Respuesta :
Answer:
a) 423.458
b) 9.53
c) 1.94
d) 424.5
e) 0.2916
Step-by-step explanation:
We are given the following in the question:
426, 433, 415, 420, 420, 438, 417, 410, 430, 434, 423, 426, 412, 434, 435, 432, 409, 426, 409, 436, 422, 430, 411, 415
a) point estimate of the mean oxide thickness
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{10163}{24} = 423.458[/tex]
b) point estimate of the standard deviation of oxide thickness
[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.
Sum of squares of differences = 2089.958
[tex]s = \sqrt{\dfrac{2089.958}{23}} = 9.53[/tex]
c) standard error of the point estimate
Standard error =
[tex]=\dfrac{\sigma}{\sqrt{n}} = \dfrac{9.53}{\sqrt{24}} = 1.94[/tex]
d) point estimate of the median oxide thickness
Sorted data: 409, 409, 410, 411, 412, 415, 415, 417, 420, 420, 422, 423, 426, 426, 426, 430, 430, 432, 433, 434, 434, 435, 436, 438
[tex]Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}[/tex]
Median =
[tex]\dfrac{12^{th}+13^{th}}{2} = \dfrac{423+426}{2} = 424.5[/tex]
e) proportion of wafers in the population that have oxide thickness greater than 430 angstrom
Sample size, n = 24
Number of wafers with oxides greater than 430 angstrom, x = 7
[tex]p = \dfrac{x}{n} = \dfrac{7}{24} = 0.2916[/tex]
