Question

To compare the efficiency of two teaching methods, a class of 24 students was divided randomly into two groups. Each group is taught according to a different method. The results at the end of the semester, on a scale of 0 to 100, are as follows: First group: 𝑛1 = 13; 𝑥̅1 = 74.5; 𝑠1 2 = 82.6 Second group: 𝑛2 = 11; 𝑥̅2 = 71.8; 𝑠2 2 = 112.6 Order: Assuming that the populations are normal and with equal and unknown variances, calculate the 95% confidence interval for the difference between the expected values of the two populations

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Answer to a math question To compare the efficiency of two teaching methods, a class of 24 students was divided randomly into two groups. Each group is taught according to a different method. The results at the end of the semester, on a scale of 0 to 100, are as follows: First group: 𝑛1 = 13; 𝑥̅1 = 74.5; 𝑠1 2 = 82.6 Second group: 𝑛2 = 11; 𝑥̅2 = 71.8; 𝑠2 2 = 112.6 Order: Assuming that the populations are normal and with equal and unknown variances, calculate the 95% confidence interval for the difference between the expected values of the two populations

Expert avatar
Jayne
4.4
102 Answers
Para calcular o intervalo de confiança (IC) a 95% para a diferença entre os valores esperados das duas populações, podemos usar a fórmula do IC para a diferença entre duas médias de populações independentes.

A fórmula é dada por:
IC = (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}

Onde:
- \bar{x}_1 e \bar{x}_2 são as médias das duas populações,
- s_1 e s_2 são os desvios padrão das duas populações,
- n_1 e n_2 são os tamanhos das amostras,
- t_{\alpha/2} é o valor crítico da distribuição t-Student com n_1 + n_2 - 2 graus de liberdade e \alpha/2 = 0.025 (para um nível de confiança de 95%).

Calculando o intervalo de confiança:

IC = (74.5 - 71.8) \pm t_{0.025} \cdot \sqrt{\frac{82.6^2}{13} + \frac{112.6^2}{11}}

Agora, precisamos encontrar o valor de t_{0.025} olhando nas tabelas da distribuição t-Student para n_1 + n_2 - 2 = 13 + 11 - 2 = 22 graus de liberdade.

Para um IC de 95%, t_{0.025} = 2.074.

Substituindo na fórmula:
IC = 2.7 \pm 2.074 \cdot \sqrt{\frac{82.6^2}{13} + \frac{112.6^2}{11}}

Calculando o intervalo de confiança:
IC = 2.7 \pm 53.8

IC = (-51.1, 56.5)

\boxed{IC = (-51.1, 56.5)}

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