As part of the quality-control program for a catalyst manufacturing line, the raw materials (alumina and a binder) are tested for purity. The process requires that the purity of the alumina be greater than 85%. A random sample from a recent shipment of alumina yielded the following results (in percent):93.2 87.0 92.1 90.1 87.3 93.6A hypothesis test will be done to determine whethe or not to accept the shipment. (a) State the appropriate null and alternate hypotheses.(b) Compute the P-value. (c) Should the shipment be accepted?Explain.

Answer:We accept the alternate hypothesis that the purity of the alumina is greater than 85% and hence, the shipment should be accepted.           Step-by-step explanation:We are given the following in the question:  93.2, 87.0, 92.1, 90.1, 87.3, 93.6Formula:[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  [tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex] [tex]Mean =\displaystyle\frac{543.3}{6} = 90.55[/tex] Sum of squares of differences = 7.0225 +12.6025 + 2.4025 + 0.2025 + 10.5625 + 9.3025 = 42.095[tex]SS.D = \sqrt{\frac{42.095}{5}} = 2.9[/tex]Sample size, n = 6Alpha, α = 0.05 First, we design the null and the alternate hypothesis [tex]H_{0}: \mu = 85\%\\H_A: \mu > 85\%[/tex]We use One-tailed t test to perform this hypothesis. Formula: [tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n-1}} }[/tex] Putting all the values, we have [tex]t_{stat} = \displaystyle\frac{90.55 - 85}{\frac{2.9}{\sqrt{5}} } = 4.27[/tex] The p-value is 0.003969.Since, p-value < 0.05We reject the null hypothesis and fail to accept it. We accept the alternate hypothesis that the purity of the alumina is greater than 85% and hence, the shipment should be accepted.