Solutions to review questions

Review 1

Please see solutions in bold.

  1. Suppose \(\overline{x} = 10\) for a negatively skewed distribution. What could be the median and mode of this distribution?
  1. median = 11 and mode = 12
  2. median = 12 and mode = 11
  1. Suppose you created a z-score distribution from a given sample. What is the mean of this distribution?
  1. 1
  2. 0
  3. There is not enough information to calculate the mean.
  1. The p-value equals:
  1. \(\alpha\)
  2. The significance level
  3. 0.05
  1. What is the percentile of a z-score \(z= 1.18\)
  1. 45%
  2. 17%
  3. 88%
  1. Using the z-table, find the values of z:
  1. That is just above 97.5% of all values in the distribution. z = 1.96
  2. That is just above 99.5% of all values in the distribution. z = 2.575
  1. Using the t-table, find the value of t for:
  1. Two-tail, \(\alpha=0.05\) (n=31), t = 2.042
  2. One-tail, \(\alpha=0.05\) (n=31), t = 1.697
  3. Two-tail, \(\alpha=0.01\) (n=31), t = 2.750
  4. One-tail, \(\alpha=0.01\) (n=31), t = 2.457
  1. Compared to a 95% confidence interval, a 99% confidence interval:
  1. Is more precise
  2. Is more accurate
  1. If \(p = 0.03\)
  1. We reject the null when \(\alpha = 0.05\)
  2. We reject the null when \(\alpha = 0.01\)
  3. We never reject the null
  1. As the absolute value of \(t_{calculated}\) increases, we are:
  1. More likely to reject the null
  2. Less likely to reject the null
  1. Given a 95% mean difference CI = (-0.12;1.2), do we reject the null that the population mean = 15 (assuming \(\alpha = 0.05)\)?
  1. No (Note: because 0 falls within our CI)
  2. Yes
  1. As sample size increases a 95% CI becomes:
  1. Narrower, more precise
  2. Less accurate

Extra questions

Consider the following sample: 1, 2, 7, 3.5, 0.5, 2, 2, 8, 3, 7.

  1. Calculate the mean, median and mode for the sample.
  • Mean = 3.6
  • Median = 2.5
  • Mode = 2.
  1. Calculate standard deviation, variance and standard error.
  • SD = 2.72
  • Variance = 7.43
  • SE = 0.86.
  1. Calculate the z-scores for the values 1 and 2 in the sample (in other words, for x=1 and x=2).
  • Z (for x = 1) = -0.95
  • z (for x = 2) = -0.587.
  1. Calculate the 95% and 99% CIs for this sample.

Using z = 2 for the 95% CI and z = 3 for the 99% CI:

  • The 95% CI = (1.88; 5.32)
  • The 99% CI = (1.02; 6.18)
  1. Test the hypothesis that the mean of the population from which this sample was drawn is 5. Interpret your results.
  • \(t_{critical} = 2.262\)
  • \(t_{calculated}= 1.64\) (absolute value).
  • Because the \(t_{calculated}\) is lower than the \(t_{critical}\), we fail to reject the null.

Review 2

Open the “school_data.sav”.

Choose the appropriate test for the following questions:

Hint: Look carefully at the variable labels in your data. Note if the variable is continuous or categorical.

  1. Did schools improve their average scores between 1999 and 2000?
  • T-test (paired sample)
  1. Does average class size influences average achievement in the year 2000?
  • Simple linear regression
  1. Is tshere a relationship between percentage of parents who have completed HS and achievement in a given school?
  • Pearson’s correlation
  1. Is there a relationship between percentage of free meals in a given school and whether the school has summer classes?
  • Chi-square test for independence
  1. Do California schools with summer classes (year round schools) have higher average achievement?
  • T-test (independent sample) or one-way ANOVA
  1. Is the mean achievement for the schools in the sample a good representative of the mean achievement for all California elementary schools?
  • T-test (one-sample)
  1. Suppose percentage of free meals are associated with achievement. Does the existence of summer courses in a school influences this relationship?
  • Two-way ANOVA
  1. Is average student achievement different between schools with different percentage of free meals?
  • One-way ANOVA