Chi-squared test for independence Problem set B Worked solutions

A school canteen records the main lunch chosen by 272 pupils from five houses. The results are shown. A chi-squared test for independence is carried out.

1. (a) State the null and alternative hypotheses.

   • H0: the lunch a pupil chooses is independent of house
   • H1: the lunch a pupil chooses is not independent of house

2. (b) Find the number of degrees of freedom.

   • 3 rows and 5 columns: df = (3 − 1)(5 − 1) = 8

The table shows the exhibition chosen by 155 visitors to a museum. If the exhibition chosen is independent of ticket type, find the expected frequency of adult visitors who chose Space.

• Row total = 72, column total = 49, grand total = 155
• E = 72 × 49 ÷ 155
• E = 22.8

Find the degrees of freedom for a chi-squared test for independence on a contingency table with: (a) 3 rows and 7 columns; (b) 6 rows and 2 columns; (c) 5 rows and 5 columns.

• (a) df = (3 − 1)(7 − 1) = 12
• (b) df = (6 − 1)(2 − 1) = 5
• (c) df = (5 − 1)(5 − 1) = 16

A shop company asks 143 customers how satisfied they were, by branch. Use your calculator to find the expected frequencies, \(\chi^2\), the p-value and the degrees of freedom for a test of independence.

• Enter the observed table as a matrix
• Expected frequencies: North: 18.9, 21.4, 19.7; South: 26.1, 29.6, 27.3
• From the calculator: χ² = 11.2, p = 0.00363, df = 2

The curve shows the \(\chi^2\) distribution with 3 degrees of freedom. The shaded right tail has area 0.10.

1. (a) Find the critical value.

   • From the calculator: critical value = 6.25

2. (b) A test gives \(\chi^2=9.1\). Find the p-value.

   • From the calculator: P(χ² > 9.1) = 0.0280
   • 9.1 > 6.25, so the test statistic is inside the shaded tail

A library asks 306 members which type of book they borrowed and which age band they are in. At the 5% significance level, test whether book type is independent of age band.

1. (a) State the hypotheses.

   • H0: book type is independent of age band
   • H1: book type is not independent of age band

2. (b) Write down the number of degrees of freedom.

   • 3 rows and 3 columns: df = (3 − 1)(3 − 1) = 4

3. (c) Use your calculator to find \(\chi^2\) and the p-value.

   • Enter the table as a matrix. From the calculator: χ² = 6.98, p = 0.137, df = 4

4. (d) State the conclusion in context.

   • 0.137 > 0.05, so do not reject H0.
   • There is insufficient evidence, at the 5% significance level, that book type is not independent of age band.

A farmer records 236 apples from two orchards by size class. Test, at the 5% significance level, whether the size of an apple is independent of the orchard.

1. (a) Find the value of \(\chi^2\).

   • Enter the table as a matrix. From the calculator: χ² = 12.2, df = 3

2. (b) The critical value is 7.81. State the conclusion in context.

   • 12.2 > 7.81, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the size of an apple is not independent of the orchard.

A café asks 323 customers which drink they usually order and how often they visit. The results are shown.

1. (a) Find the expected frequencies.

   • Enter the observed matrix. From the calculator, the expected frequencies:
   • Tea: 24.2, 25.1, 19.4, 23.4; Coffee: 29.7, 30.8, 23.8, 28.7; Juice: 31.1, 32.1, 24.8, 30.0

2. (b) Find the cell that adds most to \(\chi^2\), and give its value of \((O-E)^2/E\).

   • (O − E)²/E for each cell: Tea: 0.202, 8.90, 0.0967, 5.52; Coffee: 1.77, 3.78, 2.19, 0.474; Juice: 0.822, 0.535, 1.37, 7.55
   • The largest is 8.90, for Tea and Weekly

3. (c) Compare the observed and expected frequencies in that cell. Describe what this shows in context.

   • O = 40, E = 25.1
   • O is above E: there are more observed in the Tea, Weekly cell than independence predicts, so drink and visit frequency look linked here.

A school records whether 109 students ate breakfast and whether they arrived late.

1. (a) Find the expected frequency for students who ate breakfast and arrived late if the variables are independent.

   • Row total = 52, column total = 55, grand total = 109
   • E = 52 × 55 ÷ 109 = 26.2

2. (b) State the degrees of freedom. Use your calculator to find \(\chi^2\), then use the critical value 3.84 to state the conclusion in context, at the 5% significance level.

   • df = (2 − 1)(2 − 1) = 1
   • From the calculator: χ² = 2.08
   • 2.08 < 3.84, so do not reject H0.
   • There is insufficient evidence, at the 5% significance level, that arriving late is not independent of eating breakfast.

A test of whether the choice of pet is independent of the type of home gives a p-value of 0.018.

1. (a) State the conclusion at the 5% significance level, in context.

   • 0.018 < 0.05, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the choice of pet is not independent of the type of home.

2. (b) State the conclusion at the 1% significance level, in context.

   • 0.018 > 0.01, so do not reject H0.
   • There is insufficient evidence, at the 1% significance level, that the choice of pet is not independent of the type of home.