Sanderson M. Smith

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INFERENCE EXAMPLE USING t DISTRIBUTION
(ONE-SAMPLE PROCEDURE)

In this example, there is no population standard deviation (s) provided. This parameter is unknown. The statistic s (computed from a sample) is used in place of s.

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A machine is calibrated to produce washers with a thickness of 0.05 inches. A random sample of ten measurements yields the following thicknesses (unit = inches).

.053

.054

.049

.050

.054

.053

.048

.055

.054

.053

At the 5% level of significance, test the hypothesis that the machinery is functioning properly.

H_o: m = 0.05 (sample came from population with m = 0.05).
H_a: m is not equal to 0.05 (sample did not come from population with m = 0.05)
Type of test: 2-tail.
Degrees of freedom: 10 - 1 = 9.
Mean of sample: x(bar) = 0.0523.
Calculated s from sample: s = 0.0024.

Standard error for sample means: s/SQRT(N) = 0.0024/SQRT(10) = 0.00076.

Calculation of sample t statistic: t = (0.0523-0.05)/0.00076 = 3.03.

Critical values of t with 9 degrees of freedom: t < -2.262 and t > 2.262.

Conclusion: Calculated value of t is in the critical region. At the 5% level of significance, we reject Ho. At the 5% level of significance, there is evidence to suggest that the machinery is not functioning properly.

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Alternate approach using confidence intervals: Using the data above, the 95% confidence interval for the population mean is

0.0523 plus/minus (2.262)(0.00076) = 0.0523 plus minus 0.00172.

This interval is [0.0506,0.0540]. Since this interval does not contain 0.05, we would reject H_o at the 5% level of significance.

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Alternate approach using probability: With 9 degrees of freedom, the probability of obtaining a value t = 3.03 if H_o is true is 0.007, or about 0.7%. [This is tcdf(3.03,1E99,9) on the TI-83 calculator]. This would cause rejection of H_o at the 5% level of significance.

Note: At the 1% level of significance (2-tail), H_o would not be rejected.

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