In rhombus LMNO, diagonals top enclose L N end enclose and top enclose M O end enclose intersect at P. If the measure of angle L
1 answer:
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Answer:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
(–2975, 175)
Do not reject H₀
Step-by-step explanation:
Hello!
The objective of this experiment is to test if the salaries of elementary school teachers are equal in two districts. You can test this trough the population means of the salaries of the teachers if they are either equal or different or directly test if the difference between the salaries of the two districts is cero or not, symbolically: μ₁ - μ₂ = 0
Remember, in the null hypothesis is usually stated the known information, is the "no change" premise and always carries the = symbol.
The hypothesis is:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
You have two normally distributed variables and you are studying the difference of the means. You can use a pooled Z to make the interval, the formula is:
X[bar]₁-X[bar]₂ ± *(√(σ₁²/n₁+σ₂²/n₂)
Since the test is two-tailed and at a signification level of 5% I've made the interval at the complementary confidence level of 95% so that I can use it to decide over the hypothesis.
X[bar]₁-X[bar]₂ ± *(√(σ₁²/n₁+σ₂²/n₂)
=
= 1,96
[28900-30300 ± 1.96*(√((2300)²/15+(2100)²/15)]
[-1400 ± 1576,15]
[-2976.15;176,15]
Since I've approximated to two decimal units in the intermediate calculations, the values differ slightly, but the interval is:
(–2975, 175)Now, since the 0 is contained in the Confidence interval, the decision is to not reject the null hypothesis. In other words, the difference in average salaries between the two districts is cero.
I hope it helps!
(1)
Mean length of all fish in the sample -
<u>Mean</u> = (Sum of observations)/(Number of observations)
= (12 + 5 + 3 + 5 + 8 + 2 + 10 + 9 + 4 + 4)/(10)
= 62/10
=<em> 6.2</em>
(2)
Mean length of adult fish in the sample -
<u>Mean</u> = (Sum of observations)/(Number of observations)
= (12 + 5 + 8 + 10)/(4)
= 35/4
=<em> 8.75</em>
(3)
Mean length of juvenile fish in the sample -
<u>Mean</u> = (Sum of observations)/(Number of observations)
= (5 + 3 + 2 + 9 + 4 + 4)/(6)
= 27/6
<em>= 4.5</em>
(4)
Percentage of sample that were adult fish -
<u>Percentage</u> = (No. of adult fishes)/(Total no. of fishes) × 100
% = (4/10) × 100
<em>% = 40</em>
(5)
Percentage of sample that were juvenile fish -
<u>Percentage</u> = (No. of juvenile fishes)/(Total no. of fishes) × 100
% = (6/10) × 100
<em>% = </em><em>6</em><em>0</em>
(6)
Percentage of sample that were juveniles over 8 inches long -
<u>Percentage</u> = (No. of juveniles over 8 inches)/(Total no. of fishes) × 100
% = (1/10) × 100
<em>% = </em><em>1</em><em>0</em>