Hi, I have 3 parts to this question. Below, I have uploaded the icon to view the

Question

Hi, I have 3 parts to this question. Below, I have uploaded the icon to view the table and the icon to view the chi-square table of critical values.

Aside from Part A (the mulitple choice problems seen below) I also need:

Part B) Compute the expected count for each color. (We already have the observed count/frequency).

Part C) What is the test statistic? (Round to three decimal places)

Please answer all 3 parts. Thank you for your help! :-)

A manufacturer of colored candies states that 13% o the candies in a bag should be brown, 14% yellow 13% re 24% blue 20% orange, and I green A student and mi selected a colored candies. He counted the number of candies of each color and obtained the results shown in the table. Test whether the bag of colored candies follows the distribution stated above at = 0.05 level of significance. Using the level of significance = 0.05, test whether the color distribution is the same Click the icon to view the table Click the icon to view the chi-square table of critical values. O A. Ho: The distribution of colors is at least as uniform as stated by the manufacturer. OB. Ho: The distribution of colors is the same as stated by the manufacturer H: The distribution of colors is less uniform than stated by the manufacturer. H: The distribution of colors is not the same as stated by the manufacturer. Ho: The distribution of colors is at most as uniform as stated by the manufacturer H: The distribution of colors is more uniform than stated by the manufacturer C. D. Ho: The distribution of colors is not the same as stated by the manufacturer. H,: The distribution of colors is the same as stated by the manufacturer.

Explanation / Answer

(first part) right choice B. about the null(H0) and alternate (H1) hypothesis is

H0:distribution of color is the same as stated by the manufacturer

H1:distribution of color is not the same as stated by the manufacturer

(second part) here total frequency=n=396 and

expercted frequency for each colour would be =n*p

where p is the claimed proportion

(third part) here we use chi-square to test the hypothesis and

chi-square=sum((O-E)2/E)=15.5871 with k-1=6-1=5 df

critical chi-square (0.05,5)=11.07 is less than calculated chi-square=15.5871, so we fail to accept H0 and conclude that distribution of color is not same as stated by the manufacturer

color brown yellow red blue orange green total frequency(O) 61 65 56 63 88 63 n=396 claimed proportion(p) 0.13 0.14 0.13 0.24 0.2 0.16 1 Expected frequency (E=np) 51.5 55.4 51.5 95.0 79.2 63.4 396

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