# 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