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Ede4ka [16]
3 years ago
5

Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are

independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (σ1= 2 ), so that the standard error of the difference between means is obtained by pooling the sample variances. A paint manufacturer wanted to compare the drying times of two different types of paint. Independent simple random samples of 11 cans of type A and 9 cans of type B were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows.Type A: X1= 71.5hr, S1=3.4 hr N1=11Type B: X2=68.5 hr, S2= 3.6 hr, N2= 9Construct a 99% confidence interval for μ1-μ2 , the difference between the mean drying time for paint type A and the mean drying time for paint type B.
Mathematics
1 answer:
jasenka [17]3 years ago
5 0

Answer:

The indicated confidence interval for the difference between the two population means is  (-1.5159, 7.5159)

Step-by-step explanation:

Let the drying times of type A be the first population and the drying times of type B be the second population. Then

We have small sample sizes n_{1} = 11 and n_{2} = 9, besides \bar{x}_{1} = 71.5, s_{1} = 3.4 , \bar{x}_{2} = 68.5 and s_{2} = 3.6. Therefore, the pooled

estimate is given by  

s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(11-1)(3.4)^{2}+(9-1)(3.6)^{2}}{11+9-2} = 12.1822

The 99% confidence interval for the true mean difference between the mean drying time of type A and the mean drying time of type B is given by

(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{11}+\frac{1}{9}}, i.e.,

(71.5-68.5)\pm t_{0.005}(3.4903)\sqrt{\frac{1}{11}+\frac{1}{9}}

where t_{0.005} is the 0.5th quantile of the t distribution with (11+9-2) = 18 degrees of freedom. So

3\pm(-2.8784)(3.4903)(0.4495), i.e.,

the indicated confidence interval for the difference between the two population means is  (-1.5159, 7.5159)

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Answer:

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Step-by-step explanation:

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Need help in number 12 and 13 PLEASEE!! I don’t get it and I start school the day after tomorrow
Natasha_Volkova [10]

Answer:

The explanations for the graphs are provided down below. Please let me know if you have any questions about my answer.

12 and 13 as written on the worksheet is right.

Step-by-step explanation:

12) The answer given is correct.

The relation between x and y is given as:

y=\frac{x^2}{2}-3 with x \in \{-4,-2,0,2}.

I replaced the word domain with x since the domain is the set of x's for which the relation exists.

We are going to replace x with each of the x's given to see what y corresponds to each.

Let's begin with x=-4.

y=\frac{x^2}{2}-3 with x=-4:

y=\frac{(-4)^2}{2}-3

y=\frac{16}{2}-3

y=8-3

y=5.

So (-4,5) is an ordered pair that should be on our graph.

To find this point you move left 4 from origin then up 5. Now you put a dot where you have landed. Your graph does show this point.

Moving on.

Let's do the next x: x=-2.

y=\frac{x^2}{2}-3 with x=-2:

y=\frac{(-2)^2}{2}-3

y=\frac{4}{2}-3

y=2-3

y=-1.

So (-2,-1) is an ordered pair that should be on our graph.

To find this point you move left 2 from origin and then down 1.  Now you put a dot where you have landed. Your graph shows this point as well.

Now x=0.

y=\frac{x^2}{2}-3 with x=0:

y=\frac{0^2}{2}-3

y=\frac{0}{2}-3

y=0-3

y=-3

So (0,-3) is an ordered pair that should be on our graph.

To find this point you move left and right none and down 3.  Now you put a dot where you have landed. Your graph shows this point.

Now the last point will be at x=2.

y=\frac{x^2}{2}-3 with x=2

y=\frac{2^2}{2}-3

y=\frac{4}{2}-3

y=2-3

y=-1.

So (2,-1) is an ordered pair that should be on our graph.

To find this point you move 2 units right from the origin and then down 1 unit. Now put a dot where you landed.  The graph shows this point as well.

13) The answer given is correct.

g(x)=|x| is the parent function and makes like a V shaped graph where it's vertex is at (0,0).

If we want to move this graph right 3 it becomes:

m(x)=|x-3| \text{ or } m(x)=|(-1)(-x+3)|=|-1||-x+3|=1|-x+3|=|-x+3|=|3-x|.

If you move that up once it becomes:

n(x)=|x-3|+1 or n(x)=|3-x|+1 which is the curve given.

If you don't know about transformations you can choose a few points to plug in to see what's going on with the graph.

Let's choose x=-5,-3,-1,0,1,3,5.

x=-5

f(-5)=|3--5|+1=|3+5|+1=|8|+1=8+1=9.

There is no room for (-5,9) on our graph but if you extended the left hand side of the absolute value function there you would see that (-5,9) is reached.

x=-3

f(-3)=|3--3|+1=|3+3|+1=|6|+1=6+1=7.

(-3,7) should be a point on the graph. Same thing for this point as (-5,9).

x=-1

f(-1)=|3--1|+1=|3+1|+1=|4|+1=4+1=5.

(-1,5) is located on the graph.

x=0

f(0)=|3-0|+1=|3|+1=3+1=4.

(0,4) is also located on the graph.

x=1

f(1)=|3-1|+1=|2|+1=2+1=3.

(1,3) is located on the graph.

x=3

f(3)=|3-3|+1=|0|+1=0+1=1.

(3,1) is located on the graph.

x=5

f(5)=|3-5|+1=|-2|+1=2+1=3.

(5,3) is located on the graph.

Now if we weren't given the graph already:

I would plot the points I found which were:

(-5,9)

(-3,7)

(-1,5)

(0,4)

(1,3)

(3,1)

(5,3)

We should get a basic idea of what the function looks like from these points.

I will graph them. You will have to connect these points because the domain isn't discrete like number 12 is.  That is they didn't list out elements for the domain.

I'm going to graph one more point after x=5.

How about x=7?

f(7)=|3-7|+1=|-4|+1=4+1=5

So (7,5) is also a point on the graph.

You should see that the blue points are following the red path I made there.

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