La concentración de ingrediente activo en un detergente líquido para ropa se cree que es afectada por el tipo de catalizador utilizado en el proceso. No se tiene informacion sobre las varianzas poblacionales. Así, Se toman diez observaciones de la concentración con cada catalizador, y los datos son los siguientes: - \{Catalizador 1:\} 57.9, 66.2, 65.4, 65.4, 65.2, 62.6, 67.6, \( 63.7,67.2,71.0 \) - \{Catalizador 2:\} 66.4, 71.7, 70.3, \( 69.3,64.8,69.6,68.6,69.4,65.3,68.8 \) - (a) Definir si se debe asumir igualdad en las varianzas o no. (9p) - (b) Utilizando la información del punto (a), encuentre un intervalo de confianza del \( 95 \% \) para la diferencia en las medias de concentración activa para los dos catalizadores. (9p) - (c) La empresa debe decidir comprar uno de los dos catalizadores, y se busca aquel catalizador que genere una mayor concentración de ingrediente activo. ¿Cuál recomienda usted comprar? (5p)
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This can be done using an F-test for equality of variances. - Calculate the sample variances \( s_1^2 \) and \( s_2^2 \) for both groups: - For Catalizador 1: \( s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1} \) - For Catalizador 2: \( s_2^2 = Show more…
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Solⁱ Given that the depth of flow ranges 0.6 m < H < 0.9 m. ∴ Considering => H = (0.6 + 0.9) / 2 = 0.75 m Now verify the equation & above consideration Q = S⁰ᐧ⁵ (BH)⁵/3 / r (B+2H)%/3 => 0.0002⁰ᐧ⁵ × (20 × 0.75)⁵/3 / 0.03 × (20 + 2 × 0.75)%/3 => 5.67 m3/sec Now, Considering a slightly less depth. => (0.6 + 0.75) / 2 = 0.675 m Q = 0.0002⁰ᐧ⁵ (20 × 0.675)⁵/3 / 0.03 × (20 + 2 × 0.675)%/3 = 4.74 m3/sec Now, A new depth b/w 0.675 & 0.75 m Avg of depth will be = (0.675 + 0.75) / 2 = 0.7125 m Q = 0.0002⁰ᐧ⁵ × (20 × 0.7125)⁵/3 / 0.3 × (20 + 2 × 0.7125)%/3 => 0.014 × 84.5 / 0.03 × 7.55 => 5.22 m3/sec Now, Taking Avg depth b/w 0.675 & 0.7125 => (0.675 + 0.7125) / 2 = 0.69375 Q = 0.14 × (20 × 0.69375)⁵/3 / 0.03 × (20 + 2 × 0.69375)%/3 => 4.90 m3/sec Now Taking bisectⁱ of 0.69375 & 0.7125 m Xₘ = (0.69375 + 0.7125) / 2 = 0.703 Q = 0.014 × (20 × 0.703)⁵/3 / 0.03 × (20 × 0.703)%/3 = 4.95 m3/sec ∴ The depth will be greater than 0.703 m & less than 0.71 m Ans.
Sri K.
The accompanying data are the weights (kg) of poplar trees that were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Also shown are partial results from using the Bonferroni test with the sample data. Complete parts (a) through (c). Determine the test statistic. The test statistic is enter your response here. (Round to two decimal places as needed.) Determine the P-value. The P-value is enter your response here. (Round to three decimal places as needed.) What is the conclusion for this hypothesis test at a 0.10 significance level? A. Fail to reject H0. There is sufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight. B. Reject H0. There is insufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight. C. Fail to reject H0. There is insufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight. D. Reject H0. There is sufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight.
Areen D.
Adding Strong Acid to Buffer Part A Learning Goal: Consider a beaker with 175 mL acetic acid buffer with a pH of 5.000. A student adds 0.300 M HCl solution to the beaker. How much will the pH of this buffer change? The pKa of acetic acid is 4.740. Express your answer numerically to two decimal places. Use a minus sign (-) if the pH has decreased. A buffer is a solution containing a weak acid and its conjugate base. For example, acetic acid buffer consists of acetic acid (CH3COOH) and its conjugate base (CH3COO-). Because ions cannot simply disappear in a solution, the conjugate base is formed when a weak acid is added to the buffer solution, and the conjugate acid is formed when a base is added to the buffer solution (e.g. sodium acetate, NaCH3COO). Buffers work because the conjugate acid-base pair work together to neutralize the addition of H+ or OH- ions. For example, if H+ ions are added to the acetate buffer described above, they will be largely removed from solution by the action of the conjugate base: CH3COO- + H+ -> CH3COOH Similarly, added OH- ions will be neutralized by reaction with the conjugate acid: CH3COOH + OH- -> CH3COO- + H2O This buffer system is described by the Henderson-Hasselbalch Equation: pH = pKa + log [conjugate base]/[conjugate acid]
David C.
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