Hypothesis Testing

Have a go first. Then open the working one line at a time.

1. The totals

   Coffee row total = 115, 30 to 50 column total = 107, grand total = 290

   Take the totals from the edges of the table, not the cell itself.

2. The rule

   E = \(\dfrac{\text{row total}\times\text{column total}}{\text{grand total}}\) = \(\dfrac{115\times107}{290}\)

   If drink and age are independent, the share of coffee drinkers is the same in every age group.

3. Work it out

   E = 42.4

   An expected frequency need not be a whole number. Keep 3 significant figures.

Have a go first. Then open the working one line at a time.

1. H0

   H0: the club chosen is independent of year group.

   H0 always says the two variables are independent. Name both variables.

2. H1

   H1: the club chosen is not independent of year group.

   H1 says they are not independent. Do not write 'dependent on'.

3. Rows and columns

   4 rows (clubs), 3 columns (year groups)

   Count the rows and columns of data. Do not count a Total row or column.

4. Degrees of freedom

   df = (4 − 1)(3 − 1) = 6

   df = (rows − 1)(columns − 1).

Have a go first. Then open the working one line at a time.

1. Hypotheses

   H0: lunch chosen is independent of customer type. H1: lunch chosen is not independent of customer type.

   Both hypotheses name the two variables.

2. Degrees of freedom

   df = (3 − 1)(2 − 1) = 2

   3 rows and 2 columns.

3. Enter the observed table

   Enter the table as a 3 × 2 matrix. Expected frequencies:

   The calculator makes the expected matrix itself. Check that every expected frequency is at least 5.

4. Chi-squared

   From the calculator: χ² = 9.19, p = 0.0101, df = 2

   The calculator gives χ², the p-value and df.

5. Compare

   9.19 > 5.99, so reject H0.

   A large χ² means the observed values are far from the expected ones.

6. In context

   There is sufficient evidence, at the 5% significance level, that the lunch chosen is not independent of whether the customer is a student or staff.

   Say what it means for the lunches, not just 'reject H0'.

Have a go first. Then open the working one line at a time.

1. Expected table

   From the calculator, expected frequencies:

   Enter the observed matrix and read the expected matrix.

2. Each cell's share

   (O − E)²/E for every cell:

   These add up to χ². The cell with the biggest value is the one that departs most.

3. The biggest

   Flat and Dog: (O − E)²/E = 10.1

   Total χ² = 36.6.

4. Above or below

   O = 14, E = 32.0, so O is below E.

   In context: fewer flat households than expected have a dog.

Have a go first. Then open the working one line at a time.

1. Hypotheses

   H0: punctuality is independent of the route. H1: punctuality is not independent of the route.

   State them before the test, in context.

2. From the calculator

   χ² = 2.72, p = 0.606, df = 4

   df = (3 − 1)(3 − 1) = 4.

3. Compare p with α

   0.606 > 0.05, so do not reject H0.

   A p-value above the significance level means the data are not unusual if H0 is true.

4. In context

   There is insufficient evidence, at the 5% significance level, that punctuality is not independent of the route.

   Say 'insufficient evidence'. Do not say that the test proves they are independent.

Have a go first. Then open the working one line at a time.

1. Only the right tail

   Shaded area to the right of the critical value = 0.05

   Chi-squared tests only use the right-hand tail. The critical value cuts off the area α.

2. Critical value

   From the calculator: critical value = 9.49

   Use the inverse chi-squared with area 0.95 to the left (0.05 to the right) and df = 4.

3. p-value

   From the calculator: P(χ² > 11.2) = 0.0244

   The p-value is the area to the right of the test statistic.

4. Check

   11.2 > 9.49 and 0.0244 < 0.05

   The two ways of testing always agree.