Null hypothesis: products equally
Alternative hypothesis: #2 sells cheaper than #1
H0: µ1 = µ2, Ha: µ1 > µ2
N=20
Please assist in finding the following:
random predicted outcome.
second outcome of 16/20 statistic
difference
sample difference
standard error
df
standardized statistics
p-value
standard deviation
standard error
confidence interval within 95%
| product name | stat 1 | stat 2 | Price Comparison |
| a | $ 14.39 | $ 10.39 | $ 4.00 |
| b | $ 14.39 | $ 10.94 | $ 3.45 |
| c | $ 9.99 | $ 9.99 | $ - |
| d | $ 7.19 | $ 3.99 | $ 3.20 |
| e | $ 14.40 | $ 12.24 | $ 2.16 |
| f | $ 8.99 | $ 8.59 | $ 0.40 |
| g | $ 3.95 | $ 3.95 | $ - |
| h | $ 9.89 | $ 8.50 | $ 1.39 |
| i | $ 14.40 | $ 12.14 | $ 2.26 |
| j | $ 17.96 | $ 10.79 | $ 7.17 |
| k | $ 16.19 | $ 13.99 | $ 2.20 |
| l | $ 15.30 | $ 10.09 | $ 5.21 |
| m | $ 6.97 | $ 6.97 | $ - |
| n | $ 13.49 | $ 8.99 | $ 4.50 |
| o | $ 15.29 | $ 9.99 | $ 5.30 |
| p | $ 14.80 | $ 12.33 | $ 2.47 |
| o | $ 13.01 | $ 13.01 | $ - |
| p | $ 15.26 | $ 14.26 | $ 1.00 |
| q | $ 11.70 | $ 10.66 | $ 1.04 |
| r | $ 13.30 | $ 12.07 | $ 1.23 |
| Sample #1 | Sample #2 | difference , Di =sample1-sample2 | (Di - Dbar)² |
| 14.39 | 10.39 | 4.00 | 2.73 |
| 14.39 | 10.94 | 3.45 | 1.21 |
| 9.99 | 9.99 | 0.00 | 5.52 |
| 7.19 | 3.99 | 3.20 | 0.72 |
| 14.4 | 12.24 | 2.16 | 0.04 |
| 8.99 | 8.59 | 0.40 | 3.80 |
| 3.95 | 3.95 | 0.00 | 5.52 |
| 9.89 | 8.5 | 1.39 | 0.92 |
| 14.4 | 12.14 | 2.26 | 0.01 |
| 17.96 | 10.79 | 7.17 | 23.24 |
| 16.19 | 13.99 | 2.20 | 0.02 |
| 15.3 | 10.09 | 5.21 | 8.19 |
| 6.97 | 6.97 | 0.00 | 5.52 |
| 13.49 | 8.99 | 4.50 | 4.63 |
| 15.29 | 9.99 | 5.30 | 8.71 |
| 14.8 | 12.33 | 2.47 | 0.01 |
| 13.01 | 13.01 | 0.00 | 5.52 |
| 15.26 | 14.26 | 1.00 | 1.82 |
| 11.7 | 10.66 | 1.04 | 1.71 |
| 13.3 | 12.07 | 1.23 | 1.25 |
| Sample #1 | Sample #2 | difference , Di =sample1-sample2 | (Di - Dbar)² |
| 14.39 | 10.39 | 4.00 | 2.73 |
| 14.39 | 10.94 | 3.45 | 1.21 |
| 9.99 | 9.99 | 0.00 | 5.52 |
| 7.19 | 3.99 | 3.20 | 0.72 |
| 14.4 | 12.24 | 2.16 | 0.04 |
| 8.99 | 8.59 | 0.40 | 3.80 |
| 3.95 | 3.95 | 0.00 | 5.52 |
| 9.89 | 8.5 | 1.39 | 0.92 |
| 14.4 | 12.14 | 2.26 | 0.01 |
| 17.96 | 10.79 | 7.17 | 23.24 |
| 16.19 | 13.99 | 2.20 | 0.02 |
| 15.3 | 10.09 | 5.21 | 8.19 |
| 6.97 | 6.97 | 0.00 | 5.52 |
| 13.49 | 8.99 | 4.50 | 4.63 |
| 15.29 | 9.99 | 5.30 | 8.71 |
| 14.8 | 12.33 | 2.47 | 0.01 |
| 13.01 | 13.01 | 0.00 | 5.52 |
| 15.26 | 14.26 | 1.00 | 1.82 |
| 11.7 | 10.66 | 1.04 | 1.71 |
| 13.3 | 12.07 | 1.23 | 1.25 |
| sample 1 | sample 2 | Di | (Di - Dbar)² | |
| sum = | 250.86 | 203.88 | 46.980 | 81.080 |
Ho : µd= 0
Ha : µd > 0
Level of Significance , α =
0.05 claim:µd=0
sample size , n = 20
mean of sample 1, x̅1= 12.543
mean of sample 2, x̅2= 10.194
sample difference , D̅ =ΣDi / n
= 2.349
std dev of difference , Sd = √ [
(Di-Dbar)²/(n-1) = 2.0658
std error , SE = Sd / √n = 2.0658
/ √ 20 = 0.4619
t-statistic = (D̅ - µd)/SE = (
2.349 - 0 ) /
0.4619 = 5.085
Degree of freedom, DF= n - 1 =
19
p-value =
0.000033 [excel function: =t.dist.rt(t-stat,df) ]
sample size , n = 20
Degree of freedom, DF= n - 1 =
19 and α = 0.05
t-critical value = t α/2,df =
2.0930 [excel function: =t.inv.2t(α/2,df) ]
std dev of difference , Sd = √ [ (Di-Dbar)²/(n-1) =
2.0658
std error , SE = Sd / √n = 2.0658 /
√ 20 = 0.4619
margin of error, E = t*SE = 2.0930
* 0.4619 = 0.9668
mean of difference , D̅ =
2.349
confidence interval is
Interval Lower Limit= D̅ - E = 2.349
- 0.9668 = 1.382
Interval Upper Limit= D̅ + E = 2.349
+ 0.9668 = 3.316
so, confidence interval is ( 1.3822
< µd < 3.3158
)
Null hypothesis: products equally Alternative hypothesis: #2 sells cheaper than #1 H0: µ1 = µ2, Ha:...