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Chapter 12 Tests of a Single Mean When σ is Unknown.

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Presentation on theme: "Chapter 12 Tests of a Single Mean When σ is Unknown."— Presentation transcript:

1 Chapter 12 Tests of a Single Mean When σ is Unknown

2 A Research Question Children’s growth is stunted by a number of chemicals (lead, arsenic, mercury) The tap water in the local community contains a bit of each of these chemicals Are children in this town smaller than other children their age?

3 A Research Project 16 (n = 16) 6-yr old children are randomly selected from around town Each child’s height is measured In the US the average height of 6-yr olds is 42” (μ = 42) The variance of 6-yr-old’s height, however is not known

4 The Data The 16 kid’s heights were: 44, 38, 42, 37, 35, 41, 46, 39, 40, 42, 34, 39, 41, 42, 45, 35

5 Hypothesis Test 1.State and Check Assumptions Heights normally distributed ? - probably (n = 16 large enough) Interval level data Random Sample Population variance unknown

6 Hypothesis Test 2. Null and Alternative Hypotheses H O : μ = 42 (6-yr old’s height is 42”) H A : μ < 42 (6 yr-old’s height is less than 42”)

7 Hypothesis Test 3.Choose Test Statistic Parameter of interest - μ Number of Groups - 1 Independent Sample Normally distributed - probably Variance - unknown

8 What do we do? z-test requires that we know the population standard deviation (σ) Can we use s as a substitute for σ? Not with a z statistic, but… We can use s with a t statistic (Student’s t) and a t sampling distribution

9 Single Sample t statistic

10 Back to the Hypothesis Test 3.Choose the test statistic Parameter of interest - μ Number of Groups - 1 Independent Samples Normally distributed - probably Variance - unknown One Sample t-test

11 Hypothesis Test 4.Set significance level α =.05 critical value is found in table C What’s a df?

12 Degrees of Freedom (df) Degrees of Freedom (df) - the number of components in a statistic’s calculation that are free to vary

13 df Explained If you have a M = 10 obtained from 5 scores, what are the scores? Let’s say the first four are 15, 10, 11, and 5 –in this case the last score has to be 9, in order to have a mean of 10 As a second example, let’s say the first four are 8, 14, 3, and 11 –the last score has to be 14 in order to have a mean of 10

14 df Explained Therefore, the first 4 scores can vary, the fifth score is not free to vary - it must take on some value (in order to maintain the mean of 10) In our example, there are 4 degrees of freedom The first four scores can take on any value (they are free to vary), but that last one is fixed in order to maintain the mean

15 One Sample t test In a one sample t test the degrees of freedom are always equal to n - 1 –df = n -1

16 Back to the Hypothesis test 4.Set significance level and make decision rule α =.05 df = n -1 = 16 - 1 = 15 critical value at.05 of t(15) = 1.753 (read: “critical value at.05 of a t test with 15 degrees of freedom is 1.753”) But, since we have a directional hypothesis (< 42), then the critical value is -1.753 Thus, if our computed t ≤ -1.753, we reject H O

17 Or… If we compute the p-value associated with our t, with 15 df, we can state the decision rule as: –If p ≤ α, Reject the H O

18 Hypothesis Test 5.Compute test statistic

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22 Hypothesis Test 6. Draw conclusions Since our obtained t (-2.236) is less than the critical t (-1.753) we, Reject H O, and conclude That our town’s 6-yr olds are smaller, on average, than 6-yr olds in the US

23 Careful…a warning We have rejected the H O and concluded that our town’s 6-yr-olds are smaller, on average, than 6-yr-olds in the US But, we are not allowed, in this case, to conclude that it is because of chemicals in the water, or any other cause

24 Alternative Explanations There are likely many causes for children’s small stature, not limited to: –Genetics –Diet –Environmental contaminants –Chemicals in ground water –Etc. The hypothesis test allows us to conclude that these children are smaller, on average, but does not allow us to say why

25 Before we move on… Although we already rejected the null hypothesis, We can determine the actual probability of our results if the null hypothesis were true (p-value) We know that it is less than.05, but how much less?

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27 Ugghh!!!

28 Excel recognizes only positive values for a t distribution, but because the t is symmetrical, use the absoute value function (ABS) to find the p-value

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