Inference for a Mean

when you have a “small” sample...

As long as you have a “large” sample….

A confidence interval for a population mean is:

n

sZx *

where the average, standard deviation, and n depend on the sample. Z* depends on the confidence level.

As long as you have a “large” sample….

A test statistic for a population mean is:

where the average, standard deviation, and n depend on the sample. is the value specified in the null.

ns

xZ

/

Example

Random sample of 59 students spent an average of $273.20 on Spring 1998 textbooks. Sample standard deviation was $94.40.

09.2420.27359

4.9496.120.273

We can be 95% confident that the average amount spent by all students was between $249.11 and $297.29.

ExampleA sample of 59 students spent an average of $273.20 on textbooks with a standard deviation of $94.40. Do students spend less than $300 on average?

There is enough evidence, at 0.05 level, to conclude that, on average, students spend less than $300 on textbooks.

015.0)18.2(59/4.94

3002.273)300(

ZPZPXP

What happens if you can only take a “small” sample?

• Random sample of 15 students slept an average of 6.4 hours last night with standard deviation of 1 hour.

• What is the average amount all students slept last night?

• Is the average amount less than 7 hours?

If you have a “small” sample...Replace the Z multiplier with a t multiplier to get:

n

stx *

where “t*” comes from Student t distribution, and depends on the sample size through the degrees of freedom “n-1”.

If you have a “small” sample...

Replace the Z statistic with the t statistic:

Again, “t” follows the Student’s t distribution, which depends on the sample size through the degrees of freedom “n-1”.

ns

xt

/

Student’s t distribution versus Normal Z distribution

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Value

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T-distribution and Standard Normal Z distribution

T with 5 d.f.

Z distribution

Student t distribution

• Shaped like standard normal distribution (symmetric around 0, bell-shaped).

• But, t depends on the degrees of freedom “n-1”.

• And, more likely to get extreme t values than extreme Z values.

Graphical Comparison of t and Z Multipliers

0.90 0.92 0.94 0.96 0.98 1.00

0

1

2

3

4

5

Cumulative Probability

Z o

r T

Mul

tiplie

r T with 5 df

Z distribution

Tabular Comparison of t and Z Multipliers

Confidencelevel

t value with5 d.f

Z value

90% 2.015 1.65

95% 2.571 1.96

99% 4.032 2.58

For small samples, t value is larger than Z value.

So, t interval is longer than a Z interval, and for a given test statistic the P-value is larger.

Back to our CI example!

Sample of 15 students slept an average of 6.4 hours last night with standard deviation of 1 hour.

55.04.615

1145.24.6

n

stx

Need t with n-1 = 15-1 = 14 d.f. For 95% confidence, t14 = 2.145

That is...

We can be 95% confident that average amount slept last night by all students is between 5.85 and 6.95 hours.

Hmmm! Adults need 8 hours of sleep each night.

Logical conclusion:On average, students need more sleep.

(Just don’t get it in this class!)

T-Interval for Mean in Minitab

T Confidence Intervals

Variable N Mean StDev SE Mean 95.0 % CIComb 89 2.011 1.563 0.166 (1.682, 2.340)

We can be 95% confident that the average number of times a “Stat-250-like” student combs his or her hair is between 1.7 and 2.3 times a day.

T- interval in Minitab

• Select Stat.

• Select Basic Statistics.

• Select 1-Sample t…

• Select desired variable.

• Specify desired confidence level.

• Say OK.

And to our HT example!

New sample of 18 students slept an average of 6.01 hrs last night with standard deviation of 1.11 hrs. Do students sleep less than 7 hours on average?

H0: = 7 vs. HA: < 7

If the population mean is 7, how likely is it that a sample of 18 students would sleep an average as low as 6.01 hours?

Or, how likely is it that we’d get a t statistic as low as -3.78?

78.318/11.1

701.6

t

T-test for Mean in MinitabT-Test of the MeanTest of mu = 7.000 vs mu < 7.000

Variable N Mean StDev SE Mean T PSleepHrs 18 6.011 1.113 0.262 -3.77 0.0008

If the population mean was 7, it is not likely (P-value = 0.0008) that we’d get a sample mean as small as 6.011

Reject the null hypothesis. There is enough evidence to conclude that students sleep on average less than 7 hours.

T- test in Minitab

• Select Stat.

• Select Basic Statistics.

• Select 1-Sample t…

• Select desired variable.

• Specify the null mean in “Test mean” box.

• Select the alternative hypothesis.

• Say OK.

What happens as sample gets larger?

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dens

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T-distribution and Standard Normal Z distribution

Z distribution

T with 60 d.f.

Example

Random sample of 64 students spent an average of 3.8 hours on homework last night with a sample standard deviation of 3.1 hours.

Z Confidence Intervals The assumed sigma = 3.10

Variable N Mean StDev 95.0 % CIHomework 64 3.797 3.100 (3.037, 4.556)

T Confidence IntervalsVariable N Mean StDev 95.0 % CIHomework 64 3.797 3.100 (3.022, 4.571)

ExampleRandom sample of 139 students own an average of 12.7 pairs of shoes with a sample standard deviation of 9.6 pairs.

Z-TestTest of mu = 10.000 vs mu > 10.000The assumed sigma = 9.63

Variable N Mean StDev SE Mean Z PShoes 139 12.669 9.625 0.816 3.27 0.0006

T-Test of the MeanTest of mu = 10.000 vs mu > 10.000

Variable N Mean StDev SE Mean T PShoes 139 12.669 9.625 0.816 3.27 0.0007

One not-so-small problem!

• It is only OK to use the t interval for small samples if your original measurements are normally distributed.

Strategy

• If you have a large sample of, say, 30 or more measurements, then don’t worry about normality, and use a t-interval or do a t-test.

• If you have a small sample and your data are normally distributed, then use a t-interval or do a t-test.

• If you have a small sample and your data are not normally distributed, then use nonparametric hypothesis tests.