• t-test:
  • One-sample t-test
    • (e.g. \(H_0\): mean=5)
  • Independent two-sample t-test
    • (e.g. \(H_0\): mean of sample 1 = mean of sample 2)
  • Paired two-sample t-test
    • (e.g. \(H_0\): mean difference between pairs = 0)

• e.g. Question: Published data suggests that the failure rate for a particular piece of equipment from a supplier is 2.1%

• A research facility want to know if this holds true in their own lab?

• Null hypothesis, \(H_0\):
  • Mean monthly failure rate = 2.1%
• Alternative hypothesis: \(H_1\):
  • Mean monthly failure rate \(\ne\) 2.1%

• Tails: two-tailed

• Either reject or do not reject the null hypothesis -

       Month Monthly.failure.rate
1    January                 2.90
2   February                 2.99
3      March                 2.48
4      April                 1.48
5        May                 2.71
6       June                 4.17
7       July                 3.74
8     August                 3.04
9  September                 1.23
10   October                 2.72
11  November                 3.23
12  December                 3.40

• Observations are independent
• Observations are normally distributed

mean = \((2.9 + \dots + 3.40) / 12\) = 2.841

Standard deviation = 0.837

Hypothesised Mean = 2.1

Test statistic: \[t_{n-1} = t_{11} = \frac{\bar{x} - \mu_0} {s.d. / \sqrt{n}} = \frac{2.84 - 2.10}{s.e.(\bar{x})} = \]3.065

Test statistic: \[t_{n-1} = t_{11} = \frac{\bar{x} - \mu_0} {s.d. / \sqrt{n}} = \frac{2.84 - 2.10}{s.e.(\bar{x})} = \]3.065

df = 11 P = 0.01

Reject \(H_0\) - Evidence that mean monthly failure rate \(\ne\) 2.1%

• The mean monthly failure rate of the equipement in the lab is 2.84
• It is not equal to the hypothesized mean proposed by the company of 2.1.
• t=3.07, df=11, p=0.01

• Two types of two-sample t-test:
  • Independent:
  • e.g.the weight of two different breeds of mice
• Paired
  • e.g. a measurement of disease at two different parts of the body in the same patient / animal
  • e.g. measurements before and after treatment for the same individual

• e.g. research question: 40 male mice (20 of breed A and 20 of breed B) were weighed at 4 weeks old

• Does the weight of 4-week old male mice depend on breed?

• Null hypothesis, \(H_0\)
  • mean weight of breed A = mean weight of breed B
• Alternative hypothesis, \(H_1\)
  • mean weight of breed B \(\ne\) mean weight of breed B
• Tails: two-tailed
• Either reject or do not reject the null hypothesis -

• Observations are independent
• Observations are normally-distributed

• Equal variance in the two comparison groups
  • Use "Welch's correction" if variances are different
  • alters the t-statistic and degrees of freedom

\(t_{df} = \frac{\bar{X_A} - \bar{X_B}}{s.e.(\bar{X_A} - \bar{X_B})}\) = -1.21

df = 29.78 (with Welch's correction)

P-value: 0.24

Do not reject \(H_0\)

(No evidence that mean weight of breed A \(\ne\) mean weight of breed B)

• The difference in mean weight between the two breeds is -1.30
  • [NB as this is negative, breed B mice tend to be bigger than breed A].
• There is no evidence of a difference in weights between breed A and breed B.
• t=-1.21, df= 29.78 (Welch’s correction), p=0.24

• e.g. Research question: 20 patients with ovarian cancer were studied using MRI imaging. Cellularity was measured for each patient at two sites of disease.
• Does the cellularity differ between two different sites of disease?
  • cellularity is amount of tumour (versus normal cells)
  • high cellularity means lots of tumour

• Null hypothesis, \(H_0\):
  • Cellularity at site A = Cellularity at site B
• Alternative hypothesis, \(H_1\)
  • Cellularity at site A \(\ne\) Cellularity at site B
• Tails: two-tailed
• Either reject or do not reject the null hypothesis

• \(H_0\); Cellularity at site A = Cellularity at site B
  • or
• \(H_0\): Cellularity at site A - Cellularity at site B = 0

• Observations are independent
• The paired differences are normally-distributed

\(t_{n-1} = t_{19} = \frac{\bar{X_{A-B}}}{s.e.(\bar{X_{A-B}})} =\) 3.66

df = 19

P-value: 0.002

Reject \(H_0\) (evidence that cellularity at Site A \(\ne\) site B)

• The difference in cellularity between the two sites is 19.14 (95% CI: 8.20, 30.08).
• There is evidence of a difference in cellularity between the two sites.
• t=3.66, df=19, p=0.002.

• What if normality is not reasonable?
  • Transform your data, e.g. log transformation
  • Non-parametric tests….
• What if you have more than two groups?
  • Approaches such as ANOVA
• What if you want to look at the relationship between two continuous variables
  • Linear regression

• One-sample t-test
  • Use when we have one group.
• Independent two-sample t-test
  • Use when we have two independent groups. A Welch correction may be needed if the two groups have different spread.
• Paired two-sample t-test
  • Use when we have two non-independent groups.
• Non-parametric tests or transformations
  • Use when we cannot assume normality.

• Turn scientific question to null and alternative hypothesis

• Think about test assumptions

• Calculate summary statistics

• Carry out t-test if appropriate