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) ?

Question

Mathematics

Ruffles

30 January, 21:51

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) ?

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Answers (1)

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Aarav Oconnor · Answer 1

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