Session 18 Notes, Section 9-2

Testing the Difference Between Two Means Using the Z-Test

Special z and t tests allow researchers to compare population parameters, such as means.

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

• Test the difference between two large sample means using the z test
  • Example one tailed
  • Example two tailed
• 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 two large sample means using the z test

• Assumptions for the test to determine the difference between two means:
  • The samples must be independent of each other; that is, there can be no relationship between the subjects in each sample.
  • The populations from which the samples were obtained must be normally distributed, and the standard deviations of the variable must be known, or the sample sizes must be greater than or equal to 30.
• Formula for the z test for comparing two means from independent populations
  • .
  • When n₁≥ 30 and n₂ ≥ 30,  the sample variances can be used instead of the population variances (sigmas):
    •  
• The General Formula  Format:
  • 
  • Observed value =
  • Expected value =
  • Standard deviation =
• Confidence Interval for Difference Between Two Means for Large Samples
  • For Large Samples:
    • 

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Example:

A survey found that the average hotel room rate in a city is $86.74 and the average room rate in another city is $81.90. Assume that the data were obtained from two samples of 49 hotels each and that the standard deviations were $5.54 and $5.00, respectively. At α = 0.05, can it be concluded that there is a significant difference in the rates?

• 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 α = 0.05, the critical values are +1.96 and -1.96.
• Step 3: Compute the test value. Use:
  •  .  Then z = 4.54.
• Step 4: Make the decision.
  • Reject the null hypothesis at = 0.05, since 4.54 > 1.96.
• Summarize the results.
  • There is enough evidence to support the claim that the means are not equal.
  • Hence there is a significant difference in the rates.

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Example:

At the age 9 the average height (124.5 cm) for boys and girls is exactly the same. A random sample yielded these results. Estimate the mean difference in height between boys and girls with 95% confidence. Does your interval support the given claim?