Stat 2601

Chapter 8 Hypothesis Testing Section 8.1

Introduction

1. Basic Concepts 1). A statistical hypothesis – a statement/claim/theory about a population parameter. The null hypothesis is a theory about a population parameter. It is denoted by H0, and it is usually stated as H0: parameter = value. The alternative hypothesis is a theory that contradicts the null hypothesis. It is denoted by H1, and it is always specified as strict inequalities, such as , >, or < . Steps for Selecting the Null and Alternative Hypotheses I. Select the alternative hypothesis as that which the sampling experiment is intended to establish. The alternative hypothesis has one of the three forms: H1: μ > value (right-tailed) H1: μ < value (left-tailed) H1: μ value (two-sided) II. Select the null hypothesis which is usually specified as equality. Example 1: A researcher is interested in finding out whether the average age of all cars in use is higher than 8 years. The null and alternative hypotheses can be stated as: H0: μ = 8 and H1: μ > 8 . This test is called a right-tailed test. Example 2: According to a report, the mean monthly bill for cell phone users in the United States was $49.70 in 1999. We want to determine whether last year’s mean monthly bill for cell phone users has decreased from the 1999 mean of $49.70. The hypotheses are H0: μ = 49.70 and H1: μ < 49.70 . This test is called a left-tailed test. Example 3: A researcher wishes to find out whether the mean body temperature of humans is 98.6°F. The null and alternative hypotheses can be stated as: H0: μ = 98.6° and H1: μ 98.6°. This test is called a two-tailed test. 2). A test statistic or test value– is a numerical value obtained from a sample and is used to decide whether the null hypothesis should be rejected. 3). The critical or rejection region – the set of values for the test statistic that leads to rejection of the null hypothesis. Critical values – the values that separate the rejection from the non-rejection regions. 4). Type I error – rejecting the null hypothesis when it is in fact true. Type II error – not rejecting the null hypothesis when it is in fact false. Do not reject H0 Reject H0

H0 is true Correct decision Type I error

H0 is false Type II error Correct decision

5). Significance level - the probability of making a Type I error, denoted by α. Three common values for significance level: 0.01, 0.05, and 0.10. 1

Stat 2601 Section 8.2

z Test for a Mean (Traditional Method and P-value Method)

1. Definition The z test is a statistical test which can be used when the population standard deviation is known and the population is normally distributed or when n ≥ 30. The test statistic for a z test with null hypothesis H0: μ = μ0 is: X 0 . z / n

2. Traditional Method (Critical-Value Method) for Hypothesis Testing 1) Steps: State the hypotheses. Compute the value of the test statistic. Find the rejection region at a specified significance level α. Make the decision to reject or not reject H0. Interpret the results of the hypothesis test. 2) Finding the Rejection Region for a Specific α value when a z-test is performed. a) Right-tailed Rejection region: z > zα, where zα is a z value that will give an area of α in the right tail of the standard normal distribution. zα is also called the critical value, a value that separates the rejection region and non-rejection region. If α = .05, the rejection region is z > 1.645. Right tail area α

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C.V.

b) Left-tailed For a left-tailed test, the rejection region: z < -zα. When α=.05, the C.V. is -1.645, the rejection region is z < -1.645

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Stat 2601 c) Two-sided For a two-sided test, the rejection region: z > zα/2 or z < -zα/2. When α=.05., the rejection region is z > 1.96 or z < -1.96. Total tail area =

0 Table 8.1 Summary of rejection region at a specific significance level α

α = .01 α = .05 α = .1

Upper-tailed z > 2.33 z > 1.645 z > 1.28

Rejection Region Lower-tailed z < -2.33 z < -1.645 z < -1.28

Two-sided z > 2.575 or z < -2.575 z > 1.96 or z < -1.96 z > 1.645 or z < -1.645

Example 1: A researcher is interested in finding out whether the average age of all cars in use is higher than 8 years. A random sample of 40 cars were selected and the average age of cars was found to be 8.5 yrs, and the standard deviation was 3.5 yrs. At α=.05, can it be concluded that the average age of all cars in use is higher than 8 years?

Example 2: According to a report, the mean monthly bill for cell phone users in the United States was $49.70 in 1999. We want to determine whether last year’s mean monthly bill for cell phone users has decreased from the 1999 mean of $49.70. A random sample of 50 sample users was selected and average of the last year’s monthly cell phone bills for these 50 users was $41.40. At the 1% significance level, do we have enough evidence to conclude that last year’s mean monthly bill for cell phone users has decreased from the 1999 mean of $49.70? Assume that σ = 25.

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Stat 2601 Example 3: A researcher wishes to find out whether the mean body temperature of humans is 98.6° F. The researcher obtained the body temperature of 93 healthy humans and found that the mean body temperature is 98.1°F. At the 1% significance level, do the data provide sufficient evidence to conclude that the mean body temperature of healthy humans differs from 98.6°F. Assume that σ = .63°F.

3. P-value Method 1) Definition P-value- the P-value of a hypothesis test is the probability of observing a value of the test statistic as extreme or more extreme than that observed when the null hypothesis is true. Right-tailed test: The P-value is the probability of observing a value of the test statistic as large as or larger than the value actually observed, which is the area under the standard normal curve that lies to the right of the observed test statistic. Left-tailed test: The P-value is the probability of observing a value of the test statistic as small as or smaller than the value actually observed. Two-sided test: The P-value is the probability of observing a value of the test statistic at least as large in magnitude as the value actually observed. 2) Steps State the hypotheses. Compute the value of the test statistic, say z0. Find the P-value. For an upper-tailed test, P-value = P(z > z0). For a lower -tailed test, P-value = P(z < z0). For a two-sided test, P-value = 2*P(z > z0) if z0 > 0, or = 2*P(z < z0) if z0 < 0. If P –value ≤ α, reject H0; otherwise, do not reject H0. Interpret the results of the hypothesis test. Example 1: Use the P-value method to perform the same hypothesis test.

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Stat 2601 Example 2 Use the P-value method to perform the same hypothesis test.

Example 3: Use the P-value method to perform the same hypothesis test.

Section 8.3

t Test for a Mean

1. Definition The t test is a statistical test which can be used when the population is normally distributed, σ is unknown and n < 30. The test statistic for a t test with null hypothesis H0: μ = μ0 is: X 0 . t s/ n The degrees of freedom are d.f. = n -1. 2. Finding the rejection region 1) H1 : µ > µ0 , rejection region: t > tα. 2) H1 : µ < µ0 , rejection region: t tα/2 or t < -tα/2. Where tα.is the t-value that will give an area of α to its right. tα.and tα/2 are based on (n-1) df.

Example: The average undergraduate cost for tuition, fees, and room for two-year institutions last year was $13,252. The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500. Is there sufficient evidence at α = .05 to conclude that the mean cost has increased?

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Stat 2601 Section 8.4

z Test for a Proportion

1. The test statistic for a test of hypothesis with H0 : p = p0 is: z

pˆ

p0

, p0 q0 / n Where pˆ is the sample proportion, n is the sample size, p0 is the hypothesized value of p.

2. Rejection region (Traditional method) See Table 8.1. 3. p-value (p-value method) See Section 8.2.

Example 8.4.1: A recent survey found that 64.7% of the population owns their homes. In a random sample of 150 heads of household, 92 responded that they owned their homes. At the 0.01 level of significance, does that indicate a difference from the national proportion? Use the traditional method.

Example 8.4.2: Nationally 60.2% of federal prisoners are serving time for drug offenses. A warden feels that in his prison the percentage is even higher. He surveys 400 inmates’ records and finds that 260 of the inmates are drug offenders. At α = 0.05, is he correct? Use the P-value method.

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