Answer:
1).
99% CI for mean difference= (0.5109, 6.0891).
2).
Type of variable: independent: categorical dependent: continuous
Type of question: difference
Basis of comparison: means
Ho: µ1 = µ2 H1: µ1 > µ2
α =0.01 one tail test ( upper tail test)
Two sample students unpaired t
P= 0.0016 compare p value with α =0.01
Reject H0. ( because P value < α).
There is sufficient evidence to conclude that the mean yield of the genetically modified variety is greater than that for the standard variety.
|
Pooled-Variance t Test for the Difference Between Two Means |
|
|
(assumes equal population variances) |
|
|
Data |
|
|
Hypothesized Difference |
0 |
|
Level of Significance |
0.01 |
|
Population 1 Sample |
|
|
Sample Size |
10 |
|
Sample Mean |
24.1 |
|
Sample Standard Deviation |
2.183269719 |
|
Population 2 Sample |
|
|
Sample Size |
10 |
|
Sample Mean |
20.8 |
|
Sample Standard Deviation |
2.1499354 |
|
Intermediate Calculations |
|
|
Population 1 Sample Degrees of Freedom |
9 |
|
Population 2 Sample Degrees of Freedom |
9 |
|
Total Degrees of Freedom |
18 |
|
Pooled Variance |
4.6944 |
|
Standard Error |
0.9690 |
|
Difference in Sample Means |
3.3000 |
|
t Test Statistic |
3.4057 |
|
Upper-Tail Test |
|
|
Upper Critical Value |
2.5524 |
|
p-Value |
0.0016 |
|
Reject the null hypothesis |
|
|
Confidence Interval Estimate |
|
|
for the Difference Between Two Means |
|
|
Data |
|
|
Confidence Level |
99% |
|
Intermediate Calculations |
|
|
Degrees of Freedom |
18 |
|
t Value |
2.8784 |
|
Interval Half Width |
2.7891 |
|
Confidence Interval |
|
|
Interval Lower Limit |
0.5109 |
|
Interval Upper Limit |
6.0891 |
Ag company hayang more t congtoa om og that has developed a genetically made tomatplant h...