Respuesta :
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, 93.6
Formula:
[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 = 6
Alpha, 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 < 0.05
We 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.
