My degree of freedom is 0.18 but I don't know the p value nor the conclusion. I have the chi square distribution for the p value below but I'm very confused on what any of this means
Homework Answers
Degree of freedom can't be less than 1 . It should be always 1 or more than 1
Look at the table what u have provided in that minimum degrees of freedom is always 1 or more than 1
For example: In chi square test if you have 5 values than 5-1(N-1) = 4 is the degrees of freedom.
If more than two either coloums are rows are present then (r-1)(c-1) is the degrees of freedom.
So please check the degrees of freedom can't be less than 1
Add Answer to:
My degree of freedom is 0.18 but I don't know the p value nor
the conclusion....
please answer clearly so i can understand, thank you. 2. 10-points: The MN blood group in...
please answer clearly so i can understand, thank you.
2. 10-points: The MN blood group in humans is under the control of a pair of co-dominant alleles LM and LN. I a group of 556 Individuals, the following genotypic frequencies are found: 167 MM 280 MN 109 NN a) Calculate the frequency of the LM and LN alleles Note: the degrees of freedom is interestingly "1" since while there are genotypes.these are determined by only 2 allelic frequencies. Our book...
The type of household for the U.S. population and for a random sample of 411 households...
The type of household for the U.S. population and for a random sample of 411 households from the community Dove Creek, Montana, are shown (based on Statistical Abstract of the United States) Observed Number of Percent of U.S. Households in Type of Househol Households Dove Creek 102 26% Married with children 112 29% Married, no children Single parent 9% One person 25% 96 11% Other (e.g., roommates, siblings) 68 Use a 5% level of significance to test the claim that...
For the below output, what was the test used, the p-value and the conclusion? a. An...
For the below output, what was the test used, the p-value and
the conclusion?
a.
An F test, with 1 and 94 degrees of freedom; p
= 0.000;reject the null
b.
A chi-square test with 4 degrees of freedom; p = 0.383; do not
reject the null.
c.
A t-test with 91 degrees of freedom; p =
0.000; reject the null
d.
An F test with 4 and 91 degrees of freedom; p
= 0.000; reject the null
lSummary of...
A corn genetics lab. i got total chi-square of 14.34. And the number of degrees of...
A
corn genetics lab.
i got total chi-square of 14.34. And the number of degrees of
freedom of 1.
How can I find the probability of a Match base on the table?
please explain.
Degree of Freedom 0.99 0.000 0.020 0.115 0.297 0.554 0.872 1.239 1.647 2.088 2.558 3.053 3.571 4.107 4.660 5.229 5.812 6.408 7.015 7.633 8.260 9.542 10.856 12.198 13.565 14.953 22.164 Probability of Exceeding the Critical Value 0.95 0.90 0.75 0.50 0.25 0 .10 0.0040.016 0.1020.455 1.32...
Determine (a) the test statistic, (b) the degrees of freedom, (c) the P-alue, and (d) test...
Determine (a) the test statistic, (b) the degrees of freedom, (c) the P-alue, and (d) test the hypothesis at the a-0.05 level of significance. Outcome Obsenved Expected 25 25 25 25 A B C Do 22 28 24 26 H,: At least one of the proportions is different from the others Click the icon to wew the chi-square distribution table Round to two decimal places as needed.) (b) (True False) The number of degrees of freedom is oneless than the...
Conduct the hypothesis test and provide the test statistic, critical value and P-value, and state the conclusion A package of 100 candies are distributed with the following color percentages 13% r...
Conduct the hypothesis test and provide the test statistic, critical value and P-value, and state the conclusion A package of 100 candies are distributed with the following color percentages 13% red 20% ora ge, 15% yellow 12% br dlaimed. Use a 0.10 significance level. EEB Click the icon to view the color counts for the candy in the package Click here fo view the chi-square distribution table, 25% blue and 15% g een Use the given sa ple data to...
For the below output, what was the test used, the p-value and the conclusion? a. An F test...
For the below output, what was the test used, the p-value and
the conclusion?
a.
An F test, with 1 and 94 degrees of freedom; p
= 0.000;reject the null
b.
A chi-square test with 4 degrees of freedom; p = 0.383; do not
reject the null.
c.
A t-test with 91 degrees of freedom; p =
0.000; reject the null
d.
An F test with 4 and 91 degrees of freedom; p
= 0.000; reject the null
lSummary of...
test. Use the Chi-square table provided by the TA. In the lab write-up answer these questions:...
test. Use the Chi-square table provided by the TA. In the lab write-up answer these questions: 1. What does it mean if a locus is in Hardy- Weinberg proportions?, it 2. Can one locus be in Hardy-Weinberg proportions and the other not? 3. Why or why not? C. The Next Generation and Genetic Drift: Before you begin, do you expect allele frequencies to change from one generation to the next if the om? What about Degrees of Freedom 0.01 1.610...
HELLO, PLEASE HELP WITH QUESTIONS 8-10 PLEASE PLEASE AND THANK YOU!! PLEASE, I DON'T HAVE ANY...
HELLO, PLEASE HELP WITH QUESTIONS 8-10 PLEASE PLEASE AND THANK
YOU!!
PLEASE, I DON'T HAVE ANY MORE QUESTIONS, AND I ASKED THIS 2
TIMES ALREADY AND NO RESPONSE. THANK YOU!!! APPRECIATE IT!!
Question Completion Status: BUTTON Questions 5 to 11 are based on the following scenario Question 5: Although the following scenario is hypothetical, it is based on actual facts. African leopards live in a variety of habitats including forests, mountains, grasslands and deserts. Most leopards are light brown with...
Practice questions for BIO 340 (Exam 2) I need help with these questions Please. WILL GIVE...
Practice questions for BIO 340 (Exam 2) I need help with these
questions Please. WILL GIVE GOOD RATING
1. Wild type blue-eyed mary has blue flowers. Two genes control the pathway that makes the blue pigment: The product of gene W turns a white precursor into magenta pigment. The product of gene M turns the magenta pigment into blue pigment. Each gene has a recessive loss-of-function allele: w and m, respectively. A double heterozygote (Ww Mm) is self-pollinated. What proportion...
please answer clearly so i can understand, thank you.
2. 10-points: The MN blood group in humans is under the control of a pair of co-dominant alleles LM and LN. I a group of 556 Individuals, the following genotypic frequencies are found: 167 MM 280 MN 109 NN a) Calculate the frequency of the LM and LN alleles Note: the degrees of freedom is interestingly "1" since while there are genotypes.these are determined by only 2 allelic frequencies. Our book...
The type of household for the U.S. population and for a random sample of 411 households from the community Dove Creek, Montana, are shown (based on Statistical Abstract of the United States) Observed Number of Percent of U.S. Households in Type of Househol Households Dove Creek 102 26% Married with children 112 29% Married, no children Single parent 9% One person 25% 96 11% Other (e.g., roommates, siblings) 68 Use a 5% level of significance to test the claim that...
For the below output, what was the test used, the p-value and
the conclusion?
a.
An F test, with 1 and 94 degrees of freedom; p
= 0.000;reject the null
b.
A chi-square test with 4 degrees of freedom; p = 0.383; do not
reject the null.
c.
A t-test with 91 degrees of freedom; p =
0.000; reject the null
d.
An F test with 4 and 91 degrees of freedom; p
= 0.000; reject the null
lSummary of...
A
corn genetics lab.
i got total chi-square of 14.34. And the number of degrees of
freedom of 1.
How can I find the probability of a Match base on the table?
please explain.
Degree of Freedom 0.99 0.000 0.020 0.115 0.297 0.554 0.872 1.239 1.647 2.088 2.558 3.053 3.571 4.107 4.660 5.229 5.812 6.408 7.015 7.633 8.260 9.542 10.856 12.198 13.565 14.953 22.164 Probability of Exceeding the Critical Value 0.95 0.90 0.75 0.50 0.25 0 .10 0.0040.016 0.1020.455 1.32...
Determine (a) the test statistic, (b) the degrees of freedom, (c) the P-alue, and (d) test the hypothesis at the a-0.05 level of significance. Outcome Obsenved Expected 25 25 25 25 A B C Do 22 28 24 26 H,: At least one of the proportions is different from the others Click the icon to wew the chi-square distribution table Round to two decimal places as needed.) (b) (True False) The number of degrees of freedom is oneless than the...
Conduct the hypothesis test and provide the test statistic, critical value and P-value, and state the conclusion A package of 100 candies are distributed with the following color percentages 13% red 20% ora ge, 15% yellow 12% br dlaimed. Use a 0.10 significance level. EEB Click the icon to view the color counts for the candy in the package Click here fo view the chi-square distribution table, 25% blue and 15% g een Use the given sa ple data to...
For the below output, what was the test used, the p-value and
the conclusion?
a.
An F test, with 1 and 94 degrees of freedom; p
= 0.000;reject the null
b.
A chi-square test with 4 degrees of freedom; p = 0.383; do not
reject the null.
c.
A t-test with 91 degrees of freedom; p =
0.000; reject the null
d.
An F test with 4 and 91 degrees of freedom; p
= 0.000; reject the null
lSummary of...
test. Use the Chi-square table provided by the TA. In the lab write-up answer these questions: 1. What does it mean if a locus is in Hardy- Weinberg proportions?, it 2. Can one locus be in Hardy-Weinberg proportions and the other not? 3. Why or why not? C. The Next Generation and Genetic Drift: Before you begin, do you expect allele frequencies to change from one generation to the next if the om? What about Degrees of Freedom 0.01 1.610...
HELLO, PLEASE HELP WITH QUESTIONS 8-10 PLEASE PLEASE AND THANK
YOU!!
PLEASE, I DON'T HAVE ANY MORE QUESTIONS, AND I ASKED THIS 2
TIMES ALREADY AND NO RESPONSE. THANK YOU!!! APPRECIATE IT!!
Question Completion Status: BUTTON Questions 5 to 11 are based on the following scenario Question 5: Although the following scenario is hypothetical, it is based on actual facts. African leopards live in a variety of habitats including forests, mountains, grasslands and deserts. Most leopards are light brown with...
Practice questions for BIO 340 (Exam 2) I need help with these
questions Please. WILL GIVE GOOD RATING
1. Wild type blue-eyed mary has blue flowers. Two genes control the pathway that makes the blue pigment: The product of gene W turns a white precursor into magenta pigment. The product of gene M turns the magenta pigment into blue pigment. Each gene has a recessive loss-of-function allele: w and m, respectively. A double heterozygote (Ww Mm) is self-pollinated. What proportion...
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