One-tail Z-test is used to determine whether the sample mean is significantly greater or less than the population mean, based on the standard normal distribution. Here's an example problem with a solution in steps:

Question: A beverage company claims that the mean sugar content in its soda bottles is 12 ounces. A consumer group suspects that the mean sugar content is actually less than 12 ounces. A random sample of 36 soda bottles is selected, and the mean sugar content is found to be 11.8 ounces with a standard deviation of 0.5 ounces. Using a significance level of 0.05, is there enough evidence to support the consumer group's claim?

Solution in Steps: Step 1: State the null and alternative hypotheses.

• Null hypothesis (H0): μ = 12 (The mean sugar content is 12 ounces)
• Alternative hypothesis (H1): μ < 12 (The mean sugar content is less than 12 ounces)

Step 2: Determine the significance level (α).

• Given significance level α = 0.05

Step 3: Calculate the test statistic (Z).

• Z = (X̄ - μ) / (σ / √n) where X̄ = sample mean, μ = population mean, σ = population standard deviation, n = sample size
• Z = (11.8 - 12) / (0.5 / √36) = (-0.2) / (0.5 / 6) = (-0.2) / 0.0833 ≈ -2.40

Step 4: Determine the critical value or p-value.

• For a one-tail test at α = 0.05, the critical Z-value is -1.645 (obtained from the standard normal distribution table) or p-value can be calculated using statistical software.

Step 5: Make a decision.

• If |Z| > critical value or p-value < α, reject the null hypothesis.
• | -2.40 | > 1.645, so we reject the null hypothesis.

Step 6: Conclusion

• Since the calculated Z-value (-2.40) is less than the critical value (-1.645) and p-value < α, there is enough evidence to support the consumer group's claim that the mean sugar content is less than 12 ounces at a 5% significance level.

This represents a typical approach for conducting a one-tail Z-test and interpreting the results.