Exam-Style Problems

Forming hypotheses — problem 1 · Maths4U Original · 7 marks

17103

Info

Maths4U Original · M4U-EXT-9231-FPS-T435-P01

A null model assigns equal probabilities to four outcomes, with no fitted parameters. Observed counts are \(33,23,38,18\).

Use a 5% significance level for a written conclusion. A distribution calculator may be used for numerical p-values.

Find the expected frequency in each category. Explain your reasoning. [2]
Find the chi-squared statistic. Explain your reasoning. [2]
Find the p-value. Explain your reasoning. [2]
State the hypothesis-test conclusion at 5%, in the direction and context specified. [1]

Solution

Hint

Categories cover the whole outcome space and do not overlap. Multiply the total by each null probability.

Worked solution

(a)

\(\displaystyle E= \frac{112}{4}= 28\)

Answer: \(28\).

Answer: \(28\)

(b)

\(\displaystyle \sum \frac{(O- E)^{2}}{E}= \frac{125}{14}\)

Answer: \(\displaystyle \frac{125}{14}\).

Answer: \(\displaystyle \frac{125}{14}\)

(c)

\(\displaystyle P(\chi^2_3\ge \frac{125}{14})=0.0302557983169476\)

Answer: \(0.0302557983169476\).

Answer: \(\approx 0.0302557983169476\)

(d)

The p-value is \(0.0302557983169476\). Since it is at most 0.05, reject \(\displaystyle H_{0}\): there is evidence that the specified categorical distribution does not fit. This does not give the probability that \(\displaystyle H_{0}\) is true.

Answer:

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.

(a) · 2 marks

1 mark: Multiply the total by each null probability.
1 mark: Obtain the stated result with the required sign, units and domain.

Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.

Common error: Use expected, not observed, frequency in the denominator.

(b) · 2 marks

1 mark: Add all four squared standardised deviations.
1 mark: Obtain the stated result with the required sign, units and domain.

Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.

Common error: Subtract the number of fitted parameters when determining degrees of freedom.

(c) · 2 marks

1 mark: Use an upper tail with four minus one degrees of freedom.
1 mark: Obtain the stated result with the required sign, units and domain.

Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.

Common error: Use expected, not observed, frequency in the denominator.

(d) · 1 marks

1 mark: Compare the relevant p-value with 0.05 and state the contextual conclusion without claiming the null is proved.

Alternative methods: A comparison with the correct critical value and the same valid contextual conclusion is acceptable.

Common error: A p-value is not the probability that the null hypothesis is true.

Total: 7 marks.

Forming hypotheses — problem 2 · Maths4U Original · 7 marks

17104

Notes & worked examples

Info

Maths4U Original · M4U-EXT-9231-FPS-T435-P02

A null model assigns equal probabilities to four outcomes, with no fitted parameters. Observed counts are \(35,23,41,17\).

Use a 5% significance level for a written conclusion. A distribution calculator may be used for numerical p-values.

Find the expected frequency in each category. Explain your reasoning. [2]
Find the chi-squared statistic. Explain your reasoning. [2]
Find the p-value. Explain your reasoning. [2]
State the hypothesis-test conclusion at 5%, in the direction and context specified. [1]

Solution

Hint

Categories cover the whole outcome space and do not overlap. Multiply the total by each null probability.

Worked solution

(a)

\(\displaystyle E= \frac{116}{4}= 29\)

Answer: \(29\).

Answer: \(29\)

(b)

\(\displaystyle \sum \frac{(O- E)^{2}}{E}= \frac{360}{29}\)

Answer: \(\displaystyle \frac{360}{29}\).

Answer: \(\displaystyle \frac{360}{29}\)

(c)

\(\displaystyle P(\chi^2_3\ge \frac{360}{29})=0.00609210677782619\)

Answer: \(0.00609210677782619\).

Answer: \(\approx 0.00609210677782619\)

(d)

The p-value is \(0.00609210677782619\). Since it is at most 0.05, reject \(\displaystyle H_{0}\): there is evidence that the specified categorical distribution does not fit. This does not give the probability that \(\displaystyle H_{0}\) is true.

Answer:

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.

(a) · 2 marks

1 mark: Multiply the total by each null probability.
1 mark: Obtain the stated result with the required sign, units and domain.

Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.

Common error: Use expected, not observed, frequency in the denominator.

(b) · 2 marks

1 mark: Add all four squared standardised deviations.
1 mark: Obtain the stated result with the required sign, units and domain.

Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.

Common error: Subtract the number of fitted parameters when determining degrees of freedom.

(c) · 2 marks

1 mark: Use an upper tail with four minus one degrees of freedom.
1 mark: Obtain the stated result with the required sign, units and domain.

Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.

Common error: Use expected, not observed, frequency in the denominator.

(d) · 1 marks

1 mark: Compare the relevant p-value with 0.05 and state the contextual conclusion without claiming the null is proved.

Alternative methods: A comparison with the correct critical value and the same valid contextual conclusion is acceptable.

Common error: A p-value is not the probability that the null hypothesis is true.

Total: 7 marks.