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Goodness of fit Test

You have collected data on a variable, and you want to determine if a normal distribution is a reasonable model for these data. The following table shows how many of the values fall within certain ranges of z values for these data Category z score below -2 Count 48 67 146 248 125 score from -2 to less than -1.5 z score from -1.5 to less than -1 score from -1 to less than -0.5 z score from-0.5 to less than 0 z score from 0 to less than z score from 1 to less than 1.5 z score from 1.5 to less thar z score of 2 and above Total 47 25 19 1000 Perform a hypothesis test to determine if a normal distribution is an appropriate model for these data. Use a significance level of 5%.

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We have been asked to use 5% level of significance, so level of significance would be 0.05 We would like to test, whether the normal distribution is an appropriate model for the data or not. The null hypothesis states that a population parameter is equal to a hypothesized value, and the alternative hypothesis states that a population parameter is smaller, greater, or different than the hypothesized value in the null hypothesis Ho: The normal distribution is appropriate model Hi: The normal distribution is not appropriate model To find the chi-square statistic of the sample, we need to calculate the expected frequencies from given z score category i.e. find out the normal probabilities for each category and multiply each probability value by 813 (total count) **Note: The total of count of categories is 813 not 1000** To calculate standard norma (left tailed) probabilities z; we can use -NORM.S.DIST(z,cummulative) excel function e.g. for z score below -2 use NORM.S.DIST(-2,TRUE) and for z score from -2 to less than -1.5 use NORM.S.DIST(-1.5,TRUE)- NORM.S.DIST(-2,TRUE) for z score above 2 use 1- NORM.S.DIST(2,TRUE) [for z score above right tailed probability we need to subtract the left tailed probability from 1] Category z score below -2 z score from -2 to less than -1.5 z score from -1.5 to less than -1 z score from -1 to less than -0.5 z score from -0.5 to less than 0 z score from 0 to less than 1 z score from 1 to less than 1.5 z score from 1.5 to less than 2 z score above 2 Total Normal probabiltyExprected count 18.49586 35.8184 74.67247 121.8543 155.659 277.513;3 74.67247 35.8184 18.49586 813 Count 48 67 146 248 125 0.02275 0.044057 0.091848 0.14988.2 0.19146.2 0.34134!5 0.091848 0.044057 0.02275 1 47 25 19 813 The chi square test statistic is given by, (Oi - Ei)2 Where Oi would be observed count and Ei would be expected count Subtract each Ei from Oi, square the difference, divide it by Ei, and then find out the total(Oi - Ei)A2)/Ei Category z score below -2 z score from -2 to less than -1.5 z score from -1.5 to less than -1 z score from -1 to less than -0.5 z score from -0.5 to less than 0 z score from 0 to less than 1 z score from 1 to less than 1.5 z score from 1.5 to less than 2 z score above 2 Total Oi Ei 48 67 146 248 125 18.49586 35.8184 74.67247 121.8543 155.659 277.513;3 74.67247 35.8184 18.49586 813 47.06429 27.14505 68.13243 130.588.2 6.038669 129.418.2 10.25499 3.26753:1 0.013741 421.9232 47 25 19 813 So, x2 -421.9232 Chi-square will have k-1 9-1 (categories) 8 degrees of freedom, where k is the number of classes The P-value is the probability that a chi-square statistic having 8 degrees of freedom is more extreme than 421.9232. i.e. P( χ2 > 421.9232). we can use-CHisQDISTRT(x, deg-freedom) excel function. Type CHISQ.DIST.RT 421.9232,1). So p-value would be 3.81232E-860 Since, p-value < 0.05 (5% level of significance) we reject the null hypothesis Null hypothesis is got reject, so we accept the alternative hypothesis, Hi: The normal distribution is not appropriate model

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