This Question: 1 pt 11 of 12 (0 complete) This Quiz: 12 pts possible Conduct the

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This Question: 1 pt 11 of 12 (0 complete) This Quiz: 12 pts possible Conduct the hypothesis test and provide the test statistic and the critical value, and state the conclusion. A company claims that its packages of 100 candies are d tributed with the follow ng color percentages: 13% red, 21% orange distribution is as claimed. Use a 0.025 significance level EEB Click the icon to view the color counts for the candy in the package 14% yellow. 11% brown, 25% blue, and 16% green. Use the g en sample data to test the cla that the coor stribution tabl The test statistic is (Round to two decimal places as needed) The critical value is Round to three decimal places as needed.) State the conclusion Candy Package Counts Candy Counts . Theresfficient evidence to warrant rejection of the claim that the color distribution is as claimed Color Number in Package a Red Orange Yellow Brown Blue Green 12 26 Reject Do not reject 27 17 Print Done

Explanation / Answer

Answer according to Question format:

Test ststistic is : 3.45

Critical value is : 12.833

State the conclusion,

Do not reject Ho . There is not sufficient evidence to warrant rejection of the claim that the color distribution is as claimed.

This solution section is for your understanding

Null hypothesis: There is not sufficient evidence to warrant rejection of the claim that the color distribution is as claimed. Alternative hypothesis: There is a sufficient evidence to warrant rejection of the claim that the color distribution is as claimed. Color No. in package Red 13% Orange 21% Yellow 14% Brown 11% Blue 25% Green 16% And a random sample of 100 was taken and results are as follows. Color No. in package Red 12.00 Orange 26.00 Yellow 10.00 Brown 8.00 Blue 27.00 Green 17.00 Now, expected frequency can be obtained if we multiply believe percentage with size of sample taken that is 100. Color No. in package Red 13.00 Orange 21.00 Yellow 14.00 Brown 11.00 Blue 25.00 Green 16.00 The value of chi-square statistic is given as follows: Sno Observed Frequency(Oi) Expected Frequency(Ei) (Oi-Ei)2/Ei Red 12.00 13.00 0.08 Orange 26.00 21.00 1.19 Yellow 10.00 14.00 1.14 Brown 8.00 11.00 0.82 Blue 27.00 25.00 0.16 Green 17.00 16.00 0.06 Test statistic = Sum -> 3.45 Test statistic = 3.45 As we have 6 observations So, degree of freedom is given by n-1 = 6-1 = 5 We can calculate the critical value in MS-Excel by using the function "CHIINV()" which takes (? = 0.025, degree of freedom = 5) as inputs and outputs the critical value. Critical value = CHIINV(0.025,5) = 12.833 We observed that test statistic value is less than critical value, so we accept our null hypothesis and we conclude that Company's claim is true that is there is no significant difference between the observed and the expected value.

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