STATS FINAL PT. I

  1. To determine a person's percentile, you first need to convert the person's raw score to a:




    C) z score.
  2. What proportion of a normal distribution corresponds to z scores greater
    than 1.04?




    B) 0.1492
  3. If we know that the percentage of scores falling between the mean and a z score of 0.40 is 15.54, then what is the percentage of scores falling between the mean and a z score of –0.40?




    D) 15.54
  4. If we know that the percentage of scores falling between the mean and a z score of 0.50 is 19.15, then what is the percentage of scores falling below a z score of –0.50?




    B) 30.85
  5. What is the percentage of observations that fall between z scores of –1.2 and 0.50?




    A) 57.64
  6. In one statistics course, students reported studying an average of 9.92 hours a week, with a standard deviation of 4.54. Treating this class as the population, what is the percentile for a student in the class who studies 8 hours a week?




    B) 33.72
  7. What is the difference between the denominator of the equation for the z score and that of the z statistic?




    C) When computing a z score, we divide by the population standard deviation, but when computing a z statistic, we divide by the standard error of the sampling distribution.
  8. Under what conditions is it permissible to proceed with a hypothesis test even though the assumption that participants are randomly selected is violated?




    C) We are cautious about generalizing the results.
  9. ________ requires that all members of a population have an equal chance of being selected for a study.




    B) Random selection
  10. Because of the principle of ________, when sample sizes are at least 30, the distribution will most likely resemble a normal distribution.




    B) central limit theorem
  11. The null hypothesis states that:




    A) there are no differences between the populations being studied.
  12. A New York Times article published on April 24, 2007, reported the research of Dr. Giorgio Vallortigara, a neuroscientist at the University of Trieste, Italy, and his two colleagues. The researchers asked whether a dog wags its tail in a preferred direction in response to positive stimuli and in another direction in response to negative stimuli. To answer their question, they recruited 30 dogs that were family pets. Filming each dog from above, they allowed it to view (through a slat in its cage) three positive stimuli separately, in order of descending positivity: its owner, an unfamiliar human, and a cat. All the dogs responded by wagging their tails to the right. But when the dogs were presented with an unfamiliar, aggressive dog, a negative stimulus, all dogs wagged their tails to the left.

    (Study Description: Tail Wagging) Which of the following statements is the null hypothesis for this study?




    B) A dog's tail wagging will be the same in response to positive stimuli as to negative stimuli.
  13. The typical probability adopted by researchers to determine whether a result is extreme is:




    A) .05.
  14. Mehl (2007) published a study in the journal Science
    reporting the results of an extensive study of 396 men and women comparing the
    number of words uttered per day by each sex. If Mehl was testing the idea that
    women talk more than men do, the null hypothesis for the study would be
    ________ and the research hypothesis would be ________ (in symbolic terms).
    Image Upload 2 b
  15. If a researcher is testing the idea that women talk more than men do, he/she could use a ______ test, whereas if a researcher is testing the statement “It is hypothesized that men and women will differ on the number of words uttered per day, ” he/she must use a _____ test:




    A) one-tailed; two-tailed
  16. When hypothesis testing, a conservative approach is to use a ________ rather than a ________.




    A) two-tailed test; one-tailed test
  17. The critical value(s) associated with a p level of .05 for a one-tailed hypothesis test using the z statistic is (are):




    A) –1.65 or 1.65.
  18. The range of raw scores contained in an 80% confidence interval will be ________ the range of raw scores contained in a 95% confidence interval.




    A) smaller than
  19. It is known that the population mean for the verbal section of the SAT is 500 with a standard deviation of 100. In 2006, a sample of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511. The point estimate of the mean for this group is ________ and the 80% confidence interval for this group is ________.




    C) 511; (504.6, 517.4)
  20. The statement “The findings based on a sample of 1000 participants were statistically significant, providing evidence for our hypothesis” would be strengthened by:




    A) measuring effect sizes.
  21. Measures of effect size:




    D) are unaffected by sample size.
  22. Imagine that a study of memory and aging finds that younger participants correctly recall 55% of studied words, older participants correctly recall 42% of studied words, and the size of this effect is Cohen's d = 0.49. This effect size indicates that the memory performance of:




    D) older participants is approximately half a standard deviation below that of younger participants.
  23. To remove the adjustment for the influence of sample size, Cohen's d uses the ________ rather than the ________ as part of its formula.




    A) standard deviation; standard error
  24. If an effect is significant but the effect size for the difference between the two means is small (according to Cohen's conventions), about how much overlap will there be between the two distributions?




    B) 85%
  25. QUESTION WAS DELETED
  26. An overlap between two distributions of approximately 39% is most likely to result in a(n) ________ effect size.




    D) large
  27. A prep value of 0.67 would indicate:




    C) that an effect would replicate 67% of the time.
  28. The concepts of effect size and prep are preferred over statistical significance testing alone because they:




    A) assess the relative importance of results.
  29. Our ability to reject the null hypothesis given that the null hypothesis is false is:




    B) statistical power.
  30. We calculate a statistical power and find that it is .93. This means that if the null hypothesis is ________, we have a ________% chance of rejecting the null hypothesis.




    A) false; 93
  31. When alpha increases, both ________ and ________ increase.




    A) power; probability of a Type I error
  32. If you are correct about the expected direction of an effect, then using a one-tailed hypothesis test instead of a two-tailed hypothesis test:




    C) increases power.
  33. According to the textbook, a ________ has more statistical power; however, a ________ is more conservative.




    A) one-tailed test; two-tailed test
  34. Increasing sample size:




    C) All of these answers are correct.
  35. Why do we divide by N – 1 rather than by N when estimating a population standard deviation from the sample standard deviation?




    A) Because the sample standard deviation is likely to be an imprecise estimate, we allow the error of the estimate (the standard deviation) to be larger by dividing the sum of squares by a smaller number.
  36. The symbol representing a standard deviation calculated by using a sample to estimate the population standard deviation is:




    D) s.
  37. The t statistic indicates the:




    A) distance of a sample mean from the population mean in terms of estimated standard error.
  38. Identify the correct formula for using the
    sample standard deviation to estimate the population standard deviation.
    Image Upload 6
    B
  39. A newspaper article reported that the typical American family spent an average of $81 for Halloween candy and costumes last year. A sample of N = 16 families this year reported spending a mean of M = $85, with s = $20. What statistical test would we use to determine whether these data indicate a significant change in holiday spending?




    D) single-sample t test
  40. Many companies that manufacture lightbulbs advertise their 60-watt bulbs as having an average life of 1000 hours. A cynical consumer bought 30 bulbs and burned them until they failed. He found that they burned for an average of M = 1233, with a standard deviation of s = 232.06. What statistical test would this consumer use to determine whether the average burn time of lightbulbs differs significantly from that advertised?




    A) single-sample t test
  41. The difference between the denominator of the z test and that of the single sample t test is that in a:




    B) z test we divide by the actual population standard error (Image Upload 16M), but in a t test we divide by the estimated standard error (sM).
Author
yogalindo
ID
301189
Card Set
STATS FINAL PT. I
Description
STATS FINAL
Updated