Solution :
We know that

At least one mean is different form the others (claim)
We need to find the critical values.
We know k = 3 , N = 35, α = 0.05
d.f.N = k - 1
= 3 - 1 = 2
d.f.D = N - k
= 35 - 3 = 32
SO the critical value is 3.295
The mean and the variance of each sample :
Goust Jet red Cloudtran


The grand mean or the overall mean is(GM) :


= 52.1714
The variance between the groups

![$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B%5Cleft%5B14%2850.5-52.1714%29%5E2%2B14%2852.07143-52.1714%29%5E2%2B7%2855.71426-52.1714%29%5E2%5Cright%5D%7D%7B3-1%7D%24)

= 63.55714
The Variance within the groups



= 20.93304
The F-test statistics value is :


= 3.036212
Now since the 3.036 < 3.295, we do not reject the null hypothesis.
So there is no sufficient evidence to support the claim that there is a difference among the means.
The ANOVA table is :
Source Sum of squares d.f Mean square F
Between 127.1143 2 63.55714 3.036212
Within 669.8571 32 20.93304
Total 796.9714 34
we know that
p^2 is the the frequency of hba homozygote
we are given
frequency of hba homozygotes is 0.1
so,
p^2 =frequency of hba homozygotes
we can plug values
and we get
..............Answer
Answer:
x²
Step-by-step explanation:
x³ - 1 ÷ x + 2
the first term of the quotient is x³ ÷ x = x²
Answer:
C. with 3000 successes of 5000 cases sample
Step-by-step explanation:
Given that we need to test if the proportion of success is greater than 0.5.
From the given options, we can see that they all have the same proportion which equals to;
Proportion p = 30/50 = 600/1000 = 0.6
p = 0.6
But we can notice that the number of samples in each case is different.
Test statistic z score can be calculated with the formula below;
z = (p^−po)/√{po(1−po)/n}
Where,
z= Test statistics
n = Sample size
po = Null hypothesized value
p^ = Observed proportion
Since all other variables are the same for all the cases except sample size, from the formula for the test statistics we can see that the higher the value of sample size (n) the higher the test statistics (z) and the highest z gives the strongest evidence for the alternative hypothesis. So the option with the highest sample size gives the strongest evidence for the alternative hypothesis.
Therefore, option C with sample size 5000 and proportion 0.6 has the highest sample size. Hence, option C gives the strongest evidence for the alternative hypothesis
(c/5+c)-20 You divide the total cost by five to get the tip. You add that to the cost, and the tip and cost added together is the total cost. Subract that from a $20.00 bill and you have the answer! Hope this helps!