Testing the fit of a model Problem set B One question at a time

A commuter's journey time, in minutes, is claimed to follow \(N(25, 4^2)\). For 250 journeys the times are grouped as: below 19, 19 to 25, 25 to 29, and 29 or more. Find the expected frequency of each group.

• Probabilities from the calculator: 0.0668, 0.4332, 0.3413, 0.1587
• E = 250 × P
• 16.7, 108, 85.3, 39.7 (total 250)

A footballer takes 3 penalties in a match. Each scores with probability 0.7, independently. She plays 200 such matches. Let X be the number of penalties scored in one match.

1. (a) State the distribution of X.

   • 3 independent trials, each with probability 0.7: \(X \sim B(3, 0.7)\)

2. (b) Find the expected frequency of each value of X.

   • Probabilities from the calculator: 0.0270, 0.1890, 0.4410, 0.3430
   • E = 200 × P = 5.4, 37.8, 88.2, 68.6

A shop models the number of items X (from 1 to 4) in a customer's basket by \(P(X = x) = \frac{x+2}{18}\). There are 270 customers. Find the expected frequency of each number of items.

• Put x = 1, 2, 3, 4 into the formula: 3/18, 4/18, 5/18, 6/18
• These add up to 1
• E = 270 × P = 45, 60, 75, 90

Write \(H_0\) and \(H_1\) for each goodness of fit test, and state the degrees of freedom for (c).

1. (a) Testing whether the masses of apples follow \(N(150, 12^2)\).

   • \(H_0\): the masses of apples follow \(N(150, 12^2)\)
   • \(H_1\): they do not follow this distribution

2. (b) Testing whether the number of defective chips in a pack of 6 follows \(B(6, 0.2)\).

   • \(H_0\): the number of defective chips follows \(B(6, 0.2)\)
   • \(H_1\): it does not follow this distribution

3. (c) A table with 7 groups, no values estimated from the data.

   • df = 7 − 1 = 6

A goodness of fit test has observed frequencies 11, 58, 92, 66, 23 and expected frequencies 15, 50, 100, 60, 25.

1. (a) State the degrees of freedom.

   • 5 groups: df = 5 − 1 = 4

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

   • Observed list and expected list, df = 4
   • From the calculator: χ² = 3.75, p = 0.441

In the matches above, the numbers with 0, 1, 2, 3 penalties scored were 8, 31, 95 and 66. Test at the 5% level whether the number of penalties scored follows \(B(3, 0.7)\). The critical value of \(\chi^2\) with df = 3 at the 5% level is 7.815.

1. (a) Write \(H_0\) and \(H_1\) and find the expected frequencies.

   • \(H_0\): the number of penalties scored follows \(B(3, 0.7)\)
   • \(H_1\): it does not
   • E = 200 × P = 5.4, 37.8, 88.2, 68.6

2. (b) Find \(\chi^2\) on the calculator.

   • df = 4 − 1 = 3
   • From the calculator: χ² = 3.10, p = 0.377

3. (c) Compare \(\chi^2\) with the critical value and write the conclusion in context.

   • 3.10 < 7.815
   • Do not reject \(H_0\). There is insufficient evidence, at the 5% level, that the number of penalties scored does not follow \(B(3, 0.7)\)

The reaction times, in milliseconds, of 250 players are grouped as below 102, 102 to 114, 114 to 126, 126 to 138, and 138 or more, with frequencies 36, 70, 64, 52, 28. Test at the 5% level whether the times follow \(N(120, 15^2)\).

1. (a) Find the expected frequencies.

   • Probabilities from the calculator: 0.1151, 0.2295, 0.3108, 0.2295, 0.1151
   • E = 250 × P = 28.8, 57.4, 77.7, 57.4, 28.8

2. (b) Write \(H_0\) and \(H_1\), and state the degrees of freedom.

   • \(H_0\): the reaction times follow \(N(120, 15^2)\)
   • \(H_1\): they do not
   • df = 5 − 1 = 4

3. (c) Find \(\chi^2\) and the p-value from the calculator.

   • From the calculator: χ² = 7.54, p = 0.11

4. (d) Write the conclusion in context.

   • 0.11 > 0.05, so do not reject \(H_0\). Insufficient evidence, at the 5% level, that the reaction times do not follow \(N(120, 15^2)\).

A wheel at a school fair has six equal sectors numbered 1 to 6. It is spun 210 times. The numbers 1 to 6 come up 30, 41, 33, 39, 28 and 39 times. Test at the 2% significance level whether the wheel is fair.

1. (a) Write \(H_0\) and \(H_1\).

   • \(H_0\): the wheel is fair (each number has probability 1/6)
   • \(H_1\): the wheel is not fair

2. (b) State the expected frequency and the degrees of freedom.

   • E = 210 ÷ 6 = 35
   • df = 6 − 1 = 5

3. (c) Find \(\chi^2\) and the p-value from the calculator.

   • From the calculator: χ² = 4.17, p = 0.525

4. (d) Write the conclusion in context.

   • 0.525 > 0.02, so do not reject \(H_0\). Insufficient evidence, at the 2% level, that the wheel is not fair.

A gardener plants 4 seeds in each of 320 pots. Each seed is thought to germinate with probability 0.5, independently. The numbers of pots with 0, 1, 2, 3, 4 seeds germinating are 12, 62, 118, 94 and 34. Test at the 5% level whether the model for each seed is correct.

1. (a) State the distribution of X if \(H_0\) is true, and write \(H_0\) and \(H_1\).

   • 4 independent trials, each with probability 0.5: \(X \sim B(4, 0.5)\)
   • \(H_0\): the number germinating follows \(B(4, 0.5)\)
   • \(H_1\): it does not

2. (b) Find the expected frequencies.

   • Probabilities from the calculator: 0.0625, 0.2500, 0.3750, 0.2500, 0.0625
   • E = 320 × P = 20, 80, 120, 80, 20

3. (c) State the degrees of freedom and find \(\chi^2\) and the p-value on the calculator.

   • df = 5 − 1 = 4
   • From the calculator: χ² = 19.5, p = 6.17 × 10⁻⁴

4. (d) Write the conclusion in context.

   • 6.17 × 10⁻⁴ < 0.05, so reject \(H_0\). Sufficient evidence, at the 5% level, that the number germinating does not follow \(B(4, 0.5)\).

A café owner models the number of hot drinks X (from 1 to 5) ordered by a table by \(P(X = x) = \frac{x + 1}{20}\). At 240 tables the numbers of hot drinks ordered were 1 to 5: 30, 32, 52, 54 and 72. Test at the 5% level whether the model fits.

1. (a) Find the expected frequencies.

   • Put x = 1 to 5 into the formula and check the probabilities add up to 1
   • E = 240 × P = 24, 36, 48, 60, 72

2. (b) Write \(H_0\) and \(H_1\), and state the degrees of freedom.

   • \(H_0\): the number of hot drinks follows the model
   • \(H_1\): it does not
   • df = 5 − 1 = 4

3. (c) Find \(\chi^2\) and the p-value on the calculator and write the conclusion in context.

   • From the calculator: χ² = 2.88, p = 0.578
   • 0.578 > 0.05, so do not reject \(H_0\). Insufficient evidence, at the 5% level, that the model does not fit.