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8.2 Testing the Difference Between Means (Small, Independent Samples)

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8.2 Testing the Difference Between Means (Small, Independent Samples)

Statistics

Mrs. Spitz

Spring 2009

- How to perform a t-test for the difference between two population means, 1 and 2 using small independent samples.
Assignment: pp. 384-388 #1-24 all

- So, it is impractical and costly to collect samples of 30 or more from each of two populations. What about if both populations have a normal distribution? If so, you can still test the difference between their means. In this section, you will learn how to use a t-test to test the difference between two population means, 1 and 2 using a sample from each population.

- To use a t-Test for small independent samples, the following conditions are necessary:
- The samples must be independent, so 1st sample cannot be related to the sample selected from the second population.
- Each population must have a normal distribution.

- The sampling distribution for , the difference between he sample means, is a t-distribution with mean
The standard error and the degrees of freedom of the sampling distribution depend on whether or not the population variances and are equal.

- If the population variances are equal, information from both samples is combined to calculate a pooled estimate of the standard deviation.

Pooled estimate of

- The standard error for the sampling distribution of is:

And d.f. of n1 + n2 - 2

- The standard error for the sampling distribution of is:

And d.f. smaller of n1 – 1 or n2 - 1

- The requirements for the z-test described in 8.1 and the t-test described in this section are compared below:

If the sampling distribution for is a t-distribution, you can use a two-sample t-test to test the difference between two populations 1 and 2.

- Consumer Reports tested several types of snow tires to determine how well each performed under winter conditions. When traveling on ice at 15 mph, 10 Firestone Winterfire tires had a mean stopping distance of 51 feet with a standard deviation of 8 feet. The mean stopping distance for 12 Michelin XM+S Alpine tires was 55 feet with a standard deviation of 3 feet. Can you conclude that there is a difference between the stopping distances of the two types of tires? Use = 0.01. Assume the populations are normally distributed and the population variances are NOT equal.

- Sample Statistics for Stopping Distances
- So now that you have all your relevant data, now go back and figure it out.

- You want to test whether the mean stopping distances are different. So, the null and alternative hypotheses are:
Ho: 1 = 2 and Ha: 1 2 (Claim)

Because the variances are NOT equal, and the smaller sample size is 10, use the d.f. = 10 – 1 = 9. Because the test is a two-tailed test with d.f. = 9, and = 0.01, the critical values are?

- Okay, you got me. . . the critical values are -3.250 and 3.250. The rejection region is t < -3.250 and t > 3.250. The standard error is:

Everybody good so far? Questions?

- Using the t-test, the standardized test statistic is:

The graph following shows the location of the critical regions and the standardized test statistic, t.

- Because t is not in a rejection region, you should fail to reject the null hypothesis. At the 1% level, there is not enough evidence to conclude that the mean stopping distances of the tires are different.

- A manufacturer claims that the calling range (in miles) of its 900-MHz cordless phone is greater than that of its leading competitor. You perform a study using 14 phones from the manufacturer and 16 similar phones from its competitor. The results are shown on the next slide. At = 0.05, is there enough evidence to support the manufacturer’s claim? Assume the populations are normally distributed and the population variances are equal.

- The claim is “the mean range of our cordless phone is greater than the mean range of yours.” So, the null and alternative hypotheses are:
Ho: 1 2 and Ha: 1 > 2 (Claim)

- Because the variances are equal,
d.f. = n1 + n2 – 2

= 14 + 16 – 2

= 28

Because the test is a right-tailed test, d.f. = 28 and = 0.05, the critical value is 1.701. The rejecetion is t > 1.701.

- The standard error is:

- Using the t-Test, the standardized test statistic is:

- The graph at the left shows the location of the rejection region and the standardized test statistic, t. Because t is in the rejection region, you should decide to reject the null hypothesis. At the the 5% level, there is enough evidence to support the manufacturer’s claim that its phone has a greater calling range than its competitor’s.