# Hypothesis Testing Chapter 10.

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Hypothesis Testing Chapter 10

Chapter Goals When you have completed this chapter, you will be able to: Define null and alternative hypothesis and hypothesis testing Define Type I and Type II errors Describe the five-step hypothesis testing procedure Distinguish between a one-tailed and a two-tailed test of hypothesis and...

10 Chapter Goals Conduct a test of hypothesis about a population mean
Conduct a test of hypothesis about a population proportion Explain the relationship between hypothesis testing and confidence interval estimation Compute the probability of a Type II error, and power of a test

Terminology Hypothesis
…is a statement about a population distribution such that: (i) it is either true or false, but never both, and (ii) with full knowledge of the population data, it is possible to identify, with certainty, whether it is true or false. …the mean monthly income for all systems analysts is \$3569. Examples …35% of all customers buying coffee at Tim Horton’s return within a week.

Terminology Alternative Hypothesis H1 Null Hypothesis Ho
…is the statement that we are interested in proving . It is usually a research hypothesis. Null Hypothesis Ho …is the complement of the alternative hypothesis. We accept the null hypothesis as the default hypothesis. It is not rejected unless there is convincing sample evidence against it. Hypothesis Testing Steps

State the null and alternate hypotheses
Hypothesis Testing Step 1 State the null and alternate hypotheses Step 2 Select the level of significance Step 3 Identify the test statistic State the decision rule Step 4 Step 5 Compute the value of the test statistic and make a decision Do NOT reject H0 Reject H0 and accept H1

Keep in Mind When a decision is based on analysis of sample data and not the entire population data, it is not possible to make a correct decision all the time. Our objective is to try to keep the probability of making a wrong decision as small as possible!

But do the courts always make the “right” decision?
Two kinds of errors Let’s look at the Canadian legal system for an analogy... Two hypotheses: 1. …the accused person is innocent 2. …the accused person is guilty After hearing from both the prosecution and the defence, a decision is made, declaring the accused either: Innocent! But do the courts always make the “right” decision? or Guilty!

Person is declared “guilty” Person is declared ’not guilty’
Two kinds of errors Reality Court Decision Person is declared “guilty” Person is declared ’not guilty’ Person is “innocent” Error Correct Decision Type I Error H0 is true Person is “guilty” Error Correct Decision Type II Error H1 is true H0: person is innocent H1: person is guilty

…accepting the null hypothesis when it is actually false.
Terminology Level of Significance …is the probability of rejecting the null hypothesis when it is actually true, i.e. Type I Error Type II Error …accepting the null hypothesis when it is actually false.

Terminology Test Statistic Critical Value
…is a value, determined from sample information, used to determine whether or not to reject the null hypothesis. Critical Value …is the dividing point between the region where the null hypothesis is rejected and the region where it is not rejected.

One-Tail Vs. Two-Tail Tests

One-Tail  = rejection region 1-  = acceptance region Critical z 

Two-Tail  = rejection region /2 /2 1-  = acceptance region -z/2

One-Tailed Tests of Significance
A test is one-tailed when the alternate hypothesis, H1, states a direction. Examples H1: The mean yearly commissions earned by full-time realtors is more than \$65,000. (µ>\$65,000) H1: The mean speed of trucks traveling on the 407 in Ontario is less than 120 kilometres per hour. (µ<120) H1: Less than 20 percent of the customers pay cash for their gasoline purchase. (p<.20)

Sampling Distribution 1-  = 95% acceptance region
One-Tailed 5% Level of Significance =.05 Reject Ho when z >1.65 1-  = 95% acceptance region = 5% rejection region 1.65

Two-Tailed Tests of Significance
A test is two-tailed when no direction is specified in the alternate hypothesis, H1 Examples H1: The mean time Canadian families live in a particular home is not equal to 10 years. (µ10) H1: The average speed of trucks travelling on the in Ontario is different than 120 kph. (µ120) H1: The percentage of repeat customers within a week at Tim Horton’s is not 50%. (p .50)

Sampling Distribution
Two-Tailed 5% Level of Significance Reject Ho when z>1.96 or z< -1.96 = 5% rejection region = 95% acceptance region 0.025 0.025 1.96 & are called “critical values”

n / s m - = X z Test Statistic to be used:
Testing for the Population Mean: Large Sample, Population Standard Deviation Known Test Statistic to be used: n / s m - = X z

Solve Testing for the Population Mean:
Large Sample, Population Standard Deviation Known The processors of eye drop medication indicate on the label that the bottle contains 16 ml of medication. The standard deviation of the process is 0.5 ml A sample of 36 bottles from the last hour’s production revealed a mean weight of ml per bottle. At the .05 significance level is the process out of control? That is, can we conclude that the mean amount per bottle is different from 16 ml? Solve

Conclusion: H0: µ = 16 H1: µ  16  = 0.05 Hypothesis Test Step 1
State the null and alternate hypotheses Step 1 H1: µ  16  = 0.05 Select the level of significance Step 2 Identify the test statistic Step 3 Because we know the standard deviation, the test statistic is Z State the decision rule Step 4 Reject H0 if z > or z < -1.96 Compute the test statistic and make a decision Step 5 - = n X z s m 36 5 . 00 16 12 - = 44 . 1 = Conclusion: Do not reject the null hypothesis We cannot conclude the mean is different from 16 ml.

Solve Testing for the Population Mean:
Large Sample, Population Standard Deviation Unknown Rock’s Discount Store chain issues its own credit card Lisa, the credit manager, wants to find out if the mean monthly unpaid balance is more than \$400. The level of significance is set at .05. A random check of 172 unpaid balances revealed the sample mean to be \$ and the sample standard deviation to be \$38. Should Lisa conclude that the population mean is greater than \$400, or is it reasonable to assume that the difference of \$7 (\$407-\$400) is due to chance? Solve

Tip When the sample is large, i.e. over 30, you can use the z-distribution as your test statistic. (Just replace the sample standard deviation for the population standard deviation) Remember, use the best that you have!

Conclusion: H0: µ = 400 H1: µ > 400  = 0.05 Hypothesis Test Step 1
State the null and alternate hypotheses Step 1 H1: µ > 400  = 0.05 Select the level of significance Step 2 Identify the test statistic Step 3 Because the sample is large, we use the test statistic Z State the decision rule Step 4 Reject H0 if z > 1.645 = 172 38 \$ 400 407 - Compute the test statistic and make a decision Step 5 - = n X z s m 42 . 2 = Conclusion: Reject the hypothesis. H0 . Lisa can conclude that the mean unpaid balance is greater than \$400!

Test Statistic to be used:
Testing for the Population Mean: Small Sample, Population Standard Deviation Unknown Test Statistic to be used: n s X t / m - =

Testing for the Population Mean:
Small Sample, Population Standard Deviation Unknown The current production rate for producing 5 amp fuses at Ned’s Electric Co. is 250 per hour. A new machine has been purchased and installed that, according to the supplier, will increase the production rate! A sample of 10 randomly selected hours from last month revealed the mean hourly production on the new machine was 256 units, with a sample standard deviation of 6 per hour. At the .05 significance level, can Ned conclude that the new machine is faster?

Conclusion: H0: µ = 250 H1: µ > 250  = 0.05 Hypothesis Test Step 1
State the null and alternate hypotheses Step 1 H1: µ > 250  = 0.05 Select the level of significance Step 2 Identify the test statistic Step 3 Because the sample is small and  is unknown, we use the t-test State the decision rule Step 4 … = 9 degrees of freedom Reject H0 if t > 1.833 = 10 6 250 256 - Compute the test statistic and make a decision Step 5 - = n X t s m 162 . 3 = Conclusion: Reject the hypothesis. H0 . Ned can conclude that the new machine will increase the production rate!

p-value in hypothesis testing
A P -Value is the probability, (assuming that the null hypothesis is true) of finding a value of the test statistic at least as extreme as the computed value for the test! If the P-Value is smaller than the significance level, H0 is rejected. If the P-Value is larger than the significance level, H0 is not rejected.

Recall 42 . 2 = - n s X z m Previously determined…  = 0.05
Rock’s Discount Store chain issues its own credit card. Lisa, the credit manager, wants to find out if the mean monthly unpaid balance is more than \$ The level of significance is set at A random check of 172 unpaid balances revealed the sample mean to be \$407 and the sample standard deviation to be \$ Should Lisa conclude that the population mean is greater than \$400?  = 0.05 42 . 2 = - n s X z m P(z  2.42) = = .0078 Since P-value is smaller than  of 0.05, reject H The population mean is greater than \$400.

p-value in hypothesis testing
One-Tail Two-Tail P-Value = p(z  |computed value|) P-Value = 2p(z  |computed value|) |....| means absolute value of…

2p(z  |computed value|)
Recall Previously determined… The processors of eye drop medication indicate on the label that the bottle contains 16 ml of medication. The standard deviation of the process is 0.5 ml. A sample of 36 bottles from last hour’s production revealed a mean weight of ml per bottle. At the .05 significance level is the process out of control? That is, can we conclude that the mean amount per bottle is different from 16 ml?  = 0.05 44 . 1 = - n X z s m P-Value = 2p(z  |computed value|) = 2p(z  |1.44|) = 2( ) = 2(.0749) = .1498 Since > .05, do not reject H0.

Interpreting the Weight of Evidence against Ho
If the P-value is less than … .10 we have some evidence that Ho is not true .05 we have strong evidence that Ho is not true .01 we have very strong evidence that Ho is not true .001 we have extremely strong evidence that Ho is not true

.05 we have strong evidence .01 we have very strong evidence
If the P-value is less than… .10 we have some evidence .05 we have strong evidence .01 we have very strong evidence .001 we have extremely strong evidence that Ho is not true Since P-value is .0078 … we have very strong evidence to conclude that the population mean is greater than \$400!

Tests concerning Proportions
A Proportion … is the fraction or percentage that indicates the part of the population or sample having a particular trait of interest Sample Proportion … is denoted by p … is found by: sampled Number sample in the successes of = p

Testing a Single Population Proportion: Test Statistic to be used:
z ) 1 ( ˆ - = where p ^ … is the symbol for sample proportion p … is the symbol for population proportion p0 … represents a population proportion of interest

In the past, 15% of the mail order solicitations for a certain charity resulted in a financial contribution. A new solicitation letter that has been drafted is sent to a sample of 200 people and responded with a contribution. At the .05 significance level can it be concluded that the new letter is more effective?

Conclusion: 97 . 2 = H0: p = .15 H1: p > .15  = 0.05
Hypothesis Test H0: p = .15 State the null and alternate hypotheses Step 1 H1: p > .15  = 0.05 Select the level of significance Step 2 Identify the test statistic Step 3 We will use the z-test State the decision rule Step 4 Reject H0 if z > 1.645 Compute the test statistic and make a decision Step 5 200 ) 15 . 1 ( .15 - 45 = p z ˆ n ) 1 ( - = 97 . 2 = Reject the hypothesis. More than 15% are responding with a pledge, therefore, the new letter is more effective! Conclusion:

Our decision rule can be restated as:
Relationship Between Hypothesis Testing Procedure and Confidence Interval Estimation Two-Tail Case 1: TEST Our decision rule can be restated as: Do not reject H if 0 lies in the (1-) confidence interval estimate of the population mean, computed from the sample data

Two-Tail  = rejection region 1-  = Confidence Interval region
Do not reject Ho when z falls in the confidence interval estimate

Our decision rule can be restated as:
Relationship Between Hypothesis Testing Procedure and Confidence Interval Estimation Case 2: Lower-tailed test Our decision rule can be restated as: Do not reject H if 0 is less than or equal to the (1-) upper confidence bound for , computed from the sample data.

1-  = confidence level region
Relationship Between Hypothesis Testing Procedure and Confidence Interval Estimation Lower-Tailed 1-  = confidence level region  = rejection region Do not reject

Our decision rule can be restated as:
Relationship Between Hypothesis Testing Procedure and Confidence Interval Estimation Case 3: Upper-tailed test Our decision rule can be restated as: Do not reject H if 0 is greater than or equal to the (1-) lower confidence bound for , computed from the sample data.

Upper-Tailed 1- = acceptance region  = rejection region

…accepting the null hypothesis when it is actually false.
Type II Error Level of Significance …is the probability of rejecting the null hypothesis when it is actually true, i.e. Type I Error …accepting the null hypothesis when it is actually false. Type II Error

Calculating the Probability of a Type II Error
A batch of 5000 light bulbs either belong to a superior type, with a mean life of 2400 hours, or to an inferior type, with a mean life of 2000 hours (By default, the bulbs will be sold as the inferior type.) Both bulb distributions are normal, with a standard deviation of 300 hours.  = Suppose we select a sample of 4 bulbs Find the probability of a Type II error. Solve

Superior: =2400 Inferior: =2000
=300 =0.025 H0: µ = 2000 State the null and alternate hypotheses Step 1 H1: µ = 2400 Select the level of significance Step 2  = 0.025 Identify the test statistic Step 3 As populations are normal,  is known, we use the z-test State the decision rule Step 4 Reject H0 if the computed z > 1.96, or stated another way, If the computed value x bar is greater than xu = (300/n), REJECT H0 in favour of H1

The probability of a Type II Error
Suppose H0 is false and H1 is true i.e. the true value of µ is 2400, then x bar is approximately normally distributed with a mean of 2400 and a standard deviation of /n = 300/n The probability of a Type II Error …is the probability of not rejecting Ho …is the probability that the value of x bar obtained will be less than or equal to xu Xu X

Suppose we select a sample of 4 bulbs
Suppose we select a sample of 4 bulbs. Then x bar has a mean of 2400 and a sd of 300/4 = 150 Xu = (300/4) = 2294 = - n X z s m 4 300 2400 2294 = - 70666 . - A1 = , giving us a left tail area of 0.24

The probability of a Type II error is 0.24 i.e.=0.24

If we decrease the value of (alpha), the value z increases and the critical value xu moves to the right, and therefore the value of (beta) increases. Conversely, if we increase the value of (alpha), xu moves to the left, thereby decreasing the value of (beta) For a given value of (alpha), the value of (beta) can be decreased by increasing the sample size.

Therefore, the test’s power is 1-0.24 = 0.76
Power of a Test … is defined as the probability of rejecting H0 when H0 is false, or …the probability of correctly identifying a true alternative hypothesis …it is equal to (1-) In previous example,  = 0.24 Therefore, the test’s power is = 0.76

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This completes Chapter 10