ypothesi s, test-value, critical-value, comparison logic, comparison selection a

Question

ypothesi s, test-value, critical-value, comparison logic, comparison selection and conclusion. (B) Construct and interpret the 95% confidence interval for the difference between the population means. (6) (12.5 gum for 15 minutes after each meal were discharged from the clinic after an average of 4.4 days, compared to 5.2 days for those who did not chew gum after each meal. The group data from two (2) independent samples are as follows: Group I (chewed gum): Xi average-4.4 days, SD;= 1.3 days, ni=51 Group 2 (did not chewed gum): X2 average. = 5.2 days, SD,-1.9 days, n51 (A) Using a significance level of 0.01 and the z-test, determine whether after meals gum chewing could be a better process? For this find and show: hypothesis, test-value, critical-value, comparison logic, comparison selection and conclusion (B) Identify and interpret the p-value. points): In a medical study of stomach disorders, it was found that those in the study that chewed (7) (12.5 points): A trucking firm is considering the installation of a new, low-restriction engine air filter for its long-haul trucks, but only if the new filter can be shown to improve the fuel mileage economy of these vehicles. A test is set up in which each of 5 trucks makes the same run twice - once with the old filtration system and once with the new version. Given the mpg test results shown below, use the 0.05 level of significance in determining whether the new filtration system could be superior. For this find and show: hypothesis, test-value, critical-value, comparison logic, comparison selection and conclusion. Current Filter New Filter Truck Number 7.3 mpg 7.2 6.8 10.6 8.8 7.6 mpg 10.4 6.9 5.6

Explanation / Answer

6)

Two-Sample T-Test and CI

Sample   N Mean StDev SE Mean
1       51    4.40     1.30     0.18
2     51    5.20     1.90     0.27

Difference = (1) - (2)
Estimate for difference: -0.800
99% upper bound for difference: -0.038
T-Test of difference = 0 (vs <): T-Value = -2.48 P-Value = 0.007 DF = 100
Both use Pooled StDev = 1.6279

critical value = -2.327

TS = -2.48

-2.48 < -2.327

conclusion- reject the null hypothesis

b) p-value = 0.007 < 0.01

we reject the null hypothesis

and conclude that there is significant evidence that after meal chewing gum could be better process

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.