2. The sample standard deviation, s, provides an estimate of the unknown population standard deviation, sigma. That is, s estimates the standard distance between the mean an an individual score or X-value. The estimated standard error, s, estimates the standard distance between the population mean and the sample mean. The estimated standard error depends, in part, on the sample size; the larger the sample, the smaller the error.
4. Because the t statistic does not require tht you know the population variance, it can be used in situations where sigma may be changed by the treatment and in situations where you do not know either mu or sigma-squared.
6. The sample variance (s-squared) in the t formula is an estimated value and contributes to the variability.
8. a. Mean = 10, s-squared = 36, and s = 6.
b. Estimated standard error of the mean = 3 points.
10. a. Ho: mu less than or equal to 20 (not better)
Ha: mu less than 20 (better - fewer trials)
The null hypothesis states that watching other animals will not result in fewer trials needed to solve the problem. He states that performance will improve with watching.
b. In the distribution of t scores with df = 3, the critical region consists of values less than - 4.541.
c. t(3) = -5/5 = -1.00
d. Fail to reject Ho. These data do not provide sufficient evidence to conclude that animals perform significantly better after viewing others.
12. Ho: mu = 7.9 (no change from 10 years ago). Because df = 99 is not in the table, use df = 60 to obtain critical boundaries of plus/minus 2.660. For these data, s = .10 and t(99) - 6.00. Reject Ho.
14. a. Your sketch should show a normal distribution centered at X = 55 with a standard deviation of s = 2. The hypothesized mean, mu = 50, is located far in the tail of the distribution. It does not appear that the sample distribution is centered around mu = 50.
b. Ho: mu = 50. The critical region consists of t values beyond plus/minus 2.131. For these data the standard error is .50 and t(15) = 10.00. Reject Ho and conclude that the population mean is significantly different from mu = 50. This conclusion is consistent with the sketch in part a.
c. Your sketch should show a normal distribution centered at X = 55 with a standard deviation of s = 20. The hypothesized mean, mu = 50, is located near the center of the distribution. Because the sample is centered around mu = 50, it appears that mu = 50 is a reasonable value for the population mean.
d. Ho: mu = 50. With SS = 6000 the standard error is 5.00 and t(15) = 1.00. Fail to reject Ho and conclude that the population mean is not significantly different from mu = 50. This conclusion is consistent with the sketch in part a.
16. Ho: mu = 21. Because df = 99 is not in the table, use df = 60 to obtain critical boundaries of plus/minus 1.98. With a standard error of .50, t(99) = -4.6. Reject Ho and conclude that humidity has a signficant effect on the rats eating behavior.
18. Ho: mu less than/equal 60 (not more pleasant), Ha: mu greater than 60 (more pleasant). The critical region consists of t values greater than 1.711. For these data, s-aquared = 25, the standard error is 1.00, and t(24) = 4.30. Reject Ho and conclude that memories for older adults are significantly more pleasant than memories for college students.
20. Ho: mu less than/equal to 25 (not more than $25). The critical region consists of t values greater than 1.833. For these data, mean = $28.00, s-squared = 201.11, the standard error is 4.49, and t(9) = .67. Fail to reject Ho. The average donation is not significantly greater than $25.00
22. Ho: mu = 62. The critical region consists of t values beyond plus/minus 2.771. For these data, mean = 48.86, s-squared = 281.90, the standard error is 3.17, and the t statistic is t(27) = -4.15. Reject ho.
24. a. Ho: mu = 17.5. The critical region consists of values beyond t = plus/minus 2.145. For these data, mean = 19.73, s-squared = 10.78, the standard error is .85, and t(14) = 2.62. Reject Ho and conclude that presentation rate does affect recall.
b. With alpha = .01, the critical boundaries are plus/minus 2.977. The obtained statistic, t = 2.62, is not in the critical region. Fail to rejct Ho.
26. Ho: mu = 10. The critical region consists of values beyond t = plus/minus 2.093. For these data mean = 12.4, s-squared = 11.70, the standard error is .765, and t(19) = 3.14. Reject Ho and conclude that the time estimate is significantly longer than the actual time.