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Chapter 12
Tests of a Single Mean When σ is Unknown
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?
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
The Data
The 16 kid’s heights were:44, 38, 42, 37, 35, 41, 46, 39,
40, 42, 34, 39, 41, 42, 45, 35
Hypothesis Test
1.State and Check AssumptionsHeights normally distributed ? -
probably (n = 16 large enough)
Interval level data
Random Sample
Population variance unknown
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”)
Hypothesis Test
3.Choose Test StatisticParameter of interest - μ
Number of Groups - 1
Independent Sample
Normally distributed - probably
Variance - unknown
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
Single Sample t statistic
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
Hypothesis Test
4.Set significance levelα = .05critical value is found in table C
What’s a df?
Degrees of Freedom (df)Degrees of Freedom (df) - the number
of components in a statistic’s calculation that are free to vary
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
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
One Sample t test In a one sample t test the degrees of
freedom are always equal to n - 1– df = n -1
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
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
Hypothesis Test
5.Compute test statistic
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
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
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
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?
Ugghh!!!
Excel recognizes onlypositive values fora t distribution, but because the t is symmetrical, use the absoute value function (ABS) to find the p-value