A manufacturer produces both a deluxe and a standard model o

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

Retail Outlet        Deluxe       Standard

1                         40              27

2                         39              28

3                         43              35

4                         38              31

5                         40              30

6                         39              32

7                         36              29

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) Develop the null and alternative hypotheses

b)Calculate the value of the test statistic. If required enter negative values as negative numbers. (to 2 decimals).

c) P-value =

d) What is the 95% confidence interval for the difference between the mean prices of the two models (to 2 decimals)?

Solution

a)
Set Up Hypothesis
Null, mean difference between the prices of the two models is $10 Ho: u1 - u2 = 10
Alternate, mean difference between the prices of the two models is n\'t $10 - H1: u1 - u2 != 10
b)
Test Statistic
X(Mean)=39.286
Standard Deviation(s.d1)=2.138 ; Number(n1)=7
Y(Mean)=30.286
Standard Deviation(s.d2)=2.69; Number(n2)=7
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to = ((39.286-30.286)-10)/Sqrt((4.57104/7)+(7.2361/7))
to = -0.7699
| to | =0.7699
c)
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -0.7699 ) = 0.471
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 6 d.f is 2.447
We got |to| = 0.7699 & | t | = 2.447
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho

d)
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
CI = [ ( 39.286-30.286) ±t a/2 * Sqrt( 4.571044/7+7.2361/7)]
= [ (9) ± t a/2 * Sqrt( 1.69) ]
= [ (9) ± 2.447 * Sqrt( 1.69) ]
= [5.82 , 12.18]