chapter 12 tests of a single mean when σ is unknown
TRANSCRIPT
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Chapter 12
Tests of a Single Mean When σ is Unknown
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A Research QuestionChildren’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?
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A Research Project16 (n = 16) 6-yr old children are
randomly selected from around townEach 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
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The Data
The 16 kid’s heights were:44, 38, 42, 37, 35, 41, 46, 39,
40, 42, 34, 39, 41, 42, 45, 35
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Hypothesis Test
1.State and Check AssumptionsHeights normally distributed ? -
probably (n = 16 large enough)
Interval level data
Random Sample
Population variance unknown
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Hypothesis Test
2. Null and Alternative Hypotheses
HO : μ = 42 (6-yr old’s height is 42”)
HA : μ < 42 (6 yr-old’s height is less than 42”)
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Hypothesis Test
3.Choose Test StatisticParameter of interest - μ
Number of Groups - 1
Independent Sample
Normally distributed - probably
Variance - unknown
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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
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Single Sample t statistic
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Back to the Hypothesis Test
3.Choose the test statisticParameter of interest - μ
Number of Groups - 1
Independent Samples
Normally distributed - probably
Variance - unknown
One Sample t-test
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Hypothesis Test
4.Set significance levelα = .05critical value is found in table C
What’s a df?
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Degrees of Freedom (df)Degrees of Freedom (df) - the number
of components in a statistic’s calculation that are free to vary
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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
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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
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One Sample t test In a one sample t test the degrees of
freedom are always equal to n - 1– df = n -1
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Back to the Hypothesis test
4.Set significance level and make decision ruleα = .05df = n -1 = 16 - 1 = 15critical 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 HO
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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 HO
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Hypothesis Test
5.Compute test statistic
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Hypothesis Test
6. Draw conclusionsSince our obtained t (-2.236) is less than the critical t (-1.753) we,
Reject HO, and concludeThat our town’s 6-yr olds are smaller, on average, than 6-yr olds in the US
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Careful…a warning
We have rejected the HO 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
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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
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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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Ugghh!!!
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Excel recognizes onlypositive values fora t distribution, but because the t is symmetrical, use the absoute value function (ABS) to find the p-value
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