Session 19 Notes, Section 9-4

Testing the Difference Between Two Means with Small Independent Samples

At the end of this session you will be able to:

• Test the difference between the variations or standard deviations of two independent samples.
• Test the difference between two means for small independent samples
  • Example
  • Example
• Your Turn:  Summary

If you have learned all of these objectives, then close this window to return to where you were.

Test the difference between the variations or standard deviations of two independent samples. (Section 9-3).

In Session 19, we are learning how to test the difference between two means when we have small independent samples. Because we have small samples, then we know that we need to use the t-test.  However, the t-test to be used depends upon variances being equal of unequal. To determine whether two sample variances are equal, we can use the F-test as shown in Section 9-3 (We used Tables H to determine the critical values for our test.):

If you do not want to go through the five steps of hypotheses testing, then you can use your TI-83 to test the difference between two variances or standard deviations.  See pages 453 and 454 in your textbook for step-by-step procedures.

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Test the difference between two means for small independent samples (9-4)

• A t test is used to test the difference between means when
  • the two samples are independent,
  • when the sample sizes are small,
  • and when the samples are taken from two normally or approximately normally distributed populations.
• There are two different options for the use of t tests.
  • One option is used when the variances of the populations are not equal,
  • and the other option is used when the variances are equal.

   

• Formula for the Difference Between Two Means–Small Samples

  • When variances are assumed to be unequal:

    • See the formula on page 455 of your textbook.
    • where the degrees of freedom are equal to the smaller of n₁ – 1 or n₂ – 1

  • When variances are assumed to be equal:

    • 
    • where the degrees of freedom are equal to  n₁ +  n₂ – 2

• Confidence Intervals: for the difference of two means, small independent sample

  • Unequal Variance
    • 
    • d.f. = the smaller value of n₁ – 1 or n₂ – 1
  • Equal Variance
    • 
    • d.f. = n₁ + n₂ – 2

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Example.

The average size of a farm in one county is 185 acres. The average size of a farm in another county is 202 acres. Assume the data were obtained from two samples with standard deviations of 38 acres and 8 acres, respectively, and sample sizes of 11 and 13, respectively. Can it be concluded at α = 0.05 that the average size of the farms in the two counties is different? Assume the populations are normally distributed and variances are not equal.

• Step 1: State the hypotheses and identify the claim.
  • H₀:  µ₁ = µ ₂   and   H₁:  µ₁ not = µ ₂ (claim)
• Step 2:  Find the critical values.
  • Since the test is two-tailed, since α = 0.05, and since the variances are unequal, the degrees of freedom are the smaller of n₁ - 1 or n₂ - 1. In this case, the degrees of freedom are 11 - 1 = 10. Hence, from the t-distribution table, the critical values are +2.228 and -2.228.
• Step 3: Compute the test value.
  • Since the variances are unequal
    • 
• Step 4: decision
  • Do not reject the null hypothesis, since -1.457 > -2.228.
• Step 5: Summary
  • There is not enough evidence to support the claim that the average size of the farms is different.

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Example.

The average size of a farm in one county is 194 acres. The average size of a farm in another county is 198 acres. Assume the data were obtained from two samples with standard deviations of 32 acres and 15 acres, respectively, and sample sizes of 10 and 12, respectively. Find the 95% confidence interval for the data. Assume the populations are normally distributed.

• Find t _α/2. The test is two-tailed, α = 0.05, and the degrees of freedom are 10 - 1 = 9. Hence, from the t-distribution table, t _α/2 = 2.262.
• Since the variances are unequal use:
  • 
  • Substituting the data for this problem
    • 
  • Simplifying you calculate:
    • -28.90  <  µ₁ - µ₂  <  20.90
  • Thus the difference of the two sample means falls between -28.90 and 20.90 with 95% confidence

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Summary

• If the variances are not known or one or both sample sizes are less than 30, then the t test must be used.
• For independent samples the equality or inequality of the variances must  be know or calculated

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