QuestionUse the Grubbs test to check if $0.1189$ is an outlier in the data set $0.1123, 0.1087, 0.1139, 0.1112, 0.1189$ at $95\%$ confidence. Show your calculations.

Studdy Solution

STEP 1

Assumptions1. The data set is $0.1123 ,0.1087 ,0.1139 ,0.111 ,0.1189$ . We are testing whether the fifth result ( $0.1189$ ) is an outlier3. We are using the Grubbs test to determine this4. We are testing at the $95\%$ confidence limit

STEP 2

First, we need to calculate the mean of the data set. The formula for the mean is $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$

STEP 3

Plug in the given values to calculate the mean.
$\bar{x} = \frac{1}{5} (0.1123 +0.1087 +0.1139 +0.1112 +0.1189)$

STEP 4

Calculate the mean.
$\bar{x} = \frac{1}{} (0.1123 +0.1087 +0.1139 +0.1112 +0.1189) =0.1130$

STEP 5

Next, we need to calculate the standard deviation of the data set. The formula for the standard deviation is $s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}$

STEP 6

Plug in the given values to calculate the standard deviation.
$s = \sqrt{\frac{1}{5-1} [(0.1123 -0.1130)^2 + (0.108 -0.1130)^2 + (0.1139 -0.1130)^2 + (0.1112 -0.1130)^2 + (0.1189 -0.1130)^2]}$

STEP 7

Calculate the standard deviation.
$s = \sqrt{\frac{1}{4} [(0.1123 -0.1130)^2 + (0.1087 -0.1130)^2 + (0.1139 -0.1130)^2 + (0.1112 -0.1130)^2 + (0.1189 -0.1130)^2]} =0.0036$

STEP 8

Next, we need to calculate the Grubbs test statistic. The formula for the Grubbs test statistic is $G = \frac{|max(x_i - \bar{x})|}{s}$

STEP 9

Plug in the given values to calculate the Grubbs test statistic.
$G = \frac{|max(.1123 -.113,.1087 -.113,.1139 -.113,.1112 -.113,.1189 -.113)|}{.0036}$

STEP 10

Calculate the Grubbs test statistic.
$G = \frac{|max(0.1123 -0.1130,0.1087 -0.1130,0.1139 -0.1130,0.1112 -0.1130,0.1189 -0.1130)|}{0.0036} =.6389$

STEP 11

Now, we need to compare the calculated Grubbs test statistic to the critical value for the Grubbs test at the $95\%$ confidence limit for a sample size of5. The critical value for the Grubbs test at the $95\%$ confidence limit for a sample size of5 is approximately.71.

STEP 12

Since the calculated Grubbs test statistic (.6389) is less than the critical value (.71), we cannot reject the null hypothesis that there are no outliers in the data set at the $95\%$ confidence limit. Therefore, the fifth determination ( $0.1189$ ) cannot be discarded on the basis of the Grubbs test at the $95\%$ confidence limit.

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