Answer:
Part 1 , not significant
Part II, significant
Step-by-step explanation:
Given that a heterozygous white-fruited squash plant is crossed with a yellow-fruited plant, yielding 200 seeds. of these seeds, 110 produce white-fruited plants while only 90 produce yellow-fruited plants
![H_0: Both colours are equally likely\\H_a: Both colours are not equally likely](https://tex.z-dn.net/?f=H_0%3A%20Both%20colours%20are%20equally%20likely%5C%5CH_a%3A%20Both%20colours%20are%20not%20equally%20likely)
(Two tailed chi square test)
We assume H0 to be true and find out expected
If H0 is true expected would be 100 white and 100 yellow
Chi square = ![\Sigma \frac{(O-E)^2}{E} \\=\frac{(110-100)^2}{100} +frac{(90-100)^2}{100} \\=2](https://tex.z-dn.net/?f=%5CSigma%20%5Cfrac%7B%28O-E%29%5E2%7D%7BE%7D%20%5C%5C%3D%5Cfrac%7B%28110-100%29%5E2%7D%7B100%7D%20%2Bfrac%7B%2890-100%29%5E2%7D%7B100%7D%20%5C%5C%3D2)
df = 1
p value = 0.152799
Since p value > 0.05 at 5% level we accept that the colours are equally likely
2)
Here observed are 1100 and 900
Expected 1000 & 1000
df = 1
Chi square = ![\frac{20000}{100} =200](https://tex.z-dn.net/?f=%5Cfrac%7B20000%7D%7B100%7D%20%3D200)
p value <0.0001
These results are here statistically significant.
Answer:
$95
Step-by-step explanation:
you had to subtract the two numbers .
Answer: 0.332 < p < 0.490
Step-by-step explanation:
We know that the confidence interval for population proportion is given by :-
![\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}](https://tex.z-dn.net/?f=%5Chat%7Bp%7D%5Cpm%20z%5E%2A%5Csqrt%7B%5Cdfrac%7B%5Chat%7Bp%7D%281-%5Chat%7Bp%7D%29%7D%7Bn%7D%7D)
, where n= sample size
= sample proportion
z* = critical z-value.
As per given , we have
n= 258
Sample proportion of college students who own a car = ![\hat{p}=\dfrac{106}{258}\approx0.411](https://tex.z-dn.net/?f=%5Chat%7Bp%7D%3D%5Cdfrac%7B106%7D%7B258%7D%5Capprox0.411)
Critical z-value for 99% confidence interval is 2.576. (By z-table)
Therefore , the 99% confidence interval for the true proportion(p) of all college students who own a car will be :![0.411\pm (2.576)\sqrt{\dfrac{0.411(1-0.411)}{258}}\\\\=0.411\pm (2.576)\sqrt{0.00093829}\\\\= 0.411\pm (2.576)(0.0306315197142)\\\\=0.411\pm 0.0789=(0.411-0.0789,\ 0.411+0.0789)\\\\=(0.3321,\ 0.4899)\approx(0.332,\ 0.490)](https://tex.z-dn.net/?f=0.411%5Cpm%20%282.576%29%5Csqrt%7B%5Cdfrac%7B0.411%281-0.411%29%7D%7B258%7D%7D%5C%5C%5C%5C%3D0.411%5Cpm%20%282.576%29%5Csqrt%7B0.00093829%7D%5C%5C%5C%5C%3D%200.411%5Cpm%20%282.576%29%280.0306315197142%29%5C%5C%5C%5C%3D0.411%5Cpm%200.0789%3D%280.411-0.0789%2C%5C%200.411%2B0.0789%29%5C%5C%5C%5C%3D%280.3321%2C%5C%200.4899%29%5Capprox%280.332%2C%5C%200.490%29)
Hence, a 99% confidence interval for the true proportion of all college students who own a car : 0.332 < p < 0.490
Finding the rate of the weight gain, we divide the total weight gained with the total number of months that took to gain that weight.
Total time (months) = 0.5 + 2 + 3.5 = 6 months
Total weight (kg) = 80.6 + 88.4 + 96.2 = 265.2 kilograms
Rate = Kg ÷ Months
Rate = 265.2 ÷ 6 = 44.2
44.2 kilograms per month
Cheers!
DF =DE + EF
4x-8 = x-4 + 2x+5
4x - 8 = 3x + 1
4x - 3x = 1 + 8
x = 9