Math Problem Statement
2.1 A car manufacturer introduces a new method of assembling a particular component. The old method had a mean assembly time of 42 minutes. The manufacturer would like the assembly time to be as short as possible. A random sample of assembly times (minutes) taken after the new method had become established was: 25 39 48 41 47 42 35 34 35 Test at 5% level of significance if the manufacturer's expectation was met. [8]
Solution
To test whether the new method has significantly reduced the assembly time, we can perform a one-sample t-test. Here's a step-by-step guide:
1. State the Hypotheses
We want to test if the mean assembly time with the new method is less than the mean assembly time of 42 minutes with the old method. Therefore, we set up the hypotheses as follows:
- Null Hypothesis (H₀): The mean assembly time with the new method is 42 minutes or more.
- Alternative Hypothesis (H₁): The mean assembly time with the new method is less than 42 minutes.
2. Select the Significance Level
The significance level is given as 5%, or 0.05.
3. Calculate the Sample Mean ()
The sample data provided is: 25, 39, 48, 41, 47, 42, 35, 34, 35.
To find the sample mean:
Where is the number of samples.
4. Calculate the Sample Standard Deviation (s)
First, compute the deviations from the mean for each sample, square them, sum them, and then divide by to get the variance. Finally, take the square root to obtain the standard deviation:
Let's compute this:
Calculating the exact standard deviation:
5. Calculate the Test Statistic (t)
The test statistic for a one-sample t-test is calculated as:
Where:
- (the population mean under the null hypothesis)
Substituting the values:
6. Determine the Critical Value
For a one-tailed test at a significance level of 0.05 and degrees of freedom , the critical t-value can be found in a t-distribution table. For , the critical t-value is approximately:
7. Make the Decision
- If the calculated t-value is less than the critical t-value, we reject the null hypothesis.
- If the calculated t-value is greater than the critical t-value, we fail to reject the null hypothesis.
Here, is greater than , so we fail to reject the null hypothesis.
8. Conclusion
At the 5% significance level, there is not enough evidence to conclude that the new assembly method reduces the mean assembly time below 42 minutes. Therefore, the manufacturer's expectation was not statistically met.
Would you like a more detailed breakdown or have any questions?
Related Questions:
- How would the test change if the sample size were larger?
- What if the sample had more variance? How would that affect the test result?
- How do we interpret the p-value in this context?
- Could we use a different type of test for this problem? Why or why not?
- How does the choice of significance level () affect the conclusions?
Tip:
Always verify the assumptions of the statistical test you're using. For the t-test, check for normality and that the data is approximately symmetric, especially with small sample sizes.
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Math Problem Analysis
Mathematical Concepts
Hypothesis Testing
One-sample t-test
Significance Level
Formulas
t-test formula
Standard deviation formula
Theorems
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Suitable Grade Level
Advanced Undergraduate
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