Review 1

Measures of central tendency and measures of dispersion

Mean, median, mode

Median: The median is the middle value of the ordered sample values. It is the number which is half-way into the list of reported values

Mode: The mode is the value with highest frequency in the sample.

Mean:

$$\overline{x} = \frac{1}{n}\sum x_i$$

Easier way to calculate the mean

Sample: 0, 0, 1, 3, 3, 4, 4, 4, 5, 8.

x	f	fx
0	2	0
1	1	1
3	2	6
4	3	12
5	1	5
8	1	8
$\sum x = 21$	$\sum f=10$	$\sum fx=32$

$$\overline{x} = \frac{\sum fx}{\sum f} = \frac{32}{10} = 3.2$$

Skweness

It is important to understand how the skewness of a distribution influences the relationship between the mean, median and mode. The figure below is from here.

Standard deviation

$$sd = \sqrt{\frac{f(x - \overline{x})^2}{n-1} }$$

$x$	f	fx	$x - \overline{x}$	$(x - \overline{x})^2$	$f(x - \overline{x})^2$
…	…	…	…	…	…
.	$\sum f$	$\sum fx$			$\sum above = A$

Calculate the standard deviation: $$sd = \sqrt{\frac{A}{n-1} }$$

Variance

$$variance = sd^2$$

Z-scores

$$z=\frac{x_i-\overline{x}}{sd}$$

where $s$ is the sample standard deviation

Recall that, for the z-distribution:

• Mean = 0
• SD = 1

Confidence intervals

Standard error: $$se = \frac{sd}{\sqrt{n}}$$

CIs: $$\text{Lower bound} = \bar x - z_{(\frac{\alpha}{2})}se$$

$$\text{Upper bound} = \bar x + z_{(\frac{\alpha}{2})}se$$

• For 95% CI: $z_{(\frac{\alpha}{2})}= z_{0.025} \approx 2$
• For 99% CI: $z_{(\frac{\alpha}{2})}= z_{0.005} \approx 3$

Hypothesis testing: one-sample t-test

$$t = \frac {\bar x-\mu_0}{se}$$

• 1. We reject the null hypothesis only when the calculated $t$ is higher than the critical value ($c_\alpha$).
• 2. We reject the null hypothesis only when $p \leq \alpha$
• 3. We reject the null hypothesis if the mean difference CI does not include 0.

Practice questions

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

2. 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.

3. The p-value equals:

1. $\alpha$
2. The significance level
3. 0.05

4. What is the percentile of a z-score $z= 1.18$

1. 45%
2. 17%
3. 88%

5. Using the z-table, find the values of z:

1. That is just above 97.5% of all values in the distribution.
2. That is just above 99.5% of all values in the distribution.

6. Using the t-table, find the value of t for:

1. Two-tail, $\alpha=0.05$ (n=31)
2. One-tail, $\alpha=0.05$ (n=31)
3. Two-tail, $\alpha=0.01$ (n=31)
4. One-tail, $\alpha=0.01$ (n=31)

7. Compared to a 95% confidence interval, a 99% confidence interval:

1. Is more precise
2. Is more accurate

8. 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

9. As the absolute value of $t_{calculated}$ increases, we are:

1. More likely to reject the null
2. Less likely to reject the null

10. 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. Yes
2. No

11. As sample size increases a 95% CI becomes:

1. Narrower, more precise
2. Less accurate