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Standard Deviation

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10.21 An automobile company thinks that with new designs, it scars will last longer before having a problem. For this reason the company wishes to extend the warranty that comes with the vehicle in hopes of attracting more customers. Before making this change the idea is tested. Prior to the design changes, the cars lasted on the average 43 months before having a major problem. A sample consisting of 50 cars was tested. The cars lasted an average of 44 months before having a major problem. The standard deviation is 2 months.

a. Set up the null and alternative hypotheses to test if average time before having a major problem is longer than 43 months
b. Test your hypotheses using alpha = 0.05
c. Find the p value
d. Based on the p value, what can you conclude about the average time before having a major problem?

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Solution Summary

10.21 An automobile company thinks that with new designs, it scars will last longer before having a problem. For this reason the company wishes to extend the warranty that comes with the vehicle in hopes of attracting more customers. Before making this change the idea is tested. Prior to the design changes, the cars lasted on the average 43 months before having a major problem. A sample consisting of 50 cars was tested. The cars lasted an average of 44 months before having a major problem. The standard deviation is 2 months.

a. Set up the null and alternative hypotheses to test if average time before having a major problem is longer than 43 months
b. Test your hypotheses using alpha = 0.05
c. Find the p value
d. Based on the p value, what can you conclude about the average time before having a major problem?

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Solution. Denote the warranty by X. Then X is a random variable. And we asuume that X follows the normal distribution, X~N(a,b^2), where a is the mean and b is the standard deviation.
<br>
<br>Define the hypothesis.
<br>
<br>H0: a=43,
<br>H1: a is not equal 43.
<br>
<br>Let us consider the statistic T=(Xbar-a)*sqrt(n)/SX, where Xbar is the mean of the sample , ie., ...

Solution provided by:
Changping Wang, MA
Education
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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