A particular article reported the accompanying data on pheno

A particular article reported the accompanying data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. There are four possible phenotypes: (1) tall cut-leaf, (2) tall potato-leaf, (3) dwarf cut-leaf, and (4) dwarf potato-leaf. Mendel\'s laws of inheritance imply that ?₁ = 9/16, ?₂ = 3/16, ?₃ = 3/16, and ?₄ = 1/16. Are the data from this experiment consistent with Mendel\'s laws? Use a 0.01 significance level. (Use 2 decimal places.)

For my x² value I got 4.47 but everytime I enter that it says that it is incorrect.

Solution

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

A particular article reported the accompanying data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. There are four possible phenotypes: (1) tall cut-leaf, (2) tall potato-leaf, (3) dwarf cut-leaf, and (4) dwarf potato-leaf. Mendel\'s laws of inheritance imply that ?1 = 9/16, ?2 = 3/16, ?3 = 3/16, and ?4 = 1/16. Are the data from this experiment consistent with Mendel\'s laws? Use a 0.01 significance level. (Use 2 decimal places.)

1) 925
2)286
3) 291
4) 102

x²=?

Answer

For this problem, it looks like a chi-square goodness of fit test would be best.
A.) A chi-square test looks at whether the table of observed results matches the expected values from the ratios given.
So your null hypothesis would be:
The proportions of tomato phenotypes that result from crossing parents with tall-cut leaves and dwarf-potato leaves are p(tall cut leaf)=9/16=0.5625; p(tall pot leaf)=3/16=0.1875; etc.
Your alternate hypothesis would be:
At least one of the above proportions differs from its expected value.
B.) In order to use a chi-square test, your expected counts should all be greater than 1 and most of them should be greater than 5. Here are the expected counts of each type of phenotype based on a sample size of 1611:
tall cut leaf=1611(.5625)=906.2>5
tall pot leaf=1611(.1875)=302>5
dwarf cut leaf=1611(.1875)=302>5
dwarf pot leaf=1611(.0625)=100.7>5
Since all of your expected counts are greater than 5, you can proceed with the test.
C.) The chi-square test statistic is equal to the sum of (observed-expected)^2/expected. So in this case it would be:
(926-906.2)^2/906.2 + (288-302)^2/302 + ... =1.47
The P-value can be found by plugging the test statistic into the function X^2cdf in the distributions menu. You type in X^2cdf(1.47, 999999999, 3) where 3 is your degrees of freedom found by counting the number of observed counts which is 4 and then subtracting 1 to get 3. The P-value comes out to be 0.6895.
D.) The P-value is really big as far as P-values go. A big P-value means you fail to reject the alternate hypothesis and conclude that there is significant evidence to suggest that the proportions of offspring phenotypes that fall into each category as observed by the researcher matches that given by the genetic laws.