Two-Sample Tests of Hypothesis Notes

Comparing Two Population Means

• In comparing two populations, the goal is to determine if their means are equal.
• This involves investigating whether the distribution of the difference between the means could have a mean of 0.
• Examples:
  • Is there a difference in the mean value of residential real estate sold by male agents and female agents in south Florida?
  • Is there an increase in the production rate after music is introduced to the production area?

Comparing Two Population Means (Assume $\mu<em>1 = \mu</em>2; \sigma<em>1 = \sigma</em>2$ known)

• The z-formula can be used if:
  • The two populations follow normal distributions.
  • The samples are from independent (unrelated) populations.
  • The population standard deviations are known.
• In the z-formula, $\bar{x}<em>1 - \bar{x}</em>2$ is the difference in the sample means.
• The square root of the variance (Formula 11-1) is the standard deviation.

Comparing Two Population Means Example (Assume $\mu<em>1 = \mu</em>2; \sigma<em>1 = \sigma</em>2$ known)

• Scenario: FoodTown Supermarket wants to know if the mean checkout time using the standard checkout method is longer than using the Fast Lane (self-checkout).
• Step 1: State the null and alternate hypothesis.
  • Null hypothesis: $H<em>0: \mu</em>S \leq \mu_F$ (Mean checkout time for standard checkout is less than or equal to Fast Lane).
  • Alternate hypothesis: H1: \muS > \mu_F (Mean checkout time for standard checkout is greater than Fast Lane - keyword: “longer than”).
• Step 2: Select the level of significance: $\alpha = 0.01$.
• Step 3: Determine the test statistic: Use the z-distribution because population standard deviations are known.
• Step 4: Formulate the decision rule: Reject $H_0$ if z > 2.326.
• Step 5: Make the decision regarding $H_0$.
  • Randomly selected 50 customers using standard checkout (mean time = 5.5 minutes).
  • Randomly selected 100 customers using Fast Lane (mean time = 5.3 minutes).
  • Test statistic = 3.123 is greater than critical value of 2.326.
  • Reject the null hypothesis.
• Step 6: Interpret the result.
  • The difference of 0.20 minutes is too large to have occurred by chance.
  • Conclusion: The Fast Lane method is faster.
• Numerical values:
  • Test statistic: $z = 3.123$
  • Critical value: $z = 2.326$

Comparing Two Means Using t (Assume $\mu<em>1 = \mu</em>2; \sigma<em>1 = \sigma</em>2$ unknown)

• Two major differences from the previous test:
  • The sampled populations have equal but unknown standard deviations.
  • We use the t-distribution.
• Three requirements for the test:
  • The sampled populations are approximately normally distributed.
  • The sampled populations are independent.
  • The standard deviations of the two populations are equal.

Comparing Two Means Using t (Assume $\mu<em>1 = \mu</em>2; \sigma<em>1 = \sigma</em>2$ unknown)

• Finding the value of t requires two steps:
  • First, pool the standard deviations using a specific formula.
  • Then, compute the value of t using another formula.
• The degrees of freedom for the test are $n<em>1 + n</em>2 - 2$.

Two-Sample Pooled Test Example (Assume $\mu<em>1 = \mu</em>2; \sigma<em>1 = \sigma</em>2$ unknown)

• Scenario: Owens Lawn Care Inc. wants to know if there is a difference in the mean time to mount engines on lawnmower frames using the Welles method versus the Atkins method.
• Step 1: State the null and alternate hypothesis (two-tailed test).
  • $H<em>0: \mu</em>W = \mu_A$ (Mean time for Welles method equals mean time for Atkins method).
  • $H<em>1: \mu</em>W \neq \mu_A$ (Mean time for Welles method differs from mean time for Atkins method - keyword: “Is there a difference”).
• Step 2: Select the level of significance: $\alpha = 0.10$.
• Step 3: Determine the test statistic: Use the pooled t-test because population standard deviations are not known but assumed to be equal.
• Step 4: Formulate the decision rule: Do not reject $H_0$ if t falls between -1.833 and 1.833.
• Step 5: Make decision regarding $H_0$.
  • Compute the value of t in three steps:
    • Calculate the sample standard deviations.
    • Pool the sample variances.
    • Determine the value of t.
  • The decision is not to reject the null hypothesis because -0.662 falls between -1.833 and 1.833.
• Step 6: Interpret the result: The sample data failed to show a difference between the mean assembly times of the two methods.
• Numerical values:
  • Test statistic: $t = -0.662$
  • Critical values: $t = -1.833, 1.833$

Unequal Population Standard Deviations (Assume $\mu<em>1 = \mu</em>2; \sigma<em>1 \neq \sigma</em>2$ unknown)

• If we cannot assume the population standard deviations are equal, we adjust the degrees of freedom and the formula for finding t.
• Determine the degrees of freedom based on a specific formula.
• The value of the test statistic is computed from another formula.

Unequal Population Standard Deviations Example (Assume $\mu<em>1 = \mu</em>2; \sigma<em>1 \neq \sigma</em>2$ unknown)

• Scenario: Consumer testing laboratory compares the absorbency of store brand paper towels to name brand towels.
• A random sample of 9 store brand towels and 12 name brand towels are tested.
• Step 1: State the null and alternate hypothesis (two-tailed).
  • $H<em>0: \mu</em>1 = \mu_2$
  • $H<em>1: \mu</em>1 \neq \mu_2$
• Step 2: Select the level of significance: $\alpha = 0.10$.
• Step 3: Determine the test statistic: Use t and adjust the degrees of freedom. Round the degrees of freedom down to an integer (in this case, 10).
• Step 4: State the decision rule: Do not reject $H_0$ if t falls between -1.812 and 1.812.
• Step 5: Make decision. The decision is to reject the null hypothesis because -2.474 is smaller than -1.812.
• Step 6: Interpret: The mean absorption rate of the two types of towels is not the same.
• Numerical values:
  • Test static: $t = -2.474$
  • Critical values: $t = -1.812, 1.812$

Dependent Samples (paired, matched, related)

• Compute the mean and standard deviation of the sample differences.
• The value of the test statistic is computed with a specific formula.
• Degrees of freedom: $n - 1$
  • $d$ is the mean of the difference between the paired observations.
  • $s_d$ is the standard deviation of the differences between the paired observations.
  • $n$ is the number of paired observations (differences).

Dependent Samples Continued

• Note: The standard deviation of the differences is computed using formula 3-11, but $d$ is substituted for $x$.
• Example: Nickel Savings and Loan employs two firms, Schadek Appraisals and Bowyer Real Estate, to appraise the value of real estate for loans. They randomly select 10 homes and have both firms appraise them. This creates a pair of values for each home, related to the home selected (paired sample).

Dependent Samples Example

• Step 1: State the null and alternate hypothesis.
  • $H<em>0: \mu</em>d = 0$
  • $H<em>1: \mu</em>d \neq 0$
• Step 2: Select the level of significance: $\alpha = 0.05$.
• Step 3: Determine the test statistic: Use the t-statistic to test the difference between two population means with dependent samples.
• Step 4: State the decision rule: Reject $H_0$ if t < -2.262 or t > 2.262.
• Scenario: Nickel Savings and Loan wishes to compare the two companies it uses to appraise the value of residential homes. Nickel Savings selected a sample of 10 residential properties and scheduled both firms for an appraisal.
• At the 0.05 significance level, can we conclude there is a difference between the firm’s appraised values?
• Reject $H<em>0$ if $t > t</em>{\alpha/2, n-1}$ or $t < -t<em>{\alpha/2,n- 1}$$t > t</em>{0.025,9}$ or $t < - t_{0.025, 9}$$t > 2.262$ or t < -2.262

Dependent Samples Example Continued

• Step 5: Make your decision; reject the null hypothesis.
• Step 6: Interpret; we conclude there is a difference between the firms’ mean appraised home values.
• The mean of the sample differences is 4.6, and the standard deviation of the sample differences, $s_d$ is 4.402. Use these in formula 11-7 to compute the t value, 3.305.
• Numerical values:
  • Mean of sample differences: $d = 4.6$
  • Standard deviation of sample differences: $s_d = 4.402$
  • Test statistic: $t = 3.305$
  • Critical values: $t = -2.262, 2.262$

Dependent and Independent Samples

• Two types of dependent samples:
  • Those characterized by a measurement, an intervention, and then another measurement.
    • Example: Measuring worker output, playing music, and then measuring output again.
  • A matching or pairing of the observations.
    • Example: The Nickel Savings and Loan example illustrates dependent samples because a property is selected and both firms appraise the same property.
• A test based on dependent samples is preferred because it reduces the amount of variation and is considered a better test.

Question 1 Practice

• Scenario: A sample of 40 observations is selected from one population with a population standard deviation of 5. The sample mean is 102. A sample of 50 observations is selected from a second population with a population standard deviation of 6. The sample mean is 99. Conduct the following test of hypothesis using the .04 significance level.
• $H<em>0: \mu</em>1 = \mu_2$
• $H<em>1: \mu</em>1 \neq \mu_2$
• Is this a one-tailed or a two-tailed test? State the decision rule. Compute the value of the test statistic. What is your decision regarding $H_0$? What is the p-value?

Question 7 Practice

• Scenario: A random sample of 10 observations from one population revealed a sample mean of 23 and a sample standard deviation of 4. A random sample of 8 observations from another population revealed a sample mean of 26 and a sample standard deviation of 5. At the .05 significance level, is there a difference between the population means?
• The null and alternate hypotheses are:
  • $H<em>0: \mu</em>1 = \mu_2$
  • $H<em>1: \mu</em>1 \neq \mu_2$
• State the decision rule. Compute the pooled estimate of the population variance. Compute the test statistic. State your decision about the null hypothesis. Estimate the p-value.

Question 15 Practice

• Scenario: A recent survey compared the costs of adoption through public and private agencies. For a sample of 16 adoptions through a public agency, the mean cost was $21,045, with a standard deviation of $835. For a sample of 18 adoptions through a private agency, the mean cost was $22,840, with a standard deviation of $1,545. Assume the sample populations do not have equal standard deviations and use the .05 significance level
• Determine the number of degrees of freedom and round down to the nearest integer value. State the decision rule. Compute the value of the test statistic. State your decision about the null hypothesis

Question 17 Practice

• Scenario: Hypothesis Testing with Dependent Samples
• The null and alternate hypotheses are:
  • $H<em>0: \mu</em>d \leq 0$
  • H1: \mud > 0
• The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of 4 days last month. What is the p-value? Is the null hypothesis rejected? What is the conclusion indicated by the analysis?