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Phantasy [73]
3 years ago
14

A manufacturer produces both a deluxe and a standard model of an automatic sander designed for home use. Selling prices obtained

from a sample of retail outlets follow. Model Price ($) Model Price ($) Retail Outlet Deluxe Standard Retail Outlet Deluxe Standard 1 39 27 5 40 30 2 39 29 6 39 35 3 46 35 7 35 29 4 38 31 The manufacturer's suggested retail prices for the two models show a $10 price differential. Use a .05 level of significance and test that the mean difference between the prices of the two models is $10. a.Calculate the value of the test statistic (to 2 decimals).
b.What is the 95% confidence interval for the difference between the mean prices of the two models (to 2 decimals)?
Mathematics
1 answer:
Anarel [89]3 years ago
7 0

Answer:

Step-by-step explanation:

The data is incorrect. The correct data is:

Deluxe standard

39 27

39 28

45 35

38 30

40 30

39 34

35 29

Solution:

Deluxe standard difference

39 27 12

39 28 11

45 35 10

38 30 8

40 30 10

39 34 5

35 29 6

a) The mean difference between the selling prices of both models is

xd = (12 + 11 + 10 + 8 + 10 + 5 + 6)/7 = 8.86

Standard deviation = √(summation(x - mean)²/n

n = 7

Summation(x - mean)² = (12 - 8.86)^2 + (11 - 8.86)^2 + (10 - 8.86)^2 + (8 - 8.86)^2 + (10 - 8.86)^2 + (5 - 8.86)^2 + (6 - 8.86)^2 = 40.8572

Standard deviation = √(40.8572/7

sd = 2.42

For the null hypothesis

H0: μd = 10

For the alternative hypothesis

H1: μd ≠ 10

This is a two tailed test.

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 7 - 1 = 6

2) The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (8.86 - 10)/(2.42/√7)

t = - 1.25

We would determine the probability value by using the t test calculator.

p = 0.26

Since alpha, 0.05 < than the p value, 0.26, then we would fail to reject the null hypothesis.

b) Confidence interval is expressed as

Mean difference ± margin of error

Mean difference = 8.86

Margin of error = z × s/√n

z is the test score for the 95% confidence level and it is determined from the t distribution table.

df = 7 - 1 = 6

From the table, test score = 2.447

Margin of error = 2.447 × 2.42/√7 = 2.24

Confidence interval is 8.86 ± 2.24

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MAVERICK [17]

Answer:

Answer

sqrt(108)

6 sqrt(3)

10.3923

I don't know which answer you  want.

Step-by-step explanation:

  • Drop a perpendicular from the top angle to the base.
  • The base is cut into 2 equal parts.
  • Each part is 12/2 = 6
  • The perpendicular, as its name implies, meets the base at 90o.
  • You can use the Pythagorean Theorem to find the height.

h^2 = side^2 - (1/2 b)^2

h^2 = 12^2 - 6^2

h^2 = 144 - 36

h^2 = 108

Take the square root of both sides

h = sqrt(108)

h = sqrt(2*2 * 3 * 3 * 3)

h = 2 * 3 sqrt(3)

h = 6sqrt(3)

h = 6 * 1.7321

h = 10.3923

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Sveta_85 [38]

Answer:

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Step-by-step explanation:

5 0
3 years ago
The mall is 4.72 kilometers from Julie’s house and 1.83 kilometers from Taylor’s house. How much farther does Julie live from th
blondinia [14]
Julies lives 2.89 kilometers more from the mall than Taylor. 
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3 years ago
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I need help on understanding this concept
Sedbober [7]
So the slope formula is m= (y2-y1) /
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3 0
2 years ago
Mr. Baldwin is packing the shipping box with ring boxes that are cubes with an edge length of 3/2 inches. If he completely fills
Alinara [238K]

Answer:

10 ring boxes

Step-by-step explanation:

First, we need to calculate the total surface area of each cube ring boxes

The surface area of each square boxes = 6L²

Given that L =1.5inches

Total surface area = 6(1.5)²

Total surface area = 6(2.25)

Total surface area = 13.5in²

<em>Since the question is incomplete. Let us assume the total surface area of the shipping box is 135in²</em>

<em></em>

Number of ring boxes he can ship = 135/13.5

Number of ring boxes he can ship = 10

Hence the number of ring boxes he can ship is 10 ring boxes

<em />

<u><em>NB: The total surface area of the shipping box was assumed</em></u>

<u><em></em></u>

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