One sample Case
This case is concerned with only a single sample mean and we want to test the deviation of a sample mean from the population mean. Many hypotheses are tested using a statistical test based on the following formula: test statistic=(obtain sample mean-expected population mean)/(standard error of the mean)
Test on Large Sample (z-test)
Example 1: A manufacturer claims that the average lifetime of his lightbulbs is 3 years or 36 months. The standard deviation is 8 months. Fifty bulbs are selected, and the average lifetime is found to be 32 months. Should the manufacturer’s statement be rejected at 0.01 level of significance?
Solution: Following the steps in hypothesis testing we have to State the null and alternative hypothesis. Mathematically,
1. Ho: μ = 36 months (The average lifetime of lightbulbs is 36 months. Or the average lifetime of lightbulb is not different to 36 months)
2.H_a: μ ≠ 36 months (The average lifetime of lightbulbs is not equal to 36 months. Or the average lifetime of lightbulb is different to 36 months)
3. Level of significance α = 0.01.
4. Select an appropriate test statistic.
5.The test statistic is the z – test, the sample size is greater 30 and the formula is
z-test=(obtain sample mean-population mean)/(standard error of the mean)
Determine the critical value and critical region
Since the alternative hypothesis is nondirectional test, this illustration has two rejection regions one in each tail of the normal curve distribution of the sample mean. Thus, the equivalent critical value is z = ±2.58. Reject Ho if z > 2.58 or z < -2.58.
Compute the value of the test statistics:
Given:
Sample mean = 32 months
Population mean = 36 months
Standard deviation = 8 months
Sample size = 50 bulbs
z-test=(obtain sample mean-population mean)/(σ/√n)
z-test=(32-36)/(8/√50)=-3.54
6.Decision: Since the computed z = - 3.54 is in the rejection region, thus, reject Ho and accept H_a: μ ≠ 36 months
Conclusion: Therefore, the average lifetime of lightbulbs in a certain manufacturing is not 36 months but in fact it is less than 36 months.
Test on Small Sample (t-test)
When the population standard deviation is unknown and the sample size is small, that is, less than 30, the t test is appropriate for testing hypothesis involving means. The formula is
t=(sample mean-population mean)/(standard error of the mean)
Where;
standard error of the mean = s/√n
with df = n – 1 = the degrees of freedom.
Example 2: In order to increase customer service, a muffler repair shop claims its mechanics can replace a muffler in 12 minutes. A time management specialist selected six (6) repair jobs and found their mean time to be 11.6 minutes. The standard deviation of the sample was 2.1 minutes. At 0.025 level of significance, is there enough evidence to conclude that the mean time in changing a muffler is less than 12 minutes?
Solution: Follow the steps in hypothesis testing.
1. State the null and alternative hypothesis. Mathematically,
Ho: μ = 12 minutes
H_a: μ < 12 minutes
2. Level of significance α = 0.025.
3. Select an appropriate test statistic.
The test statistic is the t– test, the sample size is less than 30 and the formula is
t-test=(obtain sample mean-population mean)/(standard error of the mean)
4. Determine the critical value and critical region
Since the level of significance is 0.025 and df = n – 1 = 6 – 1 = 5, and the alternative hypothesis is left – tailed test, then the critical value (t_(α/2)) = -2.571. Reject Ho, if t computed is less than -2.571.
5. Compute the value of the test statistics:
Given:
6. Decision: Since the computed t = -0.47 is in the acceptance region, thus, we fail to reject Ho.
7. Conclusion: Therefore, the data does not provide enough evidence to conclude the mean time in changing a muffler is less than 12 minutes.
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