Test Concerning Means:Large Sample

Test Concerning Means

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 Concerning Means: 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

1. State the null and alternative hypothesis. Mathematically,

Ho: μ = 36 months (The average lifetime of lightbulbs is 36 months. Or the average lifetime of lightbulb is not different to 36 months)

Ha: 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)

2. Level of significance α = 0.01.

3. Select an appropriate test statistic.

The test statistic is the z – test, the sample size is greater 30 and the formula is

4. 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.

5. Compute the value of the test statistics:

Given:

Sample mean = 32 months

Population mean = 36 months

Standard deviation = 8 months

Sample size = 50 bulbs

6. Decision: Since the computed z = - 3.54 is in the rejection region, thus, reject Ho and accept Ha: μ ≠ 36 months

7. Conclusion:

Therefore, the average lifetime of lightbulbs in a certain manufacturing is not 36 months but in fact it is less than 36 months.