2. Goodness-of-fit tests Normal population Aa Aa E Manufacturing processes, such

Question

2. Goodness-of-fit tests Normal population Aa Aa E 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 sma 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 7.510 grams, and the sample standard deviation s is 0.033 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 7.510 grams (the sample mean) and a standard deviation of 0.033 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 Distributions You decide to divide the normal distribution into 10 equal-probability intervals. Some of the interval boundary points have already been computed for you; they are displayed in the table below. Three cutoff points are missing from the table. Use the preceding Distributions tool to find the missing cutoff points, and enter them in the table

Explanation / Answer

The expected frequency for category 3 is 25 and the contribution of category 3 to the chi square test statistic is 0.04

Calculations for expected frequencies...ie.(22+35+26+29+26+18+20+22+24+28)/10 = 250/10 = 25

For catergory 3 and for each category expected frequency will be 25

and Chi sq statistic for category 3 is obtained by

(Observed frequency -Expected frequency)2/ Exp. freq = (26-25)2/25 = 0.04

The combined contribution of all the categories besides category 3 is 8.76. The chi square test statistic is therefore 8.8 and its p value is 0.45594.

Calculations: chis square test statistic is = (8.76+0.04) = 8.8

The critical value is 21.66599, and the null hypothesis is there is no variations in the minting process of coin

(this is the critical value of chi square for 9df )

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