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Alex_Xolod [135]
3 years ago
14

A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores

. Marketing managers are prone to look at the estimate and ignore sampling error. An SRS of 28
stores this year shows mean sales of 64 units of a small appliance, with a standard deviation of 13.8 units. During the same point in time last year, an SRS of 28 stores had mean sales of 48.958 units, with standard deviation 15.4 units. An increase from 48.958 to 64 is a rise of about 24%.


1. Construct a 99% confidence interval estimate of the difference μ1−μ2

μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales.


(b) The margin of error is



2. At a 0.01 significance level, is there sufficient evidence to show that sales this year are different from last year?
Mathematics
1 answer:
allochka39001 [22]3 years ago
3 0

Answer:

Confidence interval = (4.61972, 25.46428)

Margin of Error = 3.9078675

Step-by-step explanation:

Given that :

This year:

Mean (μ1) = 64

Standard deviation (s1) = 13.8

Sample size (n1) = 28

Last year:

Mean (μ2) = 48.958

Standard deviation (s2) = 15.4

Sample size (n2) = 28

99% confidence interval estimate of the difference μ1−μ2

α = 1 - 99% = 0.01

(μ1−μ2) ± t0.01,27 * (√s1²/n1 + s2²/n2)

t0.01, 27 = 2.770683 (t value calculator)

√s1²/n1 + s2²/n2 = √13.8^2/28 + 15.4^2/28 = 3.9078675

(64 - 48.958) ± 2.667(3.9078675)

15.042 ± 10.42228

(15.042 - 10.42228), (15.042 + 10.42228)

(4.61972, 25.46428)

The margin of error :

√(s1²/n1) + (s2²/n2)

√(13.8^2/28) + (15.4^2/28)

√(6.8014285 + 8.47)

√15.2714285

= 3.9078675

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Answer:

Step-by-step explanation:

4(8+2n) = 4 + 4n

32 + 8n = 4+4n

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n = -7

4 0
3 years ago
Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
leonid [27]
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
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5 0
4 years ago
Find the sum of a finite geometric sequence from n = 1 to n = 7, using the expression −4(6)n − 1.
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Step-by-step explanation:

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a_n=-4(6)^{n-1}

Calculate a₁. Put n = 1:

a_1=-4(6)^{1-1}=-4(6)^0=-4(1)=-4

Calculate the common ratio:

r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=-4(6)^{n+1-1}=-4(6)^n\\\\r=\dfrac{-4(6)^n}{-4(6)^{n-1}}=6^n:6^{n-1}\\\\\text{use}\ a^n:a^m=a^{n-m}\\\\r=6^{n-(n-1)}=6^{n-n+1}=6^1=6

\text{Substitute}\ a_1=-4,\ n=7,\ r=6:\\\\S_7=-4\cdot\dfrac{1-6^7}{1-6}=-4\cdot\dfrac{1-279936}{-5}=-4\cdot\dfrac{-279935}{-5}=(-4)(55987)\\\\S_7=-223948

7 0
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Answer:

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When 2 lines are parallel there slope is same.

So, Slope of line BC =Slope of Line DE

\frac{y_2-y_1}{x_2-x_1}=\frac{y_2-y_1}{x_2-x_1}

We have:

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Putting values and finding c

\frac{6-2}{9-2}=\frac{-3-(-7)}{5-c}\\ \frac{4}{7}=\frac{-3+7}{5-c} \\ \frac{4}{7}=\frac{4}{5-c} \\Cross \ multiply:\\4(5-c)=4*7\\20-4c=28\\-4c=28-20\\-4c=8\\c=\frac{8}{-4}\\c=-2

So, If the lines BC and DE are parallel, the value of c is c=-2

8 0
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CaHeK987 [17]

Answer:

520miles per day

Step-by-step explanation:

2080/4=520

5 0
2 years ago
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