Chi-squared test for independence Problem set A Worked solutions

A survey asks 237 students how they usually travel to school and which year group they are in. The results are shown. A chi-squared test for independence is carried out.

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

   • H0: the way a student travels to school is independent of year group
   • H1: the way a student travels to school is not independent of year group

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

   • 4 rows and 3 columns: df = (4 − 1)(3 − 1) = 6

The table shows the sport chosen by 154 pupils. If the sport chosen is independent of gender, find the expected frequency of girls who chose tennis.

• Row total = 57, column total = 74, grand total = 154
• E = 57 × 74 ÷ 154
• E = 27.4

Find the degrees of freedom for a chi-squared test for independence on a contingency table with: (a) 5 rows and 3 columns; (b) 2 rows and 6 columns; (c) 4 rows and 4 columns.

• (a) df = (5 − 1)(3 − 1) = 8
• (b) df = (2 − 1)(6 − 1) = 5
• (c) df = (4 − 1)(4 − 1) = 9

A teacher records how often 175 students hand in homework on time, by gender. 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: Girls: 31.1, 30.6, 23.3; Boys: 32.9, 32.4, 24.7
• From the calculator: χ² = 11.0, p = 0.00401, df = 2

The curve shows the \(\chi^2\) distribution with 5 degrees of freedom. The shaded right tail has area 0.05.

1. (a) Find the critical value.

   • From the calculator: critical value = 11.1

2. (b) A test gives \(\chi^2=12.4\). Find the p-value.

   • From the calculator: P(χ² > 12.4) = 0.0297
   • 12.4 > 11.1, so the test statistic is inside the shaded tail

A tour company asks 287 customers which type of holiday they chose and which age group they are in. At the 5% significance level, test whether holiday type is independent of age group.

1. (a) State the hypotheses.

   • H0: holiday type is independent of age group
   • H1: holiday type is not independent of age group

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 = 8.63, p = 0.0711, df = 4

4. (d) State the conclusion in context.

   • 0.0711 > 0.05, so do not reject H0.
   • There is insufficient evidence, at the 5% significance level, that holiday type is not independent of age group.

A shop records 250 sales of two kinds of phone case, by the colour chosen. Test, at the 5% significance level, whether the case chosen is independent of the colour.

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

   • Enter the table as a matrix. From the calculator: χ² = 9.94, df = 3

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

   • 9.94 > 7.81, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the case chosen is not independent of the colour chosen.

A gym asks 328 members how many days a week they visit and which membership they hold. The results are shown.

1. (a) Find the expected frequencies.

   • Enter the observed matrix. From the calculator, the expected frequencies:
   • Basic: 27.5, 24.3, 22.6, 20.6; Standard: 34.2, 30.2, 28.1, 25.5; Premium: 33.3, 29.5, 27.3, 24.9

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: Basic: 15.3, 0.223, 2.55, 5.43; Standard: 0.510, 0.756, 0.334, 0.236; Premium: 7.98, 0.204, 4.15, 2.64
   • The largest is 15.3, for Basic and 1 day

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

   • O = 48, E = 27.5
   • O is above E: there are more observed in the Basic, 1 day cell than independence predicts, so membership type and visits per week look linked here.

A doctor records whether 90 patients took a vitamin and whether they caught a cold in winter.

1. (a) Find the expected frequency for patients who took the vitamin and caught a cold if the variables are independent.

   • Row total = 45, column total = 43, grand total = 90
   • E = 45 × 43 ÷ 90 = 21.5

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 = 1.11
   • 1.11 < 3.84, so do not reject H0.
   • There is insufficient evidence, at the 5% significance level, that catching a cold is not independent of taking the vitamin.

A test of whether favourite music is independent of age group gives a p-value of 0.032.

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

   • 0.032 < 0.05, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that favourite music is not independent of age group.

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

   • 0.032 > 0.01, so do not reject H0.
   • There is insufficient evidence, at the 1% significance level, that favourite music is not independent of age group.