To determine if there is enough evidence to support the claim, we perform a Z-test for the mean with the following hypothesis:

Step 1: Define the hypotheses

• Null Hypothesis ($H_0$): $\mu \geq 120$ minutes (mean delivery time is not less than 120 minutes)
• Alternative Hypothesis ($H_a$): $\mu < 120$ minutes (mean delivery time is less than 120 minutes)

Step 2: Identify the given values

• Population standard deviation ($\sigma$) = 30 minutes
• Sample mean ($\bar{x}$) = 100 minutes
• Sample size ($n$) = 49
• Population mean ($\mu_0$) = 120 minutes
• Significance level ($\alpha$) = 0.05

Step 3: Compute the test statistic

The formula for the Z-test statistic is: $Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}$ Substitute the values: $Z = \frac{100 - 120}{\frac{30}{\sqrt{49}}}$ $Z = \frac{-20}{\frac{30}{7}}$ $Z = \frac{-20}{4.29} \approx -4.66$

Step 4: Determine the critical value

For a one-tailed test at a significance level of $0.05$, the critical value ($Z_{\text{crit}}$) is: $Z_{\text{crit}} = -1.645$

Step 5: Decision rule

• If $Z < Z_{\text{crit}}$, reject $H_0$.
• Otherwise, fail to reject $H_0$.

Since $Z = -4.66$ is much smaller than $-1.645$, we reject $H_0$.

Step 6: Conclusion

There is sufficient evidence to support the claim that the mean delivery time is less than 120 minutes at the $0.05$ significance level.

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Would you like a detailed explanation of any step?
Here are 5 related questions you might explore:

1. How does changing the sample size affect the Z-test result?
2. What would the conclusion be if the significance level were $0.01$?
3. How is a Z-test different from a t-test?
4. How do you compute a p-value from the Z-score?
5. What are the assumptions of a Z-test?

Tip: For small sample sizes ($n < 30$), use the t-test instead of the Z-test unless the population standard deviation is known.