Answer:
<h3>p = 131.25</h3>Step-by-step explanation:
The variation p varies directly with T is written as
p = kT
where k is the constant of proportionality
To find p when T =500 we must first find the formula for the variation
That's
when p = 105 and T = 400
105 = 400k
Divide both sides by 400
<h3>So the formula for the variation is
<h2>when
T = 500
Substitute it into the above formula
That's
Simplify
The final answer is
<h3>p = 131.25</h3>Hope this helps you
Answer:
Part A
b. 14.6 ± 7.38
Part B
b. 3.43
Part C
a. P-value < 0.01
Part D
b. There is sufficient evidence to reject the null hypothesis
Step-by-step explanation:
Part A
The given data are;
The number of seedlings in the field = 20
The number of seedlings selected to receive herbicide A = 10
The number of seedlings selected to receive herbicide B = 10
The height in centimeters of seedlings treated with Herbicide A, = 94.5 cm
The standard deviation, s₁ = 10 cm
The height in centimeters of seedlings treated with Herbicide B, = 109.1 cm
The standard deviation, s₂ = 9 cm
The 90% confidence interval for μ₂ - μ₁, is given as follows;
The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;
The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18
α = 100% - 90% = 10%
∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05
= 1.734
C.I. ≈ 14.6 ± 7.37714603353
The 90% C.I. ≈ 14.6 ± 7.38
b. 14.6 ± 7.38
Part B
With the hypotheses are given as follows;
H₀; μ₂ - μ₁ = 0
Hₐ; μ₂ - μ₁ ≠ 0
The two sample t-statistic is given as follows;
The two sample t-statistic ≈ 3.43
b. 3.43
Part C
From the t-table, the p-value, we have, the p-value < 0.01
a. P-value < 0.01
Part D
Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis
b. There is sufficient evidence to reject the null hypothesis.