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PSYCHO15rus [73]
3 years ago
15

Explain how to perform a​ two-sample z-test for the difference between two population means using independent samples with sigma

1 and sigma 2 known.
Mathematics
1 answer:
Rom4ik [11]3 years ago
5 0

Answer:

Step-by-step explanation:

In the two independent samples application, it involves the test of hypothesis that is the difference in population means, μ1 - μ2. The null hypothesis is always that there is no difference between groups with respect to means.

Null hypothesis: ∪₁ = ∪₂. where ∪₁ represent the mean of sample 1 and ∪₂ represent the mean of sample 2.

A researcher can hypothesize that the first mean is larger than the second (H1: μ1 > μ2 ), that the first mean is smaller than the second (H1: μ1 < μ2 ), or that the means are different (H1: μ1 ≠ μ2 ). These ae the alternative hypothesis.

Thus for the z test:

    if n₁ > 30 and n₂ > 30

     z = X₁ - X₂ / {Sp[√(1/n₁ + 1/n₂)]}

where Sp is √{ [(n₁-1)s₁² + (n₂-1)s₂²] / (n₁+n₂-2)}

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What is the LCD of , 3/4 ,4/5 and 7/10
lesantik [10]

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20

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20 is divisible by 4, 5 and 10

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3 years ago
A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to fin
g100num [7]

Answer: (24.28,\ 27.72)

Step-by-step explanation:

Given : Sample size : n=50

Sample mean : \overline{x}=26

Standard deviation : \sigma =6.2

Significance level : \alpha=1-0.95=0.05

Critical value : z_{\alpha/2}=1.96

Formula to find the confidence interval for population mean :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=26\pm(1.96)\dfrac{6.2}{\sqrt{50}}\\\\\approx26\pm1.72\\\\=(26-1.72,\ 26+1.72)\\\\=(24.28,\ 27.72)

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6 0
3 years ago
(adapted from Ross, 2.31) Three countries (the Land of Fire, the Land of Wind, and the Land of Earth) each make a 3 person team.
DaniilM [7]

Answer:

The answer is "\frac{2}{9} \  and \ \frac{1}{9}"

Step-by-step explanation:

In point a:

The requires  1 genin, 1 chunin , and 1 jonin to shape a complete team but we all recognize that each nation's team is comprised of 1 genin, 1 chunin, and 1 jonin.

They can now pick 1 genin from a certain matter of national with the value:

\frac{1}{\binom{3}{1}}=\frac{1}{3} .

They can pick 1 Chunin form of the matter of national with the value:

\frac{1}{\binom{3}{1}}=\frac{1}{3} .

They have the option to pick 1 join from of the country team with such a probability: \frac{1}{\binom{3}{1}}=\frac{1}{3}

And we can make the country teams: 3! = 6 different forms. Its chances of choosing a team full in the process described also are:

6 \times \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}=\frac{2}{9}.

In point b:

In this scenario, one of the 3 professional sides can either choose 3 genins or 3 chunines or 3 joniners. So, that we can form three groups that contain the same ninjas (either 3 genin or 3 chunin or 3 jonin).

Its likelihood that even a specific nation team ninja would be chosen is now: \frac{1}{\binom{3}{1}}=\frac{1}{3}

Its odds of choosing the same rank ninja in such a different country team are: \frac{1}{\binom{3}{1}}=\frac{1}{3}

The likelihood of choosing the same level Ninja from the residual matter of national is: \frac{1}{\binom{3}{1}}=\frac{1}{3} Therefore, all 3 selected ninjas are likely the same grade: 3\times \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}=\frac{1}{9}

4 0
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7 0
3 years ago
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