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- Assume that a Chi-square test was conducted to test the goodness of fit to a 9:3:3:1 ratio and a Chi-square value of 10.62 was obtained. Should the null hypothesis be accepted?Assume that a Chi-square test was conducted to test the goodness of fit to a 3:1 ratio and that a Chi-square value of 2.62 was obtained (Table value is equal to 3.84). Should the null hypothesis be accepted? How many degrees of freedom would be associated with this test of significance?page - Columbus State Uni X content/2622738/viewContent/50155728/View + Calculating the Value of Chi-Square Once you've defined the null hypothesis (and know what you're testing!), you can finally calculate the chi-square value based on your observed data and the expected distribution. There are many statistical software packages and free web-based applications that will automate the calculation for you, but it's important to understand the calculation to understand the meaning of the calculated value. Here, we'll walk through it using a calculation table. This is Mendel's historic data for the F2 generation of monohybrid cross for flower color. Recall that he expected to see in the F2 generation a ratio of 3 purple : 1 white. Chi-square calculation for data from a monohybrid cross with an expected 3 purple : 1 white ratio. Outcomes Observed (0) Expected (E) O-E (0-E)2 (O-E)²/E Purple White Totals 705 224 929 X²= The first step is to calculate the expected (E) values for purple and…
- 1) Chi-square analysis generates a p value, an important indicator of statistical support for a hypothetical outcome. If a p value = 0.05 > p > 0.01, the tested hypothesis fails to be falsified. Agree/Disagree? Answer with explanation here:What does P-value indicate? In statistical analysis the results for two sets of data are presented with P < 0.01 and P<0.005. What does it mean?Which conclusion is most strongly supported by the data in the table?
- You perform a chi-square test to compare observed and expected values and obtain a chi-square value of 9.4 with 3 degrees of freedom. What do you conclude? it is not likely that the difference between observed and expected values is due to random chance, since p>0.05 it is impossible to conclude anything from this information it is likely that the difference between observed and expected values is due to random chance, since p>0.05 it is likely that the difference between observed and expected values is due to random chance, since p<0.05 the experiment was done incorrectly and must be repeated it is not likely that the difference between observed and expected values is due to random chance, since p<0.05Which conclusion is supported by the results shown by the Punnett square?Considering that I have a chi-square test critical value of 3.84 what is the minimum X2 value for me to reject the null hypothesis? Is it 3.84 or 3.85?
- The mean weight of loaves of bread produced at the bakery where you work is supposed to be 1 pound. You are the supervisor of quality control at the bakery, and you are concerned that new employees are producing loaves that are too light. Suppose you weigh an SRS of bread loaves and find that the mean weight is 0.975 pound. What are the appropriate hypotheses for performing this significance test? Ho : x = 1; H. : x < 1 where i = the sample mean weight (in pounds) of bread loaves produced at the bakery. Ho : u = 1; H. : µ < 1 where u = the true mean weight (in pounds) of bread loaves produced at the bakery. Ho : µ < 1; H, : H = 1 where u = the true mean weight (in pounds) of bread loaves produced at the bakery. Ho : u = 1; H. : p = 0.975 where u= the true mean weight (in pounds) of bread loaves produced at the bakery. Ho : u = 1; H. : u # 1 where u= the true mean weight (in pounds) of bread loaves produced at the bakery.The confidence interval is reported as follows: Lower 95% CI < p < Upper 95% CI For this question you will be calculating the confidence interval but only reporting the Upper 95% CI using the Agresti-Coull Method. ( (you'll need to know the whole 95% CI for a future question) Do NOT use calculations from previous questions, this is a new scenario. In the Invisible Gorilla Experiment, 24 students were watching the video, only 7 noticed the gorilla. Calculate the 95% CI using the Agresti-Coull method, but only report the Upper 95% CI. Report your answer to 3 decimals.List and describe the five conditions of the Hardy–Weinberg principle.