Z.TEST

Application:

Manufacturing of Lightbulbs

A lightbulb manufacturer claims that their bulbs have an average lifespan of 1,000 hours. From extensive historical data and quality control measures, the company knows that the population standard deviation for the lifespan of their lightbulbs is 80 hours. A new batch of bulbs is produced, and the company wants to verify if the new manufacturing process is still meeting the claimed average lifespan.

To do this, a random sample of 30 lightbulbs is selected from the new batch, and their lifespans are recorded.

• Hypothesis: The company wants to test if the mean lifespan of the new batch is still 1,000 hours.
• • Null Hypothesis (H0​): The true mean lifespan is 1,000 hours.
  • Alternative Hypothesis (Ha​): The true mean lifespan is not 1,000 hours.
• Given Values:
• • Population Mean (μ0​): 1,000 hours (The value we are testing against).
  • Population Standard Deviation (σ): 80 hours (Known from historical data).
  • Sample Size (n): 30
  • Significance Level (α): 0.05 (This is a common value, representing a 5% risk of incorrectly rejecting the null hypothesis).
• Sample Data: The quality control team collects the following lifespan data from the 30 bulbs:

• Calculations: First, we calculate the sample mean ($\bar{x}$) from the collected data.
  $\bar{x}=\frac{1}{n}{\sum }_{i=1}^n{x}_i$
  Sum = 985 + 1050 + 975 + ... + 1020 = 30200
  $\bar{x}$=$\frac{30200}{30}$​=1006.67 hours
• Using the Z.TEST Function: The Z.TEST function in a spreadsheet program would be used with the following syntax: Z.TEST(data_range, population_mean, [population_std_dev]) =Z.TEST({985, 1050, ..., 1020}, 1000, 80)
• Results: The Z.TEST function calculates the one-tailed P-value. Let's calculate the z-score manually to show the result. The formula for the z-score is:
  $z=\frac{\bar{x}-{\mu }_0}{\sigma /\sqrt{n}}$
  Plugging in the values:
  $z=\frac{1006.67-1000}{80/\sqrt{30}}$
  $z=\frac{6.67}{80/5.477}$
  $z=\frac{6.67}{14.606}\approx 0.457$

The Z.TEST function returns the one-tailed P-value. For a z-score of 0.457, the one-tailed P-value is 0.303804338.

• Conclusion: Since our alternative hypothesis is "not equal to" (a two-tailed test), we need to double the one-tailed P-value returned by Z.TEST. Two-tailed P-value = 2×0.303804338=0.607608676.
• Decision Rule: We compare the P-value to the significance level (α).
• If P-value ≤α, we reject the null hypothesis.
• If P-value >α, we fail to reject the null hypothesis.

In this case, 0.607608676>0.05. Therefore, we fail to reject the null hypothesis.

Final Summary Table: