Solution:-
39)
a) p-value = 0.01
D.F = 5
X2 = 20.36
P(X2 > 30.6) = 0.01

b) p-value = 0.504
D.F = 4
X2 = 3.33
P(X2 > 3.33) = 0.504

c) P(- 30.76 < t < 30.76) = less than 0.00001
D.F = 6
t-value = 30.76
P(- 30.76 < t < 30.76) = less than 0.00001
d) P( t > 0.85) = 0.207
D.F = 11
t-value = 0.85
P( t > 0.85) = 0.207

e) P(F> 30) = 0.001
DF1 = 1
DF2 = 8
F-value = 30
P(F> 30) = 0.001

f) P(F> 2.5) = 0.199
DF1 = 3
DF2 = 4
F-value = 2.5
P(F> 2.5) = 0.199

39. Answer the following questions in a few words without doing any calculations a. Using the...
1) After doing so, find DF
2)using the df you determine previously, include your Chi-Square
Critical value in this response
3) Indicate below (written in APA format) the outcome of the
Chi-Square test you just performed.
Given the following data, determine the Chi-Square obtained statistic: 8 8 4
Given the following data, determine the Chi-Square obtained statistic: 8 8 4
RuPaul is using a Chi Square Goodness of Fit test to determine whether there are any preferences among four fan-favorite drag queens (Monét X Change, Farrah Moan, Monique Heart, and Manila Luzon). If she computed a Chi Square value of 4.6 from a sample size of 60, what would be Cohen's w?
a) true b) false 42. For a chi-square distributed random variable with 10 degrees of freedom and a level of sigpificanoe computed value of the test statistics is 16.857. This will lead us to reject the null hypothesis. a) true b) false 43. A chi-square goodness-of-fit test is always conducted as: a. a lower-tail test b. an upper-tail test d. either a lower tail or upper tail test e. a two-tail test 44. A left-tailed area in the chi-square distribution...
(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?...
A testcross with a heterozygote b+bc+c produces a total of 700 offspring. The following are the resulting phenotypes of the progeny and corresponding numbers: b+c+ = 270; bc = 255; b+c=83; bc+=92 1) Based on the data, do the genes b and c appear to be linked or are they assorting independently? How did you determine your answer? If the genes are linked, which of the above genotypes are the non-recombinant progeny? Also, is the original configuration of the loci...
Test the claim that your coin is fair, using a 5% level of significance. Use the Chi-Square Goodness of Fit Test. Toss a coin at least 12 times (why?). a) What is n? What are the number of Tails and Heads? These are the Observed frequencies. b) What are the Expected frequencies? c) What is the Null Hypothesis H0? d) What is the Alternative Hypothesis H1? e) Is this a left, right, or two-tailed test? f) Chi-Square Test Statistic =?...
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im confused about using the Chi square value to answer these
questions
Complete the Chi Squared analysis below by filling in the empty white cells. Do not round numbers up until the very end (the final chi squared value) which you can round to one decimal place. It's best to do this in Excel or another spreadsheet program. Phenotypes Observed # Expected # O-E d d /E LO) d) Black, no horns 44 44.25 -0.25 0.0625 0.00141243 Black, homs 144.25...
Does the data meet the conditions for the chi-square test?
StatCrunch Instructions: Test of Independence Using
Technology
Next we will use StatCrunch to calculate the expected
counts:
Enter Yes and No in column var1.
Enter the observed counts as they appear in the table above
(not including the totals) into columns var2 and var3.
Rename: var1 as "911", var2 as "No Risk" and var3 as "M to S
Risk"
Choose Stat -> Tables -> Contingency -> with
summary
Select the...
For the following experiment/question, pick the most appropriate statistical test. You have the following statistical tests as choices: some may be used more than once, others not at all. Assume homogeneity of variance (where applicable) and the validity of parametric tests (where applicable), unless something is directly stated (e.g., “the data are not at all normal”) or otherwise indicated (viz., by the inspection of the data) which would indicate a strong and obvious violation of an assumption. This means you must...