Interval Estimates for Small Samples:

The Student's t-distribution

When the sample size is small, the critical values for confidence intervals are determined by the Student's t-distribution, so they are called t-values rather than z-values.

The probabilities for this distribution are defined strictly by "degrees of freedom" or
the number of data values available to estimate the population's standard deviation.

A sample of size n has n – 1 degrees of freedom.

The t-value formula is identical to the one for the z-value

...... s = sample standard deviation

Confidence Interval Estimate for µ when .

Estimation with Larger Samples and Student's t -distribution:

The Student's t -distribution is generally used on small samples with n < 30. An increase in the sample size affects both the stardard error and the number of degrees of freedom. As the degrees of freedom increase, the t-value approaches the z-value for the same level of confidence. When the sample size becomes extremely large, the t-distribution converges to the z-distribution. If n > 30 use the Normal Distribution.

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Example:

To test the durability of a new paint used for the white lines on the highway, the company paints 8 strips of it on a busy road and counts the number of "crossings" it takes to begin deteriorating the paint surface. Rounded to the nearest hundreth, here's the data:

142,600

167,800

136,500

108,300

126,400

133,700

162,000

149,400

Construct a 95% confidence interval for the average number of crossings it takes to deteriorate the paint surface. Round to the nearest hundred.

Solution: n = 8 ...... ...... s = 19, 200 (by formula for sample st. dev.)

Since n = 8, there are 8 – 1 = 7 degrees of freedom at 95%

The critical t-value for 95% and 7 degrees of freedom = 2.365.

140,800 ± 16,000 =

the confidence interval is 124,700 < µ < 156, 900.

The surface of the paint will begin to deteriorate after the lines have been crossed between 124,700 and 156,900 times.

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Formulae for Small Sample Confidence Intervals

Property	Confidence Interval for Mean ( )
Sample Size when use n – 1 degrees of freedom when is unknown
Error when .
Parameters use s when is unknown

Practice (view student's t-table)

1.

Tim weighed himself once a week for several years. Last month his four measurements (in pounds) were: ...... 190.5 ......189.0 ......195.5 ......187.0

a) Construct a 90% confidence interval for his mean weight for last month.

b) Suppose that Tim wants 99% confidence rather than 90%. Reconstruct the confidence interval for all 4 measurements.

c) Tim now wants to estimate his monthly weight accurate to within 2 pounds, with 95% confidence. What sample size does he need to achieve this?

2.

The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan.
A random sample of ten employees reveals that the family dental expenses for the preceding year had a mean of $261 and a standard deviation of $139.
Set up a 90% confidence interval estimate of the mean family dental expenses for all employees of this corporation.

3.

A random sample of 16 summer days in Montreal, showed the mean level of CO was 4.9 ppm. The standard deviation was 1.2 ppm. Based on this, construct a 95% confidence interval estimate of the true mean level of CO in summer in Montreal.

Solutions

1.

Tim weighed himself once a week for several years. Last month his four measurements (in pounds) were: ...... 190.5 ......189.0 ......195.5 ......187.0

a) Construct a 90% confidence interval for his mean weight for last month.

Solution: with 4 sample statistics, we'll use the t-distribution.

data:

with 3 degrees of freedom and alpha = 10%, the t-value is 2.353

the 90% confidence interval for his mean weight is (186.23, 194.77).

b) Suppose that Tim wants 99% confidence rather than 90%. Reconstruct the confidence interval for all 4 measurements.

Solution:

Now t = 5.841 for a 99% interval:

So the 99% confidence interval is (179.9, 201.1)

c) Tim now wants to estimate his monthly weight accurate to within 2 pounds, with 95% confidence. What sample size does he need to achieve this?

Solution:

Now t = 3.182 for a 95% with 3 dof :

a sample of 34 measurements is required.

2.

The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan.
A random sample of ten employees reveals that the family dental expenses for the preceding year had a mean of $261 and a standard deviation of $139.
Set up a 90% confidence interval estimate of the mean family dental expenses for all employees of this corporation.

Solution:

data: n = 10 ............ s = $139 ......9 dof ...... t-value = 1.833,

= 261 ± 80.57

$ 180.43 < µ < $ 341.57

3.

A random sample of 16 summer days in Montreal, showed the mean level of CO was 4.9 ppm. The standard deviation was 1.2 ppm. Based on this, construct a 95% confidence interval estimate of the true mean level of CO in summer in Montreal.

Solution:

Since n = 16, this a small sample t-distribution situation.

We know that , s = 1.2, and

degrees of freedom = 15, so t_a/2, 15 = 2.131,

so our confidence limits are 4.9 ± 2.131(0.3) = 4.9 ± 0.639

the confidence interval for the mean level of CO is [ 4.26, 5.54 ] parts per million.

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Statistics MathRoom Index

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Student's t Distribution Probabilities (t-scores)

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Statistics MathRoom Index

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