ONLY ANSWER QUESTION NUMBER 3 Answer for question 3:3. chi-square = 7.997, Do not reject Ho; no evidence that the variance is significantly less than 175. chi-square11.250, Do not reject Ho; no evidence that the variance is significantly greater than 175. 2. In the past, students in a particular course have achieved a mean Xmas exam score of 72.6, with a standard deviationof 12.5. This year, 15 students in one section achieved a mean Xmas exam score of 72.2, while 10 students inanother section achieved a mean Xmas exam score of 77.3. Is the overall (grand) mean of the 25 students in thisyear's two sections significantly different from the mean obtained in previous years (a ≤.05)?3. A new pollution control device for automobiles is claimed not only to reduce carbon monoxide emission but also toincrease a car's miles per gallon. To test the latter claim, the mileage per gallon of 10 randomly selected cars ismeasured first without and then with the device, yielding the following data:Miles per gallonCar Without Device12820447202345678910111134948With Device3221361020932141446Typically, the variance of the mileage per gallon of cars not fitted with the device is 175. Is the variance of the 10cars in this study significantly less than this when they were driven with the device (a ≤.05)? Is their variancesignificantly greater than 175 when they were driven without the device? (a <.05)?4. The lower value of a 90% confidence interval for the variance in maze-running times of a sample of 31 rats is 4.62.Based on this, form a 98% confidence interval for o².Answers:1.a) t = 3.033, reject Ho: there is evidence for a significant increase in the number of errorslb) 377.35 to 423.992. t = .656, do not reject Ho: there is no evidence that this year's students are performing differently than students inprevious years.3. chi-square= 7.997, Do not reject Ho; no evidence that the variance is significantly less than 175. chi-square=11.250, Do not reject Ho; no evidence that the variance is significantly greater than 175.4.3.973 to 13.5219

We have to test hypothesis for the variance of milenge per gallon of cars not fitted with device 175…

ONLY ANSWER QUESTION NUMBER 3 Answer for question 3: 3. chi-square = 7.997, Do not reject Ho; no evidence that the variance is significantly less than 175. chi-square 11.250, Do not reject Ho; no evidence that the variance is significantly greater than 175.

ONLY ANSWER QUESTION NUMBER 3 Answer for question 3: 3. chi-square = 7.997, Do not reject Ho; no evidence that the variance is significantly less than 175. chi-square 11.250, Do not reject Ho; no evidence that the variance is significantly greater than 175.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

100%

2. In the past, students in a particular course have achieved a mean Xmas exam score of 72.6, with a standard deviation
of 12.5. This year, 15 students in one section achieved a mean Xmas exam score of 72.2, while 10 students in
another section achieved a mean Xmas exam score of 77.3. Is the overall (grand) mean of the 25 students in this
year's two sections significantly different from the mean obtained in previous years (a ≤.05)?
3. A new pollution control device for automobiles is claimed not only to reduce carbon monoxide emission but also to
increase a car's miles per gallon. To test the latter claim, the mileage per gallon of 10 randomly selected cars is
measured first without and then with the device, yielding the following data:
Miles per gallon
Car Without Device
1
28
20
44
7
20
2
3
4
5
6
7
8
9
10
11
11
34
9
48
With Device
32
21
36
10
20
9
32
14
14
46
Typically, the variance of the mileage per gallon of cars not fitted with the device is 175. Is the variance of the 10
cars in this study significantly less than this when they were driven with the device (a ≤.05)? Is their variance
significantly greater than 175 when they were driven without the device? (a <.05)?
4. The lower value of a 90% confidence interval for the variance in maze-running times of a sample of 31 rats is 4.62.
Based on this, form a 98% confidence interval for o².
Answers:
1.a) t = 3.033, reject Ho: there is evidence for a significant increase in the number of errors
lb) 377.35 to 423.99
2. t = .656, do not reject Ho: there is no evidence that this year's students are performing differently than students in
previous years.
3. chi-square= 7.997, Do not reject Ho; no evidence that the variance is significantly less than 175. chi-square=
11.250, Do not reject Ho; no evidence that the variance is significantly greater than 175.
4.3.973 to 13.5219

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