Math Problem Statement

A software development company claims that its new project management software has reduced the average time required to complete projects. According to past data, the population standard deviation for project completion time is known to be 15 days. The company wants to test this claim using a 5% significance level. They collect a sample of 64 completed projects and find a sample mean completion time of 82 days. Is there enough evidence to support the company's claim that the new software has reduced the average project completion time?

Solution

This is a hypothesis testing problem, where the company claims that the new project management software has reduced the average project completion time. We will perform a one-tailed z-test to test this claim, as we are interested in seeing if the mean project completion time has decreased.

Steps for hypothesis testing:

  1. State the hypotheses:

    • Null hypothesis (H0H_0): The mean project completion time has not reduced. H0:μ85H_0: \mu \geq 85
    • Alternative hypothesis (HaH_a): The mean project completion time has reduced. Ha:μ<85H_a: \mu < 85 Here, μ\mu represents the true population mean project completion time, and 85 days is the past average project completion time.
  2. Known values:

    • Population standard deviation (σ\sigma) = 15 days
    • Sample size (nn) = 64
    • Sample mean (xˉ\bar{x}) = 82 days
    • Significance level (α\alpha) = 0.05
  3. Test statistic: Since the population standard deviation is known and the sample size is large (n>30n > 30), we can use the z-test formula: z=xˉμ0σnz = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} Where:

    • xˉ\bar{x} = sample mean = 82
    • μ0\mu_0 = population mean under the null hypothesis = 85
    • σ\sigma = population standard deviation = 15
    • nn = sample size = 64
  4. Compute the test statistic: Let's calculate the z-value.

z=82851564z = \frac{82 - 85}{\frac{15}{\sqrt{64}}}

We can now compute this value.The computed test statistic (z-value) is 1.6-1.6.

  1. Determine the critical value: Since this is a one-tailed test (lower tail), and the significance level (α\alpha) is 0.05, we need to find the z-critical value corresponding to a 5% significance level. The critical z-value for a left-tailed test at α=0.05\alpha = 0.05 is approximately 1.645-1.645.

  2. Compare the test statistic with the critical value:

    • The computed z-value is 1.6-1.6.
    • The critical z-value is 1.645-1.645.
  3. Conclusion: Since the computed z-value (1.6-1.6) is greater than the critical z-value (1.645-1.645), we fail to reject the null hypothesis.

Final Decision:

There is not enough evidence at the 5% significance level to support the company's claim that the new software has reduced the average project completion time.


Do you need any further details or have questions?

5 Related Questions:

  1. What is the difference between one-tailed and two-tailed tests?
  2. How is the significance level (α\alpha) chosen in hypothesis testing?
  3. What is the importance of the population standard deviation in z-tests?
  4. How would the result change if the sample size were smaller?
  5. What is the impact of increasing the significance level?

Tip:

Always check the assumptions for a z-test, such as whether the population standard deviation is known and if the sample size is large enough for the Central Limit Theorem to apply.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
One-Tailed Z-Test
Statistics

Formulas

z = (x̄ - μ₀) / (σ / √n)

Theorems

Central Limit Theorem
Z-Test

Suitable Grade Level

College level, Statistics course