# Chapter Ten McGraw-Hill/Irwin

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Chapter Ten McGraw-Hill/Irwin

One-Sample Tests of Hypothesis
GOALS WHAT Define a hypothesis and hypothesis testing. WHY Reasons behind hypothesis testing. HOW Describe the five step hypothesis testing procedure. Distinguish between a one-tailed and a two-tailed test of hypothesis. Conduct a test of hypothesis about a population mean. Define Type I and Type II errors.

The mean monthly income for systems analysts is \$6,325.
A statement about the value of a population parameter developed for the purpose of testing. What is a Hypothesis? Twenty percent of all customers at Bovine’s Chop House return for another meal within a month. The mean monthly income for systems analysts is \$6,325. What is a Hypothesis?

Hypothesis: A statement about the value of a population parameter developed for the purpose of testing. A particular brand of rice imported to United States contains the arsenic at the level allowable by the EPA. What is a Hypothesis?

What is Hypothesis Testing?
Used to determine whether the hypothesis is a reasonable statement and should not be rejected, or is unreasonable and should be rejected Based on sample evidence and probability theory What is Hypothesis Testing?

Why Hypothesis Testing?
Because we want to make sure beyond a certain level of doubt and we take into account the sampling error Why can’t we just conclude from a sample of rice that has arsenic of 9.5 parts per billion Why Hypothesis Testing?

How to Conduct Hypothesis Testing?

Step One: State the null and alternate hypotheses
Null Hypothesis H0 A statement about the value of a population parameter Alternative Hypothesis H1: A statement that is accepted if the sample data provide evidence that the null hypothesis is false

3 hypotheses about means
Step One: State the null and alternate hypotheses H0: m = 0 H1: m = 0 Three possibilities regarding means The null hypothesis always contains equality. H0: m < 0 H1: m > 0 H0: m > 0 H1: m < 0 3 hypotheses about means

Step Two: Select a Level of Significance.
Type I Error Rejecting the null hypothesis when it is actually true (a). Level of Significance The probability of rejecting the null hypothesis when it is actually true; the level of risk in so doing. Type II Error Accepting the null hypothesis when it is actually false (b). Step Two: Select a Level of Significance.

Step Two: Select a Level of Significance.
Researcher Null Accepts Rejects Hypothesis Ho Ho Ho is true Ho is false Correct decision Type I error (a) Type II Error (b) Risk table

Step Three: Select the test statistic.
z Distribution as a test statistic Test statistic A value, determined from sample information, used to determine whether or not to reject the null hypothesis. The z value is based on the sampling distribution of X, which is normally distributed when the sample is reasonably large (recall Central Limit Theorem). Examples: z, t, F, c2 Step Three: Select the test statistic.

Step Four: Formulate the decision rule.
Critical value: The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected. Sampling Distribution Of the Statistic z, a Right-Tailed Test, .05 Level of Significance

Reject the null hypothesis and accept the alternate hypothesis if
Decision Rule Reject the null hypothesis and accept the alternate hypothesis if Computed -z < Critical -z or Computed z > Critical z Decision Rule

Using the p-Value in Hypothesis Testing
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 p-Value Decision Rule If the p-Value is larger than or equal to the significance level, a, H0 is not rejected. If the p-Value is smaller than the significance level, a, H0 is rejected. Calculated from the probability distribution function or by computer Using the p-Value in Hypothesis Testing

Interpreting p-values
SOME evidence Ho is not true VERY STRONG evidence Ho is not true STRONG evidence Ho is not true

Step Five: Make a decision.
Movie

One-Tailed Tests of Significance
The alternate hypothesis, H1, states a direction H1: The mean yearly commissions earned by full-time realtors is more than \$35,000. (µ>\$35,000) H1: The mean speed of trucks traveling on I-95 in Georgia is less than 60 miles per hour. (µ<60) H1: Less than 20 percent of the customers pay cash for their gasoline purchase. (p<.20) One-Tailed Tests of Significance

One-Tailed Test of Significance
Sampling Distribution Of the Statistic z, a Right-Tailed Test, .05 Level of Significance One-Tailed Test of Significance .

Two-Tailed Tests of Significance
No direction is specified in the alternate hypothesis H1. H1: The mean price for a gallon of gasoline is not equal to \$1.54. (µ = \$1.54). H1: The mean amount spent by customers at the Wal-mart in Georgetown is not equal to \$25. (µ = \$25). Two-Tailed Tests of Significance

Two-Tailed Tests of Significance
Regions of Nonrejection and Rejection for a Two-Tailed Test, .05 Level of Significance

Test for the population mean from a large sample with population standard deviation known
Testing for the Population Mean: Large Sample, Population Standard Deviation Known

The processors of Fries’ Catsup indicate on the label that the bottle contains 16 ounces of catsup. The standard deviation of the process is 0.5 ounces. A sample of 36 bottles from last hour’s production revealed a mean weight of ounces 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 ounces? Example 1

State the decision rule. Reject H0 if z > 1.96 or z < -1.96
Step 4 State the decision rule. Reject H0 if z > 1.96 or z < -1.96 or if p < .05. Step 5 Make a decision and interpret the results. Step 3 Identify the test statistic. Because we know the population standard deviation, the test statistic is z. Step 1 State the null and the alternative hypotheses H0: m = 16 H1: m = 16 Step 2 Select the significance level. The significance level is .05. EXAMPLE 1

Step 5: Make a decision and interpret the results.
The p(z > 1.44) is for a two-tailed test. Computed z of 1.44 < Critical z of 1.96, p of > a of .05, Do not reject the null hypothesis. We cannot conclude the mean is different from 16 ounces. Example 1

As long as the sample size n > 30, z can be approximated using
Testing for the Population Mean: Large Sample, Population Standard Deviation Unknown As long as the sample size n > 30, z can be approximated using Here s is unknown, so we estimate it with the sample standard deviation s. Testing for the Population Mean: Large Sample, Population Standard Deviation Unknown

Roder’s Discount Store chain issues its own credit card
Roder’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 A random check of 172 unpaid balances revealed the sample mean to be \$407 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? Example 2

Make a decision and interpret the results.
Step 4 H0 is rejected if z > 1.65 or if p < .05. Step 5 Make a decision and interpret the results. Step 3 Because the sample is large we can use the z distribution as the test statistic. Step 1 H0: µ < \$400 H1: µ > \$400 Step 2 The significance level is .05. Example 2

Make a decision and interpret the results.
Step 5 Make a decision and interpret the results. Computed z of 2.42 > Critical z of 1.65, p of < a of .05. Reject H0. The p(z > 2.42) is for a one-tailed test. Lisa can conclude that the mean unpaid balance is greater than \$400.

The test statistic is the t distribution.
Testing for a Population Mean: Small Sample, Population Standard Deviation Unknown The test statistic is the t distribution. The critical value of t is determined by its degrees of freedom equal to n-1. Testing for a Population Mean: Small Sample, Population Standard Deviation Unknown

The current rate for producing 5 amp fuses at Neary Electric Co
The current rate for producing 5 amp fuses at Neary Electric Co. is 250 per hour. A new machine has been purchased and installed that, according to the supplier, will increase the production rate. The production hours are normally distributed. A sample of 10 randomly selected hours from last month revealed that 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 Neary conclude that the new machine is faster? Example 3

State the decision rule. There are 10 – 1 = 9 degrees of freedom.
The null hypothesis is rejected if t > or, using the p-value, the null hypothesis is rejected if p < .05. Step 4 State the decision rule. There are 10 – 1 = 9 degrees of freedom. Step 3 Find a test statistic. Use the t distribution since s is not known and n < 30. Step 1 State the null and alternate hypotheses. H0: µ < 250 H1: µ > 250 Step 2 Select the level of significance. It is .05.

Make a decision and interpret the results.
Step 5 Make a decision and interpret the results. The p(t >3.162) is for a one-tailed test. Computed t of 3.162 >Critical t of p of < a of .05 Reject Ho The mean number of fuses produced is more than 250 per hour. Example 3

The fraction or percentage that indicates the part of the population or sample having a particular trait of interest. Proportion The sample proportion is p and p is the population proportion. Test Statistic for Testing a Single Population Proportion

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 45 responded with a contribution. At the .05 significance level can it be concluded that the new letter is more effective? Example 4

Find a test statistic. The z distribution is the test statistic.
Step 3 Find a test statistic. The z distribution is the test statistic. Step 4 State the decision rule. The null hypothesis is rejected if z is greater than 1.65 or if p < .05. Step 5 Make a decision and interpret the results. Step 1 State the null and the alternate hypothesis. H0: p < .15 H1: p > .15 Step 2 Select the level of significance. It is .05. Example 4

Step 5: Make a decision and interpret the results.
p( z > 2.97) = Because the computed z of 2.97 > critical z of 1.65, the p of < a of .05, the null hypothesis is rejected. More than 15 percent responding with a pledge. The new letter is more effective. Example 4

"Being a statistician means never having to say you are certain.“
A statistician confidently tried to cross a river that was 1 meter deep on average. He drowned. "If you torture data enough it will confess" A biologist, a mathematician, and a statistician are on a photo-safari in Africa. They drive out into the savannah in their jeep, stop, and scour the horizon with their binoculars. The biologist: “Look! There’s a herd of zebras! And there, in the middle: a white zebra! It’s fantastic! There are white zebras! We’ll be famous!” The mathematician: “Actually, we know there exists a zebra which is white on one side.” The statistician: “It’s not significant. We only know there’s one white zebra.” The computer scientist: “Oh no! A special case!”

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