Statistics Practice Exam — Free

The population standard deviation is unknown and the sample is small ($n = 16$), so the appropriate procedure is a one-sample $t$-test with $n - 1 = 15$ degrees of freedom; the normality assumption is what makes the $t$ distribution valid at this sample size.

Compute the standard error of the mean, which measures how much sample means vary from sample to sample:

The test statistic measures how many standard errors the sample mean lies from the hypothesized mean:

For a two-sided test at $\alpha = 0.05$ with 15 degrees of freedom, the rejection region is $|t| > t_{0.025,\,15} = 2.131$. Here $|t| = 2.00 < 2.131$, so the observed mean is not far enough from 250 mL to be statistically significant: we fail to reject $H_0$. (Equivalently, the two-sided $p$-value is about 0.064, which exceeds 0.05.) There is insufficient evidence at the 5% level that the machine's mean fill volume has drifted from 250 mL.

The tempting wrong answer is B, which compares $t$ with the standard normal critical value 1.96; that is only appropriate when $\sigma$ is known or $n$ is large, and with 15 degrees of freedom the $t$ critical value is larger, which flips the decision. Option C uses $s$ itself as the standard error (forgetting to divide by $\sqrt{n}$), and option D uses $s/n = 0.5$ instead of $s/\sqrt{n}$.

$t = -2.00$; since $|t| = 2.00 < 2.131$, we fail to reject $H_0$ (option A).

The sample proportion is the point estimate:

Check the conditions for the normal approximation: $n\hat{p} = 180 \ge 10$ and $n(1 - \hat{p}) = 220 \ge 10$, and the data are a random sample from a large population, so the sampling distribution of $\hat{p}$ is approximately normal.

The standard error of a sample proportion is

A 95% confidence interval uses $z_{0.025} = 1.96$, so the margin of error is

The interval is $\hat{p} \pm E$:

Interpretation: we are 95% confident that between about 40.1% and 49.9% of all students at the university commute by public transit. (The confidence is in the method: in repeated sampling, about 95% of intervals built this way would contain the true proportion.)

The 95% confidence interval is approximately $(0.401,\ 0.499)$, or $0.45\pm 0.049$.

Start with the summary sums ($n = 5$): $\sum x = 15$, $\sum y = 150$, $\sum x^2 = 55$, $\sum y^2 = 5500$, $\sum xy = 10 + 60 + 60 + 200 + 200 = 530$. The means are $\bar{x} = 3$ and $\bar{y} = 30$.

Convert to the centred sums of squares, which measure spread about the means:

(a) The least-squares slope and intercept are

so the regression line is $\hat{y} = 6 + 8x$. (The line passes through $(\bar{x}, \bar{y}) = (3, 30)$, as every least-squares line must.)

(b) The correlation coefficient is

a fairly strong positive linear association, and $r^2 = 0.64$. Interpretation: about 64% of the variation in test-score improvement among these students is explained by the linear relationship with the number of tutoring sessions; the remaining 36% is due to other factors.

(c) At $x = 3.5$: $\hat{y} = 6 + 8(3.5) = 34$ points. This is interpolation within the observed range of 1 to 5 sessions, so the prediction is reasonable. The slope means that each additional weekly tutoring session is associated with an average improvement of about 8 more points; it describes association in these data, not a guaranteed causal effect.

$\hat{y} = 6 + 8x$; $r = 0.8$ and $r^2 = 0.64$; the predicted improvement for 3.5 sessions is 34 points.