T-Test worksheet

1. p-value: 0.999373

1. Find each sample's mean, spread (standard deviation, dividing by n − 1) and size

   $\text{x̄₁ = 26.5, s₁ = 2.4495, n₁ = 8; x̄₂ = 22, s₂ = 2, n₂ = 8}$

2. Combine the two variances (spread squared) into one, giving the bigger sample more weight

   ${\mathrm{sₚ}}^2$$=\frac{7\times {2.44949}^2+7\times 2^2}{14}$$=5$

3. Find the standard error of the difference, the usual amount it wobbles by chance

   $\mathrm{SE}$$=\sqrt{5\times \left(\frac{1}{8}+\frac{1}{8}\right)}$$=1.118034$

4. Find the degrees of freedom: both sample sizes added, minus 2 (n₁ + n₂ − 2)

   $\mathrm{df}$$=14$

5. Find t: how many standard errors your result is from the value you are testing against

   $t$$=\frac{4.5-0}{1.118034}$$=4.024922$

6. Find the p-value, the chance of a result this far out: the area below t on the t curve with 14 df

   $p$$=\text{P(T < t) =}0.9993733461$

7. The p-value is above α, so the result is not statistically significant

   $\text{Not significant: fail to reject H₀ (p > 0.05)}$

2. p-value: 0.601515

1. Subtract each 'after' value from its 'before' value. The test then checks these differences as one sample

   $\text{d = 2, -4, -10, 0, 6, 5, 1, 4}$

2. Find the differences' mean, spread (standard deviation, dividing by n − 1) and size n

   $\text{d̄ = 0.5, s = 5.2915, n = 8}$

3. Find the standard error, the usual amount a mean wobbles by chance: the standard deviation s divided by the square root of n

   $\mathrm{SE}$$=\frac{5.291503}{\sqrt{8}}$$=1.870829$

4. Find the degrees of freedom: the number of values minus 1 (n − 1)

   $\mathrm{df}$$=7$

5. Find t: how many standard errors your result is from the value you are testing against

   $t$$=\frac{0.5-0}{1.870829}$$=0.267261$

6. Find the p-value, the chance of a result this far out: the area below t on the t curve with 7 df

   $p$$=\text{P(T < t) =}0.6015145036$

7. The p-value is above α, so the result is not statistically significant

   $\text{Not significant: fail to reject H₀ (p > 0.05)}$

3. p-value: 0.035276

1. Find each sample's mean, spread (standard deviation, dividing by n − 1) and size

   $\text{x̄₁ = 26.5, s₁ = 2.4495, n₁ = 8; x̄₂ = 22, s₂ = 6.0238, n₂ = 8}$

2. Combine the two variances (spread squared) into one, giving the bigger sample more weight

   ${\mathrm{sₚ}}^2$$=\frac{7\times {2.44949}^2+7\times {6.023762}^2}{14}$$=21.142857$

3. Find the standard error of the difference, the usual amount it wobbles by chance

   $\mathrm{SE}$$=\sqrt{21.142857\times \left(\frac{1}{8}+\frac{1}{8}\right)}$$=2.299068$

4. Find the degrees of freedom: both sample sizes added, minus 2 (n₁ + n₂ − 2)

   $\mathrm{df}$$=14$

5. Find t: how many standard errors your result is from the value you are testing against

   $t$$=\frac{4.5-0}{2.299068}$$=1.957315$

6. Find the p-value, the chance of a result this far out: the area above t on the t curve with 14 df

   $p$$=\text{P(T > t) =}0.0352762603$

7. The p-value is at or below α, so the result is statistically significant

   $\text{Significant: reject H₀ (p ≤ 0.07)}$

4. p-value: 0.967512

1. Find the sample's mean, spread (standard deviation, dividing by n − 1) and size n

   $\text{x̄ = 26.25, s = 5.4968, n = 8}$

2. Find the standard error, the usual amount a mean wobbles by chance: the standard deviation s divided by the square root of n

   $\mathrm{SE}$$=\frac{5.496752}{\sqrt{8}}$$=1.943395$

3. Find the degrees of freedom: the number of values minus 1 (n − 1)

   $\mathrm{df}$$=7$

4. Find t: how many standard errors your result is from the value you are testing against

   $t$$=\frac{26.25-22}{1.943395}$$=2.186894$

5. Find the p-value, the chance of a result this far out: the area below t on the t curve with 7 df

   $p$$=\text{P(T < t) =}0.967511994$

6. The p-value is above α, so the result is not statistically significant

   $\text{Not significant: fail to reject H₀ (p > 0.04)}$

5. p-value: 1

1. Find the differences' mean, spread (standard deviation, dividing by n − 1) and size n

   $\text{d̄ = 22.5, s = 2.7, n = 7}$

2. Find the standard error, the usual amount a mean wobbles by chance: the standard deviation s divided by the square root of n

   $\mathrm{SE}$$=\frac{2.7}{\sqrt{7}}$$=1.020504$

3. Find the degrees of freedom: the number of values minus 1 (n − 1)

   $\mathrm{df}$$=6$

4. Find t: how many standard errors your result is from the value you are testing against

   $t$$=\frac{22.5-0}{1.020504}$$=22.047928$

5. Find the p-value, the chance of a result this far out: the area below t on the t curve with 6 df

   $p$$=\text{P(T < t) =}0.9999997155$

6. The p-value is above α, so the result is not statistically significant

   $\text{Not significant: fail to reject H₀ (p > 0.05)}$

6. p-value: 0.713564

1. Subtract each 'after' value from its 'before' value. The test then checks these differences as one sample

   $\text{d = -4, -2, 4, 12, 2, -4, 2, -4}$

2. Find the differences' mean, spread (standard deviation, dividing by n − 1) and size n

   $\text{d̄ = 0.75, s = 5.5485, n = 8}$

3. Find the standard error, the usual amount a mean wobbles by chance: the standard deviation s divided by the square root of n

   $\mathrm{SE}$$=\frac{5.548488}{\sqrt{8}}$$=1.961687$

4. Find the degrees of freedom: the number of values minus 1 (n − 1)

   $\mathrm{df}$$=7$

5. Find t: how many standard errors your result is from the value you are testing against

   $t$$=\frac{0.75-0}{1.961687}$$=0.382324$

6. Find the p-value, the chance of a result this far out in either direction: double the area beyond |t| on the t curve with 7 df

   $p$$=2\times 0.3567820004$$=0.7135640009$

7. The p-value is above α, so the result is not statistically significant

   $\text{Not significant: fail to reject H₀ (p > 0.04)}$

7. p-value: 0.012777

1. Subtract each 'after' value from its 'before' value. The test then checks these differences as one sample

   $\text{d = 3, 1, 4, 10, -2, 11, 2, 7}$

2. Find the differences' mean, spread (standard deviation, dividing by n − 1) and size n

   $\text{d̄ = 4.5, s = 4.504, n = 8}$

3. Find the standard error, the usual amount a mean wobbles by chance: the standard deviation s divided by the square root of n

   $\mathrm{SE}$$=\frac{4.503967}{\sqrt{8}}$$=1.592393$

4. Find the degrees of freedom: the number of values minus 1 (n − 1)

   $\mathrm{df}$$=7$

5. Find t: how many standard errors your result is from the value you are testing against

   $t$$=\frac{4.5-0}{1.592393}$$=2.825936$

6. Find the p-value, the chance of a result this far out: the area above t on the t curve with 7 df

   $p$$=\text{P(T > t) =}0.012777348$

7. The p-value is at or below α, so the result is statistically significant

   $\text{Significant: reject H₀ (p ≤ 0.06)}$

8. p-value: 1.552213e-10

1. Find the differences' mean, spread (standard deviation, dividing by n − 1) and size n

   $\text{d̄ = 25, s = 2.5, n = 10}$

2. Find the standard error, the usual amount a mean wobbles by chance: the standard deviation s divided by the square root of n

   $\mathrm{SE}$$=\frac{2.5}{\sqrt{10}}$$=0.790569$

3. Find the degrees of freedom: the number of values minus 1 (n − 1)

   $\mathrm{df}$$=9$

4. Find t: how many standard errors your result is from the value you are testing against

   $t$$=\frac{25-0}{0.790569}$$=31.622777$

5. Find the p-value, the chance of a result this far out in either direction: double the area beyond |t| on the t curve with 9 df

   $p$$=2\times 7.7610669491e-11$$=0.0000000002$

6. The p-value is at or below α, so the result is statistically significant

   $\text{Significant: reject H₀ (p ≤ 0.05)}$

9. p-value: 0.005991

1. Find each sample's mean, spread (standard deviation, dividing by n − 1) and size

   $\text{x̄₁ = 18, s₁ = 2.4, n₁ = 10; x̄₂ = 22, s₂ = 3, n₂ = 9}$

2. Find the standard error of the difference, using each sample's own spread

   $\mathrm{SE}$$=\sqrt{\frac{{2.4}^2}{10}+\frac{3^2}{9}}$$=1.255388$

3. Find the degrees of freedom with the Welch–Satterthwaite formula. It is usually not a whole number

   $\mathrm{df}$$=\frac{{\left(0.576+1\right)}^2}{\frac{{0.576}^2}{9}+\frac{1^2}{8}}$$=15.344833$

4. Find t: how many standard errors your result is from the value you are testing against

   $t$$=\frac{\mathrm{-4}-0}{1.255388}$$=\mathrm{-3.186265}$

5. Find the p-value, the chance of a result this far out in either direction: double the area beyond |t| on the t curve with 15.3448 df

   $p$$=2\times 0.0029955943$$=0.0059911887$

6. The p-value is at or below α, so the result is statistically significant

   $\text{Significant: reject H₀ (p ≤ 0.05)}$

10. p-value: 0.729245

1. Find the sample's mean, spread (standard deviation, dividing by n − 1) and size n

   $\text{x̄ = 24, s = 4.4078, n = 8}$

2. Find the standard error, the usual amount a mean wobbles by chance: the standard deviation s divided by the square root of n

   $\mathrm{SE}$$=\frac{4.407785}{\sqrt{8}}$$=1.558387$

3. Find the degrees of freedom: the number of values minus 1 (n − 1)

   $\mathrm{df}$$=7$

4. Find t: how many standard errors your result is from the value you are testing against

   $t$$=\frac{24-23}{1.558387}$$=0.641689$

5. Find the p-value, the chance of a result this far out: the area below t on the t curve with 7 df

   $p$$=\text{P(T < t) =}0.7292453623$

6. The p-value is above α, so the result is not statistically significant

   $\text{Not significant: fail to reject H₀ (p > 0.04)}$