Given the following observed phenotypic ratios, calculate the expected phenotypic ratios and chi square values. Then evaluate the chi square value using a chi square table to accept or reject the null hypothesis at a P value of 0.05. Place the appropriate boxes in correct position in the table
Problem 1
Null hypothesis: The observed values are not deviating from the 9:3:3:1 ratio.
Observed values (O) |
250 |
106 |
104 |
40 |
500 |
Exptected Ratio (ER) |
9 |
3 |
3 |
1 |
16 |
Exprected Values (E) |
281.25 |
93.75 |
93.75 |
31.25 |
|
Deviation (O-E) |
-31.25 |
12.25 |
10.25 |
8.75 |
|
D^2 |
976.5625 |
150.0625 |
105.0625 |
76.5625 |
|
D^2/E |
3.472222 |
1.600667 |
1.120667 |
2.45 |
8.643556 |
X^2 |
8.643556 |
||||
Degrees of freedom |
4-1=3 |
Inference: The calculated chisquar value i.e. 8.6 is greater than the table value i.e.7.82 at 3 DF and 0.05 probability, hence the null hypothesis is rejected.
Problem 2
Null hypothesis: The observed values are not deviating from the 1:1:1:1 ratio.
Observed values (O) |
130 |
122 |
140 |
108 |
500 |
Exptected Ratio (ER) |
1 |
1 |
1 |
1 |
4 |
Exprected Values (E) |
125 |
125 |
125 |
125 |
|
Deviation (O-E) |
5 |
-3 |
15 |
-17 |
|
D^2 |
25 |
9 |
225 |
289 |
|
D^2/E |
0.2 |
0.072 |
1.8 |
2.312 |
4.384 |
X^2 |
4.384 |
||||
Degrees of freedom |
4-1=3 |
` |
Inference: The calculated chisquar value i.e. 4.38 is less than the table value i.e.7.82 at 3 DF and 0.05 probability, hence the null hypothesis is accepted.
Problem 3
Null hypothesis: The observed values are not deviating from the 3:1 ratio.
Observed values (O) |
350 |
150 |
500 |
Exptected Ratio (ER) |
3 |
1 |
4 |
Exprected Values (E) |
375 |
125 |
|
Deviation (O-E) |
-25 |
25 |
|
D^2 |
625 |
625 |
|
D^2/E |
1.666667 |
5 |
6.666667 |
X^2 |
6.67 |
||
Degrees of freedom |
- |
1 |
Inference: The calculated chisquar value i.e. 6.67 is greater than the table value i.e. 6.67 at 1 DF and 0.05 probability, hence the null hypothesis is rejected.
Given the following observed phenotypic ratios, calculate the expected phenotypic ratios and chi square values.
for these Table S. Best-of-Fit Chi-Square Calculations for Sex Ratios. The observed numbers families can be found on this supplement page 6. Deviation (O-E/E Observed Expected 497 Deviatic Results O-E All boys 3 B:1G 2 B:2G 1B:3G All girls G99 Total e. Degrees of Freedom- d. Range of probability that deviations are due to chance- e. Accept or reject hypothesis? 5. Next, determine whether the overall ratio of boys to girls in the above data is consistent with the hypothesis...
A distribution and the observed frequencies of the values of a variable from a simple random sample of the population are provided below. Use the chi-square goodness-of-fit test to decide, at the specified significance level, whether the distribution of the variable differs from the given distribution Distribution: 0.1875, 0.1875, Observed frequencies: 16, 20, 24, 36 Significance level 0.05 0.3125, 0.3125 Determine the null and alternative hypotheses. Choose the correct answer below. OA. H: The distribution of the variable differs from...
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The type of plant favored by deer is shown in the following table. Volunteers observed the feeding habits of a random sample of 320 deer. Use a 0.05 level of significance to test the claim that the natural distribution of browse fits the deer feeding pattern. Type of Browse Plant Composition in Study Area Observed Number of Deer Feeding on Plant Sage Brush 32% 102 Rabbit Brush 38% 125 Salt Brush 12% 43 Service Berry 10% 27 Other 8% 23...
The table to the right contains observed values and expected values in parentheses for two categorical variables, X and Y, where variable X has three categories and variable Y has two categories. Use the table to complete parts (a) and (b) below. Y1 X1 X3 32 45 49 (32.72) (47.34) (45.94) 15 23 17 (14.28) 20.66) (20.06) Y2 (a) Compute the value of the chi-square test statistic. x=(Round to three decimal places as needed.) (b) Test the hypothesis that X...
(If any are cut off answer what you can see) 3A) 3b) 3C) 3D) You are conducting a multinomial Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies: H:PA = 0.25; PB = 0.4; Pc = 0.1; PD 0.25 Complete the table. Report all answers accurate to three decimal places. Observed Expected Category Residual Frequency Frequency А 17 B 57 с 8 D 23 What is the chi-square test-statistic for this data?...
55 sn car sn cart sn* car sn+Car+ 200 TABLE 5.2 Critical Chi-Square Values Values 0.99 0.90 0.50 0.10 0.05 0.01 0.001 Degrees of Freedom Values w - 0.02 0.45 2.71 0.02 0.21 1.39 4.61 0.11 0.58 2.37 6.25 0.30 1.06 3.36 778 0.55 1.61 4.35 9.24 3.84 6.64 10.83 5.99 9.21 13.82 7.81 11.35 16.27 9.49 13.28 18.47 11.07 15.09 20.52 In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive X-linked alleles. The wild-type...
What does it mean if the Chi-square test statistic has a value of zero? There is an error in the calculations as the test statistic must be positive. We will reject the null hypothesis. The test is right-tailed so that the result is inconclusive. All of the observed values equal the expected values for the corresponding cell.
(If some are cut off just answer what you can see) 2A) 2B) 2C) 2D) You are conducting a multinomial hypothesis test (a = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table. Observed Expected Squared Category Pearson Frequency Frequency Residual A 25 B 14 с 9 D 22 E 20 Report all answers accurate to three decimal places. But retain unrounded numbers for future calculations. What is the chi-square test-statistic for...
use problem above to answer Complete the following chi-square problems. (Show all work!!) In the cross AaBb x AaBb O-E (O-E) (O-E)?/E 311 105 1. Phenotype A-B- A-bb aaB- aabb Totals 25 calculated X2= mith, Beatty and Ware 7/14 revised by for Fall 2015, BIOL 251 Genetics Lab Manual, Benedictine Univ page 3 a. What are your null and alternative hypotheses? b. How many degrees of freedom? c. What is your table X2 value? d. What is your region of...