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A quality control check on the batch production of wooden skewers involves sampling and measuring the average length of the skewers per batch.For Batch A, a random sample of 20 skewers were found to have an average length of 21.6 cm.For Batch B, 10 random skewers  gave an average length of 20.4 cm.The standard error in the difference was found to be 0.766 cm.Calculate the 90% Confidence interval on the difference in length between the two batches (A-B).Assume that the lengths are normally distributed, do not assume the population variances are equal, and use the conservative "by hand" estimate for the degrees of freedom (ie lesser n-1).

Question

A quality control check on the batch production of wooden skewers involves sampling and measuring the average length of the skewers per batch.For Batch A, a random sample of 20 skewers were found to have an average length of 21.6 cm.For Batch B, 10 random skewers  gave an average length of 20.4 cm.The standard error in the difference was found to be 0.766 cm.Calculate the 90% Confidence interval on the difference in length between the two batches (A-B).Assume that the lengths are normally distributed, do not assume the population variances are equal, and use the conservative "by hand" estimate for the degrees of freedom (ie lesser n-1).

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Solution

Para calcular el intervalo de confianza del 90% sobre la diferencia en la longitud entre los dos lotes (A-B), seguimos estos pasos:

  1. Determinar la diferencia de medias: XˉA−XˉB=21.6 cm−20.4 cm=1.2 cm \bar{X}_A - \bar{X}_B = 21.6 \, \text{cm} - 20.4 \, \text{cm} = 1.2 \, \text{cm}

  2. Identificar el error estándar de la diferencia: SEdiff=0.766 cm SE_{\text{diff}} = 0.766 \, \text{cm}

  3. Determinar el valor crítico t para un intervalo de confianza del 90%:

    • Grados de libertad (df) conservadores: min⁡(nA−1,nB−1)=min⁡(20−1,10−1)=9 \min(n_A - 1, n_B - 1) = \min(20 - 1, 10 - 1) = 9
    • Para un intervalo de confianza del 90%, el nivel de significancia α\alpha es 0.10, y α/2\alpha/2 es 0.05.
    • Buscamos el valor crítico t en la tabla t de Student para df=9df = 9 y α/2=0.05\alpha/2 = 0.05. El valor crítico t es aproximadamente 1.833.

  4. Calcular el margen de error: Margen de error=tα/2×SEdiff=1.833×0.766 cm≈1.404 cm \text{Margen de error} = t_{\alpha/2} \times SE_{\text{diff}} = 1.833 \times 0.766 \, \text{cm} \approx 1.404 \, \text{cm}

  5. Calcular el intervalo de confianza: Lıˊmite inferior=(XˉA−XˉB)−Margen de error=1.2 cm−1.404 cm=−0.204 cm \text{Límite inferior} = (\bar{X}_A - \bar{X}_B) - \text{Margen de error} = 1.2 \, \text{cm} - 1.404 \, \text{cm} = -0.204 \, \text{cm} Lıˊmite superior=(XˉA−XˉB)+Margen de error=1.2 cm+1.404 cm=2.604 cm \text{Límite superior} = (\bar{X}_A - \bar{X}_B) + \text{Margen de error} = 1.2 \, \text{cm} + 1.404 \, \text{cm} = 2.604 \, \text{cm}

  6. Resultado final: El intervalo de confianza del 90% para la diferencia en la longitud entre los dos lotes (A-B) es: (−0.204 cm,2.604 cm) (-0.204 \, \text{cm}, 2.604 \, \text{cm})

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