Question

4. Perform the following postulates using the appropriate hypothesis test:

Use a = 0.10 (level of significance)

a. Is there statistical evidence that the average of Plant # 1 is less than Plant2, µ1 <H2?

b. Estimate a confidence interval of 90% for the true value of the difference between averages

4. Realice las siguientes postulados utilizando, la prueba de hipótesis apropiada: Use a = 0.10 (nivel de significación) Planta#1 Xbarra= 6 S, 2 |N, -30 Planta#2 Xbarra 8 S =1 N, =36 a. Existe evidencia estadística que el promedio de la Planta1 es menor que Planta2, 1412 b. Estime un intervalo de confianza de un 90% para el valor verdadero de la di8ferencia entre promedios

Answer 1

Let Ily and My be the population means and let x, and my be the means of the respective samples. The summary statistics of the Two-sample t-test output of the Minitab18: Samplel: 14 = 30, 4 = 6, = 2 Sample2: ng = 36, X3 = 8, 5, =1. The level of significance is a =0.10. Step1: State the null and alternative hypotheses: The null and alternative hypotheses are, respectively. We need to use left-tailed test. H.: 4-4 = 0 The population means is equal to zero) versus H:4-1 <0 (The population mean 4 is less than 4) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step 2: Select the distribution to use: Here, the two samples are independent, g, and , are unknown but equal, and the sample sizes are small but both populations are normally distributed. Hence, all conditions are satisfied. Consequently, we will use the i distribution Step 3: Determine the rejection and non-rejection regions. The < sign in the alternative hypothesis indicates that the test is left-tailed. The significance level is 0.10. From the t distribution table, the critical value of t for df = 64 and 0.10 area in the left-tail of the t distribution is – 1.29492. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Step 4: Calculate the value of the t-test statistic: The value of the test statistic is t for 27 - xz is computed as follows. The pooled standard deviation: (n-1) +(n-1) V + (30-1) (2)* +(36-1)(1) V 30+36-2 (29)4 +(35)1 V 66-2 116+35 V64 V 64 = 2.359375 = 1.536026 S, = 1.536026 -------------------------------------- Step 5: The standard error of the difference between the means: 1 1 SES, Syksyn 2 = 1.536026, le tout = 1.536026J0.033333+0.027778 = 1.536026V0.061111 = 1.536026 (0.247207) = 0.379716 Si-a = 0.379716 Therefore, the standard error of the difference between the means is sz-e = 0.379716 --------------- -------------------------- ---- Step 6: The formula for calculating the t-statistic is (15,-)-(4- Hy) (6-8)-(0) 0.379716 - -2 0.379716 =-5.2671 to * +5.2671 -------- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The P-value approach: P(t <6) = {P{t <to)) =P(t<-5.2671) =(0.999999134) P-value = 0.999999134 <a=0.10. We reject the null hypothesis H, Step7: The conclusion is based on the critical value method: Since, the test statistic is (1=5.27 > 20.10.64 = 1.29491982) We reject the null hypothesis #. There is sufficient evidence to decide that the data provides moderately suggestive evidence that the population meanl is less than population mean2 at the 10% level of significance.

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(6) Calculate the value of the t-test statistic: The value of the test statistic ist for x; – xg is computed as follows. The pooled standard deviation: (n-1) s +(n-1) V + -2 (30-1)(2)* +(36-1)(1) V 30+36-2 (29)4 +(35)1 V 66-2 116 +35 151 = 64 = 12.359375 = 1.536026 S, = 1.536026 ------- - The standard error of the difference between the means: - - - - - - - - - - - - - - - - - - - - - - - - * * * * = 1.536026, le tout = 1.536026 0.033333+0.027778 = 1.536026V0.061111 = 1.536026(0.247207) = 0.379716 S T = 0.379716 Therefore, the standard error of the difference between the means is sx -z = 10.379716 The margin of error is E = tax2.A +22-287- = 10.102.30+36-2559 = 10.05.06-287- = 10.05.0647 = 1.669013(0.379716) = 0.633751 E=0.633751 The 90% confidence interval for the true difference in the population means is (27-x)+E =(6-8) +0.633751 =-2+0.633751 -2.6337554-15-1.36625 (4, - x)+E = -2.6337 544 – 5-1.3663 Interpretation: We can be 95% confident that the difference in the two population means is between -2.6337 and [-1.3663) Lower bound (7; – x3)- 8 = 1+2.6337 Upper bound: (7,- x))+E = -1.3663)