Astudy was conducted to determine if the salaries of elementary school teachers from two neighboring districts were equal. a sample of 15 teachers from each district was randomly selected. test the claim that the salaries from both districts are equal. use significance 0.05 and assume that the data are distributed normally. n x¯ s district 1: 15 $28,900 $2,300 district 2: 15 $30,300 $2,100 3a. what is the appropriate set of hypotheses (h0, h1)?
• µ1 – µ2 = 0, µ1 – µ2 ≠ 0
• µ1 – µ2 ≠ 0, µ1 – µ2 = 0
• µ1 – µ2 = 0, µ1 – µ2 < 0
• µ1 – µ2 = 0, µ1 – µ2 > 0 3b.
what is the correct confidence interval?
• (–2475, –325)
• (–2100, –700)
• (–2975, 175)
• (–3125, 325) 3c.
are the salaries of the teachers from the two districts different?
• yes.
• no.
• sometimes.
• depends on distribution.

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!

d (5 x (-20)) * (100 x (0.7))

step-by-step explanation:

5÷1/100 * -.7 *-20

copy dot flip

5*100 * -.7 *-20

rearrange

5*-20   * 100 *-.7

-100   *-70

7000