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