α is the significance level.

Quality wireless case

For α = 0.05, critical value is 1.96. We fail to reject the nullhypothesis because | − 0.95| < 1.96.

Conclusion from the statistical analysis

At significance level of 5%, we fail to reject the null hypothesisthat the performace of the call center was significantlydifferent than 79.5 seconds.

Conclusions

System was not out of control.

What if significance level is changed?

Conclusions

System was not out of control.

What if significance level is changed?

Conclusions

System was not out of control.

What if significance level is changed?

Conclusions

System was not out of control.

What if significance level is changed?

Exercise 1

A Vice President in charge of sales for a large corporationclaims that salesperson are averaging no more than 15 salesper week. Ideally the number should be higher. In order tocheck the claim, 36 salespersons are selected at random, andthe number of contacts made by each is recorded for a week.The mean and the variance of the 36 measurements were 17and 9, respectively. Does the evidence contradict the vicepresidents’ claim? Use a test with level α = 0.05.

p-value

Probability of observing a data that is more extreme than theobserved data is known as p-value.We reject the null hypothesis if p-value < significance level.

Calculating the p-value

p-value is 2 ∗ P(Z ≥ |z |) when alternative is 6=. Z is thestandard normal random variable and z is the test statistic.

Quality Wireless Example

p-value=2*P(Z ≥ | − 0.95|)=0.34.You will fail to reject the Null Hypothesis at both 1% and 5%significance level because p-value is much bigger than 0.01and 0.05.

Exercise 1...

A Vice President in charge of sales for a large corporationclaims that salesperson are averaging no more than 15 salesper week. Ideally the number should be higher. In order tocheck the claim, 36 salespersons are selected at random, andthe number of contacts made by each is recorded for a week.The mean and the variance of the 36 measurements were 17and 9, respectively. Does the evidence contradict the vicepresidents’ claim? Answer the question using p-value.

Understanding the z-test

Under null hypothesis, X̄n−µ0

σ/√n∼ N(0, 1) as long as n is

large.

Now if significance level is α then any x̄ (sample mean)that is more than zα/2 or less than −zα/2 providesevidence against µ = µ0.

What if σ is unknown?Ans. Old friend t-distribution will help.

Understanding the z-test

Under null hypothesis, X̄n−µ0

σ/√n∼ N(0, 1) as long as n is

large.

Now if significance level is α then any x̄ (sample mean)that is more than zα/2 or less than −zα/2 providesevidence against µ = µ0.

What if σ is unknown?Ans. Old friend t-distribution will help.

Understanding the z-test

Under null hypothesis, X̄n−µ0

σ/√n∼ N(0, 1) as long as n is

large.

Now if significance level is α then any x̄ (sample mean)that is more than zα/2 or less than −zα/2 providesevidence against µ = µ0.

What if σ is unknown?Ans. Old friend t-distribution will help.

Understanding the z-test

Under null hypothesis, X̄n−µ0

σ/√n∼ N(0, 1) as long as n is

large.

Now if significance level is α then any x̄ (sample mean)that is more than zα/2 or less than −zα/2 providesevidence against µ = µ0.

What if σ is unknown?

Ans. Old friend t-distribution will help.

Understanding the z-test

Under null hypothesis, X̄n−µ0

σ/√n∼ N(0, 1) as long as n is

large.

Now if significance level is α then any x̄ (sample mean)that is more than zα/2 or less than −zα/2 providesevidence against µ = µ0.

What if σ is unknown?Ans. Old friend t-distribution will help.

t-test

Under null hypothesis, X̄n−µ0

s/√n∼ tn−1 for any n.

Now if significance level is α then any x̄ (sample mean)that is more than t(α/2,n−1) or less than −t(α/2,n−1)

provides evidence against µ = µ0.

t-test

Under null hypothesis, X̄n−µ0

s/√n∼ tn−1 for any n.

Now if significance level is α then any x̄ (sample mean)that is more than t(α/2,n−1) or less than −t(α/2,n−1)

provides evidence against µ = µ0.

Remembering Quality Wireless example

Test statistic in Quality Wireless case

Assuming s = 16.86, z = (74.4−79.5)

(16.86/√

10)= −0.95

Quality wireless case: critical value using normal

distribution

For α = 0.05, critical value is 1.96. We fail to reject the nullhypothesis because | − 0.95| < 1.96.

Quality wireless case: critical value using

t-distribution

For α = 0.05, critical value is t(α/2,9) = 2.262. We fail toreject the null hypothesis because | − 0.95| < 2.262.

Quality wireless case: What is the p-value using

t-distribution?

p-value>0.2

Exercise 1...

A Vice President in charge of sales for a large corporationclaims that salesperson are averaging no more than 15 salesper week. Ideally the number should be higher. In order tocheck the claim, 36 salespersons are selected at random, andthe number of contacts made by each is recorded for a week.The mean and the variance of the 36 measurements were 17and 9, respectively. Does the evidence contradict the vicepresidents’ claim? Calculate p-value using t-distribution.