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rosijanka [135]
2 years ago
14

Solve by completing the square:

Mathematics
1 answer:
laiz [17]2 years ago
8 0
The answer is 4 trust me bro
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Cooper was out at a restaurant for dinner when the bill came. His dinner came to $18. After adding in a tip, before tax, he paid
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Answer: Copper paid 13%

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Who was the best president in history
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<em>Abraham Lincoln </em>

Step-by-step explanation:

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2 years ago
A sample of 41 observations is selected from one population with a population standard deviation of 3.4. The sample mean is 100.
Marizza181 [45]

Answer:

a) If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

b) (-1.9886, 1.9886)

c) t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617

d) p_v =2*P(t_{84}>1.617) =0.1096

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.  

e) p_v =2*P(t_{84}>1.617) =0.1096

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 = 0

Alternative hypothesis: \mu_1 -\mu_2\neq 0

Our notation on this case :

n_1 =41 represent the sample size for group 1

n_2 =45 represent the sample size for group 2

\bar X_1 =100 represent the sample mean for the group 1

\bar X_2 =98.4 represent the sample mean for the group 2

s_1=3.4 represent the sample standard deviation for group 1

s_2=5.6 represent the sample standard deviation for group 2

Part a

If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

Part b

On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution \alpha/2 = 0.025 of the area.

The distribution on this cas since we don't know the population deviation for both samples is the t distribution with df=n_1+n_2 -2= 41+45-2=84 degrees of freedom.

We can use the following excel codes in order to find the critical values:

"=T.INV(0.025,84)", "=T.INV(1-0.025,84)"

And we got: (-1.9886, 1.9886)

Part c

The statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}

And now we can calculate the statistic:

t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617

The degrees of freedom are given by:

df=41+45-2=84

Part d

And now we can calculate the p value using the altenative hypothesis:

p_v =2*P(t_{84}>1.617) =0.1096

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.  

Part e

p_v =2*P(t_{84}>1.617) =0.1096

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

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Step-by-step explanation:

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The answer is 2. a)1.5n = 13.65 b) n = 9.1
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