Fc 2 2. Goodness-of-fit tests Normal population Aa Aa Manufacturing processes, s
ID: 3235661 • Letter: F
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Fc 2
2. Goodness-of-fit tests Normal population Aa Aa Manufacturing processes, such as coin minting, are subject to small variations due to variations in materials, temperature, and humidity. The variations in materials, temperature, and humidity from their norms are just as likely to be positive as negative but are more likely to be small than large. Consider the € 1 coin issued by Belgium. Realizing that there are small variations in the minting process and random error in the weighing process, it might be reasonable to assume that the population of coin weights is normally distributed. Let's confront this assumption with sample data and see how it fares. A random sample of 250 Belgian € 1 coins was selected. Each of the 250 coins was weighed and its weight (in grams) recorded The sample mean X is 7.520 grams, and the sample standard deviation s is 0.036 grams. The questions that follow walk you through the steps of a test of the hypothesis that the population of weights of Belgian € 1 coins has a normal distribution with a mean of 7.520 grams (the sample mean) and a standard deviation of 0.036 grams (the sample standard deviation). Note that you are using the sample mean as an estimate of the population mean and the sample standard deviation as an estimate of the population standard deviation [Data source: A sample of size n 250 was randomly selected from the sample of size n 2,000 in the Journal of Statistics Education data archive, euroweight.dat data set.] Select a Distribution DistributionsExplanation / Answer
1st cutoff point is where the left tail has a probability of 1*(100%/10) = 10%, p = 0.1, cutoff point = 7.474
5th cutoff point is where the left tail has a probability of 5*(100%/10) = 50%, p = 0.5, cutoff point = 7.520
8th cutoff point is where the left tail has a probability of 8*(100%/10) = 80%, p = 0.8, cutoff point = 7.550
The expected frequency for category 9 is 0.1*250 = 25 and the contribution of category 9 to chi square statistic is
(17-25)2/25 = 2.56
The combined contribution of all the categories besides catrgory 9 is 9.04. The chi square test statistic is therefore
9.04 + 2.56 = 11.6
The critical value is 21.666 and the null hpothesis is not rejected.
Note: Expected frequency counts. The expected frequency counts at each level of the categorical variable are equal to the sample size times the hypothesized proportion from the null hypothesis
Test statistic. The test statistic is a chi-square random variable (2) defined by the following equation.
2 = [ (Oi - Ei)2 / Ei ], where Oi is the observed frequency count for the ith level of the categorical variable, and Ei is the expected frequency count for the ith level of the categorical variable.
Degrees of freedom. The degrees of freedom (DF) is equal to the number of levels (k) of the categorical variable minus 1: DF = k - 1
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