Q22. Suppose a person has 8 red , 5 green, 15 orange and 12 blue balls.Test the null hypothesis that the colors of the balls occure with equal frequency. What is Chi Square value you get?*5.65.685.865.8
Question
Q22. Suppose a person has 8 red , 5 green, 15 orange and 12 blue balls.Test the null hypothesis that the colors of the balls occure with equal frequency. What is Chi Square value you get?*5.65.685.865.8
Solution
To test the null hypothesis that the colors of the balls occur with equal frequency, we can use the Chi-Square test. Here are the steps:
-
First, calculate the total number of balls. This is 8 (red) + 5 (green) + 15 (orange) + 12 (blue) = 40 balls.
-
If the colors occur with equal frequency, each color would have the same number of balls. So, we would expect each color to have 40/4 = 10 balls.
-
Now, calculate the Chi-Square statistic. This is done by summing the squares of the differences between the observed (actual) and expected frequencies, divided by the expected frequency, for each category (color).
Chi-Square = Σ [ (Observed-Expected)^2 / Expected ]
For red balls: (8-10)^2 / 10 = 0.4 For green balls: (5-10)^2 / 10 = 2.5 For orange balls: (15-10)^2 / 10 = 2.5 For blue balls: (12-10)^2 / 10 = 0.4
-
Add these all up to get the Chi-Square statistic: 0.4 + 2.5 + 2.5 + 0.4 = 5.8
So, the Chi-Square value is 5.8.
Similar Questions
Q25. A dice is tossed 120 times with following results. No. turned up: 1 2 3 4 5 6 Frequency: 30 25 18 10 22 15 Test the hypothesis that the dice is unbiased (χ^2=11.7) . Calculate the frequency observed for chi-square distribution*Dice is unbiased ,11.3Dice is biased ,12.9Dice is unbiased ,10.9Dice is biased ,12.3
What does a chi-square test compare?Question 4Answera.the frequency of incorrect responses for each variableb.correct and incorrect reaction timesc.the observed and total frequencies for each individual participantd.the observed and expected frequencies for each cell
3. The Chi Square test is applicable for a given data, if*
In a population of flowers, red is dominant to white. A true-breeding white flower is crossed with a heterozygous flower, and the observed values were 63 white flowers, 37 red flowers. To determine the chi square value, _____________ degrees of freedom should be used :1234
In the chi-square test, the null hypothesis states:a.There is an association between the variablesb.There is no association between the variablesc.The means of the groups are equald.The variances of the groups are equal
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.