- The document discusses hypothesis testing and the p-value approach, which involves specifying the null and alternative hypotheses, calculating a test statistic, determining the p-value, and comparing it to the significance level α to determine whether to reject or accept the null hypothesis. - It also discusses type I and type II errors, degrees of freedom as the number of independent pieces of information, and chi-square and t-tests as statistical tests.

2. • determines "likely" or "unlikely" by
determining the probability
• Assumes the null hypothesis were true of
observing a more extreme test statistic in the
direction of the alternative hypothesis than the
one observed.
• Small P-value, ≤ α, then it is "unlikely.“null
hypothesis is rejected in favor of the alternative
hypothesis
• Large P-value ≥ α, then it is "likely."
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• P-value approach involves 4 steps
• Specify the null and alternative hypotheses.
• Using the sample data and assuming the null hypothesis is true,
calculate the value of the test statistic.
• Using the known distribution of the test statistic, calculate the P-
value: "If the null hypothesis is true, what is the probability that
we'd observe a more extreme test statistic in the direction of the
alternative hypothesis than we did?" (Note how this question is
equivalent to the question answered in criminal trials: "If the
defendant is innocent, what is the chance that we'd observe such
extreme criminal evidence?")
• Set the significance level, α, Compare the P-value to α to accept or
reject the Ho.

4. Type I error Or False positive is the incorrect
rejection of a true null hypothesis
Force one to conclude that a supposed effect or
relationship exists when in fact it doesn't
A type II error is the failure to reject a false null
hypothesis.
Examples of type II errors would be a blood test
failing to detect the disease it was designed to
detect, in a patient who really has the disease
A fire breaking out and the fire alarm does not
ring
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7. • df refers to the ability to make independent
choices, or take independent actions
• Three tasks you wish to accomplish, for example
that you want to go shopping, plan a vacation, and
workout at the gym.
• Three tasks and three days with one task a day
• 2 df
• Bcoz the third event is fixed
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8. • Statistically, the df are the number of scores that are free
to vary when calculating a statistic
• The number of pieces of independent information
available when calculating a statistic.
• Suppose you are told that a student took three quizzes,
each worth a total of 10 points. You are asked to guess
what her scores were.
• In this scenario, you may guess any three numbers as
long as they are in the range from 0 to 10. In this
example, you have 3 df, for each score is free to vary.
Each score is an independent piece of information.
Choosing the score for one quiz has no effect on either
of the other two scores that you may choose.
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9. • If total score was 27
• A scenario with 2 df
• Suppose you guess 10 for the first score
• Does choosing this score place any limitation on what
you might guess for a score on the second, given that
the total of the scores must be 27?
• No, for your choice of a second score is still free to
vary from 0 to 10. You guess 9 for a second score.
• What about your choice of a third score?
• Third choice must be 8 for a total of 27 to be obtained.
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10. • Suppose U want to calculate mean of 5 observations
• Df is 5
• But suppose you know the mean for the scores and you
want to calculate the standard deviation (s) for the
scores
• 9 df for these scores
• For if you know the mean, you need to know only 9 of
the scores, the 10th score is in a sense “determined” for
you by the value of the other 9 scores.
• So, for a set of n scores, there are n – 1 df when
calculating the standard deviation.
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11. • This test only works for categorical data (data in
categories), such as Gender {Men, Women} or
color {Red, Yellow, Green, Blue} etc, but not
numerical data such as height or weight.
• The numbers must be large enough. Each entry
must be 5 or more.
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O = the Observed (actual) value
E = the Expected value

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13. • Assume that we have crossed pure-breeding
parents of genotypes A/A · B/B and a/a · b/b,
and obtained a dihybrid A/a · B/b, which we
have testcrossed to a/a · b/b. A total of 500
progeny are classified as follows (written as
gametes from the dihybrid):
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14. • Recombinant frequency is 225/500 = 45
percent
• RF value is below 50
• Which mean genes are linked
• However recombinant classes may be in
minority only by chance
• Chi square comes in
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15. • How do proceed
• Can we use 1:1:1:1 backcross ratio for independent
assortment to find the expected no of genotype?
• No
• Because to get such ratio 2 things must b true
• 1. Independent assortment between A and B locus
• 2. Equal chance of survival from sperm/egg to adult for
different genotype
• Homozygous a/a and b/b genotype low p of survival
than heterozygous a/A and B/b
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16. • We might reject the true H0
• Therefore, we need a method insensitive to
survival differences
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17. • We expect the frequency of genotype a.b to be the
product of frequency of alleles a and b
• Proportion of allele a from given data =
(135+115)/50 = 50% or 0.5
• Proportion of allele b = (135+110)/500 = 49% or
0.49
• Expected proportion of a.b genotype = 0.5* 0.49
= 0.245
• No. Of a.b genotype in 500 genotype = 500*0.245
= 122.5
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19. • The obtained value of χ2 is converted into a
probability by using a χ2 tab
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20. • Df is the number of independent deviations of
observed from expected that have been
calculated.
• We notice that, because of the way that the
expectations were calculated in the contingency
table from the row and column totals, all
deviations are identical in absolute magnitude,
12.5, and that they alternate in sign and so they
cancel out when summed in any row or column.
• Thus, there is really only one independent
deviation, so there is only one degree of freedom.
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22. • The probability of obtaining a deviation from
expectations this large (or larger) by chance
alone is 0.025 (2.5 percent).
• Because this probability is less than 5 percent,
the hypothesis of independent assortment must
be rejected.
• Genes linked
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23. • t-tests are a type of hypothesis test that allows
you to compare means.
• They are called t-tests because each t-test boils
your sample data down to one number, the t-
value.
• If you understand how t-tests calculate t-
values, you’re well on your way to
understanding how these tests work.
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24. • The population from which the sample has
been drawn should be normal
• It has however been shown that minor
departures from normality do not affect this
test - this is indeed an advantage.
• The population standard deviation is not
known.
• Sample observations should be random.
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25. • One sample
• Paired samples (2 samples)
• Independent samples (2 samples)
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26. • Formula shows that t test is a ration
• A common analogy is that the t-value is the
signal-to-noise ratio
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29. • Suppose a new variety of maize is claimed to
have yield higher than the existing yield of 4
tonnes/hectare
• Field trials give the following results
1. 4.5
2. 5.1
3. 4.2
4. 4.1
5. 4.7
• Test if the claim is true
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n–1" degrees of freedom

30. • Effect of Cr on
Photosynthesis rate
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before After
Photosynthesis/g FW
0.027 0.02
0.031 0.025
0.04 0.027
0.037 0.029
0.033 0.022
0.029 0.025
n/2–1" degrees of freedom

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•Where: x1 is the mean of sample 1
•s1 is the standard deviation of sample 1
•n1 is the sample size of sample 1
•x2 is the mean of sample 2
•s2 is the standard deviation of sample 2
•n2 is the sample size in sample 2
2n-2" degrees of freedom

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33. • A statistical procedure used to test an alternative hypothesis
against null hypothesis
• Used to determine whether two sample’s means are
different when variances are known and sample is large
(n≥30)
• The knowledge of population standard deviation is
compulsory.
• Comparison of means of two independent groups so
samples, taken from one population with known variance.
• Z= X-μ0/σ/Under root n
• Μ0=Population mean
• X is sample mean
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34. • When samples are drawn at random
• When samples are taken from population are
independent
• When standard deviation is known
• When number of observation is large
• When
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35. • A newly developed variety of wheat has better
yield than the previous variety
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36. • Used for two normally distributed, but
independent populations
• σ is known
• Formula
• Z = (x1-x2)-(μ1- μ2)/ Sqrt (σ1
2/n1+ σ2
2/n2)
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