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Eva8 [605]
3 years ago
11

Consider the following hypothesis test: H0: LaTeX: \mu_1-\mu_2=0μ 1 − μ 2 = 0 Ha: LaTeX: \mu_1-\mu_2\ne0μ 1 − μ 2 ≠ 0 The follow

ing results are for two independent samples taken from the two populations. Sample 1: LaTeX: n_1n 1 = 80; LaTeX: \overline{x}_1x ¯ 1 = 104; LaTeX: \sigma_1σ 1 = 8.4 Sample 2: LaTeX: n_2n 2 = 70; LaTeX: \overline{x}_2x ¯ 2 = 106; LaTeX: \sigma_2σ 2 = 7.6 a. What is the value of the test statistic?
Mathematics
1 answer:
Pavel [41]3 years ago
7 0

Answer:

z=\frac{104-\bar 106}{\sqrt{\frac{8.4^2}{80}+\frac{7.6^2}{70}}}}=-1.53

Step-by-step explanation:

Data given and notation

\bar X_{1}=104 represent the mean for the sample 1

\bar X_{2}=106 represent the mean for the sample 2

\sigma_{1} =8.4 represent the population standard deviation for the sample 1

\sigma_{2}=7.6 represent the population standard deviation for the sample 2

n_{1}=80 sample size for the group 1

n_{2}=70 sample size for the group 2

z would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

H0:\mu_{1}-\mu_{2} =0

H1:\mu_{1} -\mu_{2} \neq 0

If we analyze the size for the samples both are greater than 30 and we know the population deviations so for this case is better apply a z test to compare means, and the statistic is given by:

z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

First we need to calculate the mean and deviation for each sample, after apply the formulas (2) and (3) we got the following results:

And with this we can replace in formula (1) like this:

z=\frac{104-\bar 106}{\sqrt{\frac{8.4^2}{80}+\frac{7.6^2}{70}}}}=-1.53

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Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up
Artyom0805 [142]

Answer:

In the long run, ou expect to  lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X \sim geom(p = 1/2)

the probability function P can be computed as:

P (X = n) = p(1-p)^{n-1}

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

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\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2        ∵  X \sim geom(p = 1/2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =6

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to  lose $4 per game

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