Explanation Consider the following scenario, “A potato chip manufacturer produces bags of potato chips that are supposed to have a net weight of 326 grams. Because the chips vary in size, it is difficult to fill the bags to the exact weight desired. However, the bags pass inspection so long as the standard deviation of their weights is no more than 3 grams. A quality control inspector wished to test the claim that one batch of bags has a standard deviation of more than 3 grams, and thus does not pass inspection. If a sample of 25 bags of potato chips is taken and the standard deviationis found to be 3.4 grams” Step 1: State the null and alternative hypothesis. A maunfacurer think standard deviation of their weights is no more than 3 grams. That is the research hypothesis H a is that the one batch of bags has a standard deviation of more than 3 grams which is wrriten mathematically as σ > 3 . The logical opposite is σ ≤ 3 . Thus the hypotheses are tested as follows, The Null hypothesis: H 0 : σ ≤ 3 The altenative hypotheis: H a : σ > 3 Step 2: Determine which distribution to use for the test statistic, and state the level of significance. Since we are testing a population variance and we are told we can safely assume that all necessary conditions are met, we can use the chi- square distribution and thus the χ 2 -test statistic with the given level of significance of α = 0.05 . Step 3: Gather data and calculate the necessary sample statistics. The test statistic formula: Test statistic for a hypothesis test for a population variance or population stadard deviation When the sample taken is a simple random sample and the population distribution is approximately normal, the test statistic for a hypothesis test for a population variance or population standard deviation is given by, χ 2 = ( n − 1 ) s 2 σ 2 Where n is the smple size, s 2 is the sample variance, and σ 2 is the presumed value of the population variance (or square of the prsumed value of the population standard deviation) from the null hypothesis The given information is, n = 25 , s = 3.4 g a n d σ = 3 . Substitute these values in the test statistic formula to get the following, χ 2 = ( n − 1 ) s 2 σ 2 = ( 25 − 1 ) × 3.4 2 3 2 = 24 × 11.56 9 = 277.44 9 = 30.8267 Step 4: Draw a conclusion and interpret the decision

2nd Edition

ISBN: 9781932628678

Author: Carolyn Warren; Kimberly Denley; Emily Atchley

Publisher: Hawkes Learning Systems

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30. An individual who has automobile insurance from a certain company is randomly selected. Let Y be the num- ber of moving violations for which the individual was cited during the last 3 years. The pmf of Y isy | 1 2 4 8 16p(y) | .05 .10 .35 .40 .10 a.Compute E(Y).b. Suppose an individual with Y violations incurs a surcharge of $100Y^2. Calculate the expected amount of the surcharge.

24. An insurance company offers its policyholders a num- ber of different premium payment options. For a ran- domly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows: F(x)=0.00 : x < 10.30 : 1≤x<30.40 : 3≤ x < 40.45 : 4≤ x <60.60 : 6≤ x < 121.00 : 12≤ x a. What is the pmf of X?b. Using just the cdf, compute P(3≤ X ≤6) and P(4≤ X).

Chapter 10 Solutions

Beginning Statistics, 2nd Edition

Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

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Chapter 10 Solutions

To find:

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