Foundations Exam Workshop
The formulas in Foundations for Experimental Research answer different questions. Before calculating anything, identify the observational unit, the design, the parameter, and the source of uncertainty. The problems below require that choice before arithmetic.
Worked problem 1: three-arm experiment and ANOVA
An attention study randomly assigns 36 students equally to a control condition, a silent-study condition, or a guided-focus condition. Their mean test scores are 74, 80, and 86. The within-group sum of squares is 1,188.
Tasks. Test the null hypothesis that all three population means are equal. Construct the ANOVA table and explain what a rejection establishes.
Because the groups have equal size, the grand mean is
\[\bar Y=\frac{74+80+86}{3}=80.\]The between-group variation is
\[SSB=12(74-80)^2+12(80-80)^2+12(86-80)^2=864.\]There are $k-1=2$ between-group degrees of freedom and $N-k=33$ within-group degrees of freedom. Therefore,
\[MSB=\frac{864}{2}=432, \qquad MSE=\frac{1188}{33}=36,\]and
\[F=\frac{432}{36}=12.\]| Source | Sum of squares | df | Mean square | $F$ |
|---|---|---|---|---|
| Between groups | 864 | 2 | 432 | 12.00 |
| Within groups | 1,188 | 33 | 36 | |
| Total | 2,052 | 35 |
An $F_{2,33}$ value of 12 is strong evidence against the joint null. It establishes that the three means are not all equal. It does not identify which pairs differ; that requires planned contrasts or multiplicity-adjusted comparisons. Because assignment was random, the comparison can be interpreted causally for these interventions and this study population.
Worked problem 2: paired measurements
A crossover interface study records completion time under designs A and B for the same ten participants. Define $D_i=A_i-B_i$. The sample gives $\bar D=-4.8$ seconds and $s_D=6.0$ seconds.
Tasks. Test $H_0:\mu_D=0$ against a two-sided alternative and construct a 95% confidence interval.
Pairing makes the participant the block. The analysis uses the ten within-person differences, not two independent samples:
\[SE(\bar D)=\frac{6.0}{\sqrt{10}}=1.897, \qquad t=\frac{-4.8}{1.897}=-2.53.\]With 9 degrees of freedom, the two-sided $p$-value is about $0.032$. Using $t^*_{0.025,9}=2.262$,
\[-4.8\pm2.262(1.897)=(-9.09,-0.51).\]Design A is estimated to reduce mean completion time by between 0.51 and 9.09 seconds relative to B. Treating the observations as independent would discard the covariance created by observing the same people twice and would use the wrong standard error.
Worked problem 3: proportion test versus interval
In a randomized reminder experiment, 126 of 200 treated participants and 102 of 200 controls complete a form.
For the null test $H_0:p_1=p_0$, use the pooled proportion:
\[\hat p=\frac{126+102}{400}=0.57,\] \[SE_0=\sqrt{0.57(0.43)\left(\frac1{200}+\frac1{200}\right)}=0.0495,\] \[z=\frac{0.63-0.51}{0.0495}=2.42.\]The two-sided $p$-value is about $0.015$. For a confidence interval, use the two separate estimated variances rather than the null-pooled variance:
\[SE_{CI}=\sqrt{\frac{0.63(0.37)}{200}+\frac{0.51(0.49)}{200}}=0.0492,\] \[0.12\pm1.96(0.0492)=(0.024,0.216).\]The estimated increase is 12 percentage points. The distinction between the test standard error and interval standard error is a frequent exam target.
Exercises
- Four randomized teaching methods have $n_i=15$, group means $(68,72,75,77)$, and $SSE=2{,}184$. Construct the ANOVA table, test the joint null at 5%, and state what follow-up analysis is justified.
- A paired experiment reports $n=16$, $\bar D=3.2$, and $s_D=5.6$. Test $H_0:\mu_D=0$, construct a 99% interval, and explain how reversing the definition of $D_i$ changes the result.
- In two independent groups, $n_1=24$, $\bar X_1=18.4$, $s_1=4.8$, $n_2=31$, $\bar X_2=15.1$, and $s_2=7.2$. Compute the Welch statistic. Explain why a pooled test needs an additional assumption.
- A sample proportion is $0.08$ with $n=40$. Evaluate whether the ordinary Wald interval is trustworthy and name a better interval procedure.
- A study doubles each observation and then subtracts 7. Derive the transformed mean, variance, and standard error without recomputing deviations from the raw data.
Solution checks
- $\bar Y=73$, $SSB=690$, $MSB=230$, $MSE=39.0$, and $F\approx5.90$ with $(3,56)$ degrees of freedom.
- $t=2.286$ with 15 degrees of freedom. Use $t^*_{0.005,15}$ for the 99% interval; reversing $D_i$ changes the signs, not the evidence.
- $SE\approx1.72$ and $t\approx1.92$. Pooling additionally assumes equal population variances.
- $n\hat p=3.2$ is too small for a reliable Wald approximation; use a score or exact interval.
- If $Z=2X-7$, then $\bar Z=2\bar X-7$, $s_Z^2=4s_X^2$, and $SE(\bar Z)=2SE(\bar X)$.