Psychology 2211 Assignment 1

1. (Previous exam question) Over the past 50 years, an average of 370 minor earth tremors a day has been recorded in Northern California. Recently, over a period of nine days, the following number of minor tremors were recorded:

386, 394, 362, 423, 456, 370, 385, 430, 400

a) Is there evidence for a recent increase in the number of tremors ( .05)?

b) Based on the 9-day data, form a 95% confidence interval to estimate the mean number of minor tremors that might be expected each day over the next several weeks.

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 ( .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 With Device

1 28 32

2 20 21

3 44 36

4 7 10

5 20 20

6 11 14

7 11 9

8 34 32

9 9 14

10 48 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 ( .05)? Is their variance significantly greater than 175 when they were driven without the device? ( .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 ².

Answers:

1.a) t = 3.033, reject Ho: there is evidence for a significant increase in the number of errors

1b) 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 H₀; no evidence that the variance is significantly less than 175. chi-square = 11.250, Do not reject H₀; no evidence that the variance is significantly greater than 175.

4. 3.973 to 13.5219