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scZoUnD [109]
3 years ago
13

Given the following observed and expected data (total of 1000), using chi-squared calculate a p-value that corresponds with this

data:
Phenotype Observed Expected
cinnabar, vestigial 384 390
roof 408 390
cinnabar, roof, vestigial 63 70
Wild type 72 70
vestigial 32 35
cinnabar, roof 34 35
roof, vestigial 4 5
cinnabar 3 5

a. 025> p> 0.1
b. 0.9 > p >0.75
c. 0.05 > p
d. 0.75 > p>0.5
e. 0.95 >p >0.9
Mathematics
1 answer:
finlep [7]3 years ago
7 0

Answer:

The answer is "0.90>p>0.75."

Step-by-step explanation:

\text{Cinnabar vestigial} \ \ \ \ \ \ \ \ \ \ \ 384 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \  -6 36 \ \ \ \ \ \ \ \ \ \ \  0.092308\\\\

roof \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \  408 \ \ \ \ \ \ \ \ \ \ \  390  \ \ \ \ \ \ \ \ \ \ \  18 \ \ \ \ \ \ \ \ \ \ \  324  \ \ \ \ \ \ \ \ \ \ \ 0.830769\\\\\text{Cinnabar vestigial roof} \ \ \ \ \ \ \ \ \ \ \ \ \ 63 \ \ \ \ \ \ \ \ \ \ \ \ \  70\ \ \ \ \ \ \ \ \ \ \ \ -7 \ \ \ \ \ \ \ \ \ \ \  49 \ \ \ \ \ \ \ \ \ \ \  0.7\\\\\text{wild type} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 72  \ \ \ \ \ \ \ \ \ \ \  70  \ \ \ \ \ \ \ \ \ \ \ 2  \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \  0.057143\\\\

vestigial \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 32 \ \ \  \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  0.257143\\\\

\text{Cinnabar roof} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 34\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  -1 \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \  1  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  0.028571\\\\\text{roof vestigial}  \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  0.2\\\\

\text{cinnabar}  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  3 \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \  5 \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \  -2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  0.8 \\\\

Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1000 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ 2.965934

Eight phenotypes were present.  

Df is provided also by a number of phenotypes -1 The degree of freedom

\to df = 8-1= 7

For p-value 0,9, Chi-square is 2.83;

The p-value of 0.75 is 4.5. Chi-square

Chi-sqaure value is observed at 2.965.

That means 0.90>p-value>0.75.

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Answer:

10.20% probability that a randomly chosen book is more than 20.2 mm thick

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

250 sheets, each sheet has mean 0.08 mm and standard deviation 0.01 mm.

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This is 1 subtracted by the pvalue of Z when X = 20.2. So

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