Home / IB Mathematics SL 4.11 Formulation of null and alternative hypotheses AI HL Paper 1- Exam Style Questions

IB Mathematics SL 4.11 Formulation of null and alternative hypotheses AI HL Paper 1- Exam Style Questions

IB Mathematics SL 4.11 Formulation of null and alternative hypotheses AI HL Paper 1- Exam Style Questions- New Syllabus

Question

A company that owns many restaurants wants to determine if there are differences in the quality of the food cooked for three different meals: breakfast, lunch and dinner.

Their quality assurance team randomly selects 500 items of food to inspect. The quality of this food is classified as perfect, satisfactory, or poor. The data is summarized in the following table.

 PerfectSatisfactoryPoorTotal
Breakfast1011247232
Lunch68815154
Dinner356910114
Total20427422500

A \(\chi^2\) test at the 5% significance level is carried out to determine if there is significant evidence of a difference in the quality of the food cooked for the three meals.

The critical value for this test is 9.488.

The hypotheses for this test are:

\( H_0 \): The quality of the food and the type of meal are independent.

\( H_1 \): The quality of the food and the type of meal are not independent.

(a) Find the \(\chi^2\) statistic [2]

(b) State, with justification, the conclusion for this test [2]

▶️ Answer/Explanation
Markscheme

(a)
11.0

Chi-squared test for independence

Expected frequency: \( E = \frac{\text{row total} \times \text{column total}}{\text{grand total}} \)

Breakfast: Perfect \( \frac{232 \times 204}{500} = 94.656 \), Satisfactory \( \frac{232 \times 274}{500} = 127.184 \), Poor \( \frac{232 \times 22}{500} = 10.208 \)

Lunch: Perfect \( \frac{154 \times 204}{500} = 62.832 \), Satisfactory \( \frac{154 \times 274}{500} = 84.392 \), Poor \( \frac{154 \times 22}{500} = 6.776 \)

Dinner: Perfect \( \frac{114 \times 204}{500} = 46.512 \), Satisfactory \( \frac{114 \times 274}{500} = 62.424 \), Poor \( \frac{114 \times 22}{500} = 5.016 \)

\( \chi^2 = \sum \frac{(O – E)^2}{E} \)

Contributions: Breakfast, Perfect \( \frac{(101 – 94.656)^2}{94.656} \approx 0.4246 \)

Satisfactory \( \frac{(124 – 127.184)^2}{127.184} \approx 0.0797 \), Poor \( \frac{(7 – 10.208)^2}{10.208} \approx 1.0072 \)

Lunch, Perfect \( \frac{(68 – 62.832)^2}{62.832} \approx 0.4248 \), Satisfactory \( \frac{(81 – 84.392)^2}{84.392} \approx 0.1364 \)

Poor \( \frac{(5 – 6.776)^2}{6.776} \approx 0.4656 \)

Dinner, Perfect \( \frac{(35 – 46.512)^2}{46.512} \approx 2.8519 \), Satisfactory \( \frac{(69 – 62.424)^2}{62.424} \approx 0.6926 \)

Poor \( \frac{(10 – 5.016)^2}{5.016} \approx 4.9504 \)

Sum: \( \approx 0.4246 + 0.0797 + 1.0072 + 0.4248 + 0.1364 + 0.4656 + 2.8519 + 0.6926 + 4.9504 \approx 11.0332 \)

Rounded to three significant figures: 11.0

Result: 11.0 [2]

(b)
Reject \( H_0 \), quality of food and type of meal are not independent

Chi-squared test for independence, \( df = (3-1)(3-1) = 4 \)

Significance level: \( \alpha = 0.05 \)

Critical value: 9.488

Method 1
\( \chi^2 \approx 11.0 > 9.488 \)

Reject \( H_0 \)

Method 2
p-value \( \approx 0.0263 \)

\( p < 0.05 \), reject \( H_0 \)

Conclusion: Evidence suggests quality and meal type are dependent

Result: Reject \( H_0 \), not independent [2]

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