Two types of plastic are suitable for an electronics component manufacturer to use. the breaking strength of this plastic is important. it is known that σ1 =σ2 = 1.0 psi. from a random sample of size n1 =10 and n2 =12, you obtain ¯x1 =162.5 and ¯x2 = 155.0. the company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi (inclusive). (a) based on the sample information, should it use plastic 1? test this using α=0.05. write formal null and alternate hypotheses statements for the test. (b) find the p-value for this test.

(a) Null Hypothesis (H₀) = Mean of x₁ > Mean of x₂

Alternative Hypothesis (H₁) = Mean of x₁ ≠ Mean of x₂

(b) We will use here t-test for two sample means which is given by,

" alt="t = \frac{\bar_{x_{1}}-\bar_{x_{2}}}{\sqrt{s^2(\frac{1}{n_{1}}+\frac{1}{n_{2}} )}}" />" />

Putting all values we get, t = 17.5 with 20 degree of freedom

At α = 0.05

p-value = 95% Confidence Interval for the Difference ( 6.6068 , 8.3932 )

phosphate, calcium ion, chloride ion, neon atom

step-by-step explanation:

a 4/6 = 9/12   not proportional

step-by-step explanation:

to find which are not proportional, we can use cross products

if it is true, they are proportional.   if it is not true, they are not proportional.

a 4/6 = 9/12

4* 12 = 6*9

48 = 54

not this is not proportional  

b 2/3 = 12/18  

2* 18 = 3 * 12

36 = 36

this is proportional

c 5/4 = 10/8  

5*8 = 4*10

40 = 40

this is proportional

d 7/5 = 35/25

7*25 = 5 * 35

175 = 175

is proportional