Steele Electric Products Inc. assembles cell phones. For the last 10 days, Mark Nagy completed a mean of 39 phones per day, with a standard deviation of 2 per day. Debbie Richmond completed a mean of 38.5 phones per day, with a standard deviation of 1.5 per day. At the .05 significance level, can we conclude that there is more variation in Mark’s daily production?
![Check Mark](/static/check-mark.png)
Find whether there is more variation in Person M’s daily production.
Answer to Problem 1SR
There is no more variation in Person M’s daily production.
Explanation of Solution
Here,
The null and alternative hypotheses are stated below:
That is, the variation in Person M’s daily production is less than or equal to the variation in Person R’s production.
That is, the variation in Person M’s daily production is more than that of Person R’s production.
Step-by-step procedure to obtain the test statistic using MINITAB software:
- Choose Stat > Basic Statistics > 2 Variance.
- Under Data, choose Sample standard deviation.
- In First, enter 10 under Sample size.
- In First, enter 2 under Standard deviation
- In Second, enter 10 under Sample size.
- In Second, enter 1.5 under Standard deviation
- Check Options, enter Confidence level as 95.0.
- In Hypothesized ratio StDev 1 / StDev 2
- Choose greater than in alternative.
- Click OK in all dialog boxes.
- Output obtained using MINITAB is represented as follows:
- From the above output, the F test statistic value is 1.78 and the p-value is 0.202.
Decision Rule:
If the p-value is less than the level of significance, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.
Conclusion:
The level of significance is 0.05.
From the output, the p-value is 0.202.
The p-value is greater than the level of significance 0.05. Hence, one is failed to reject the null hypothesis at the 0.05 significance level.
Therefore, there is no more variation in Person M’s daily production.
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Chapter 12 Solutions
STATISTICAL TECHNIQUES FOR BUSINESS AND
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