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Confidence interval of two sample tests

  1. Mar 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Some car tires can develop what is known as "heel and toe" wear if not rotated after a certain mileage. To assess this issue, a consumer group investigated the tire wear on two brands of tire, A and B, say. Fifteen cars were fitted with new brand A tires and thirteen with brand B tires, the cars assigned to brand at random. (Two cars initially assigned to brand B suffered serious tire faults other than heel and toe wear, and were excluded from the study.) The cars were driven in regular driving conditions, and the mileage at which heal and toe wear could be observed was recorded on each car. For the cars with brand A tires, the mean mileage observed was 24.99 (in 103 miles ) and the variance was 7.75 (in 106 miles2). For the cars with brand B, the corresponding statistics were 32.92 (in 103miles) and 6.47 (in 106 miles2 ) respectively. The mileage before heal and toe wear is detectable is assumed to be Normally distributed for both brands.

    Calculate the pooled variance s2 to 3 decimal places. During intermediate steps to arrive at the answer, make sure you keep as many decimal places as possible so that you can achieve the precision required in this question.

    Determine a 95% confidence interval for μA−μB, the difference in the mean 103 mileages before heal and toe wear for the two brands of tire. Leave your answer to 2 decimal places. (

    2. Relevant equations


    3. The attempt at a solution

    I am fairly sure I know how to do this question, the only issue I have is I am not sure what value to use for n for tire 2. I am assuming we would need to use n = 13 when calculating pooled variance, and n = 11 when calculating the confidence interval? Or would I use n = 11 for everything?

    Any help is appreciated, thanks
     
  2. jcsd
  3. Mar 29, 2015 #2

    RUber

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    I read that to say they initially had 15 and 15, but since 2 were excluded they have 15 and 13. Use 13 for B.
     
  4. Mar 29, 2015 #3

    Ray Vickson

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    I cannot figure out what you mean in parts of the problem statement. The statement that the mean is 24.99 is simple enough, but where does the "103 miles" come into it? The statement that the variance is 7.75 is simple enough, but where does the "106 miles2" come in? Do you mean that a car makes an average of 24.99 103-mile trips (in other words, drives an average of 2,573.97 miles)? Do you mean that the variance is (7.75)(106) = 821.5 miles2?
     
  5. Mar 30, 2015 #4

    vela

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    I think "103" and "106" are supposed to be 103 and 106.
     
  6. Mar 30, 2015 #5

    Ray Vickson

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    Thank you; that makes sense. Too bad the OP could not have been as helpful.
     
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