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vichka [17]
2 years ago
11

Solve by substitution

Mathematics
1 answer:
Alekssandra [29.7K]2 years ago
3 0
In second equation add 0.74 and y to both sides: 3.25x+0.74=y;
Substitute y in the first equation: -4.5x-2(3.25x+0.74)=-12.5;
Distribute multiplication over addition:
-4.5x-6.50x -1.48=-12.5;
Combine like terms and add 1.48 to both sides:
-11x= -12.5+1.48;
-11x= 11.02 divide both sides by -11;
X= -11.02/11= -1.00(18)
Y= 3.25• -1.00(18)+.74= -2.515(90)
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Null hypothesis:\mu_{1}-\mu_{0}\leq 0

Alternative hypothesis:\mu_{1}-\mu_{2}>0

SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705

b) 2.70

t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850

b. 1.85

p_v =P(Z>1.85)=0.032

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a. We can reject H(0) in favor of H(a)

Step-by-step explanation:

Data given and notation

\bar X_{1}=164 represent the mean for the sample 1

\bar X_{2}=159 represent the mean for the sample 2

\sigma_{1}=12.5 represent the population standard deviation for the sample 1

s_{2}=9.25 represent the population standard deviation for the sample B2

n_{1}=35 sample size selected 1

n_{2}=30 sample size selected 2

\alpha=0.05 represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the expenditure of households in City 1 is more than that of households in City 2, the system of hypothesis would be:

Null hypothesis:\mu_{1}-\mu_{0}\leq 0

Alternative hypothesis:\mu_{1}-\mu_{2}>0

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})-0}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)

t-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".

Standard error

The standard error on this case is given by:

SE=\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}

Replacing the values that we have we got:

SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705

b. 2.70

Calculate the statistic

We can replace in formula (1) the info given like this:

t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850  

P-value

Since is a one side right tailed test the p value would be:

p_v =P(Z>1.85)=0.032

b. 0.03

Conclusion

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

a. We can reject H(0) in favor of H(a)

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