Math Problem Statement
An electronics manufacturing process has historically had a mean completion time of 70 minutes. It is claimed that, due to improvements in the process, the mean completion time,, is now less than 70 minutes. A random sample of 19 completion times using the new process is taken. The sample has a mean completion time of 65 minutes, with a standard deviation of 12 minutes.
Assume that completion times using the new process are approximately normally distributed. At the 0.10 level of significance, can it be concluded that the population mean completion time using the new process is less than 70 minutes?
Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (a)State the null hypothesis H0 and the alternative hypothesis H1. (b)Determine the type of test statistic to use. (c)Find the value of the test statistic. (Round to three or more decimal places.) (d)Find the critical value. (Round to three or more decimal places.) (e)Can it be concluded that the mean completion time using the new process is less than 70 minutes?
Solution
To address this problem, we'll perform a one-tailed t-test to determine if the mean completion time of the new process is statistically less than 70 minutes.
(a) State the Hypotheses
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Null Hypothesis (H₀): The population mean completion time is equal to 70 minutes.
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Alternative Hypothesis (H₁): The population mean completion time is less than 70 minutes.
(b) Determine the Type of Test Statistic
Since the sample size is small (n = 19), and the population standard deviation is unknown, we will use the t-statistic for this hypothesis test.
(c) Find the Value of the Test Statistic
The test statistic is calculated using the formula: where:
- = sample mean = 65 minutes
- = hypothesized population mean = 70 minutes
- = sample standard deviation = 12 minutes
- = sample size = 19
Substituting the values:
Calculating this:
(d) Find the Critical Value
For a one-tailed test at the 0.10 significance level with degrees of freedom, we need to find the critical value .
Using a t-distribution table or calculator, the critical value is approximately:
(e) Conclusion
Compare the calculated test statistic to the critical value:
- If , reject the null hypothesis .
- If , fail to reject the null hypothesis .
Here, is less than the critical value of , so we reject the null hypothesis.
Conclusion: Yes, at the 0.10 level of significance, there is sufficient evidence to conclude that the mean completion time using the new process is less than 70 minutes.
Would you like a detailed explanation of any step, or do you have any further questions?
Here are some related questions to consider:
- How would the conclusion change if the significance level were 0.05 instead of 0.10?
- What assumptions underlie the use of the t-test in this context?
- How does the sample size affect the reliability of the test?
- What would the test statistic be if the sample mean was 68 minutes instead of 65?
- What is the difference between a one-tailed and two-tailed t-test?
- Why is the critical value different for different degrees of freedom?
- How would the test result differ if the standard deviation were 8 minutes instead of 12?
- What steps would you take to ensure that the process improvements are statistically significant over time?
Tip: Understanding the context of the significance level () is crucial. A lower level (e.g., 0.01) makes it harder to reject the null hypothesis, reducing the risk of a Type I error but increasing the chance of a Type II error.
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Math Problem Analysis
Mathematical Concepts
Hypothesis Testing
T-Distribution
One-Tailed Test
Formulas
T-statistic formula: t = (x̄ - μ) / (s / √n)
Theorems
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Suitable Grade Level
Advanced Undergraduate
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