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Question 6(07) 8.57 points Exercise 12-25 Algo Consider the tolowing sample data

ID: 3375421 • Letter: Q

Question

Question 6(07) 8.57 points Exercise 12-25 Algo Consider the tolowing sample data with mean and standard deviation of 22 0 and 68, respectively Use Table 3 Less than 10 10 up to 20 20 up to 30 30 or more 35 87 n 210 fit test for normality, specity the competing hypotheses in order to determine whether or not the data are normaly distributed Ho The data are normally distributed with a mean of 22 0 and a standard deviation or 68 Ha The data are not normally distributed with a mean of 22 0 and a standard deviation of 6 8 O Ho The data are not normally distributed with a mean of 22 0 and a standard deviation of 6.8, HA The data are normally distributed with a mean of 22 0 and a standard deviation of 6 8 b. At the 1 0% sggnificance level, what is the decision rule? (Round your answer to 3 decimal places.) c. Calculiate the value of the test statistic (Round the z value to 2 decimal places, all other intermediate values to at least 4 decimal places and final answer to 2 decimal places) Test statistic d. What is the conclusion? Reject Ho the data are not normally distnibuted Do not reject Ho the data are normally distr O Type here to search DOLL FT F2 F3 F6 F7 F8

Explanation / Answer

Solution: Solving using Excel

The chi-square test statistic is 105.708 which is less than the critical value of CHIINV(.01,3) = 0.1148, and so we can conclude that there is not good fit for normal distribution.

(a) (A)

(b) 0.1148

(c)105.71

(d) Reject H0, data are not normally distributed.

Class P(a<x<b) Observed Freq.(O) Expected_Freq (E) (O-E)^2/E Less than 10 0.038806605 35 8.149386993 88.46744 10 up to 20 0.345527399 87 72.5607537 2.873342 20 up to 30 0.495962557 66 104.152137 13.97557 30 or more 0.119703439 22 25.13772227 0.391654 Total 210 210 105.708 Degree of Freedom(n-1) 3 Chi-square critical value 0.114832
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