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Refer to the data of Exercise 30 and compute a 95% CI for the difference between true average concentrations for treatments II and III.

Short Answer

Expert verified

The 4th and 22th ordered difference are -5.9 and -3.8, respectively; a 94.5% confidence interval is

\(\left( {{d_{ij\left( 4 \right)}},{d_{ij\left( {22} \right)}}} \right)\)=(-5.9, -3.8)

Step by step solution

01

solving to get values:

Let,

\({d_{ij}} = {x_i} - {y_i}\)

Where\({x_1},{x_2},....,{x_m}\)and\({y_1},{y_2},...{y_n}\)are observed values of continuous distribution that only in the location but not in shap the general form of a\(100(1 - \alpha )\)confidence interval for\({\mu _1} - {\mu _2}\)

Is,

\(\left( {{d_{ij\left( {mn - c + 1} \right)}},{d_{ij\left( c \right)}}} \right)\)

Where\({d_{ij\left( 1 \right)}},{d_{ij\left( 2 \right)}}...,{d_{ij\left( {mn} \right)}}\)are the ordered difference, use appendix values of c.

For m = 5, n = 5 and for 94.5% confidence interval, c = 22, the confidence interval is the form of

\(\left( {{d_{ij\left( {mn - c + 1} \right)}},{d_{ij\left( c \right)}}} \right) = \left( {{d_{ij\left( {5.5 - 22 + 1} \right)}},{d_{ij\left( {22} \right)}}} \right) = \left( {{d_{ij\left( 4 \right)}},{d_{ij\left( {22} \right)}}} \right)\)

02

computing values:

Compute all difference, find 4th and 23th ordered difference values. The four smallest ordered difference are

\( - 7.1, - 6.5, - 6.1. - 5.9\)

And four large ordered differences are

\( - 3.8, - 3.7, - 3.4. - 3.2\)

The 4th and 22th ordered difference are -5.9 and -3.8, respectively; a 94.5% confidence interval is

\(\left( {{d_{ij\left( 4 \right)}},{d_{ij\left( {22} \right)}}} \right)\)=(-5.9, -3.8)

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Most popular questions from this chapter

Compute the\(90\% \)rank-sum \(CI\) for\({\mu _1} - {\mu _2}\)using the data in Exercise\(11\).

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1 6.4 5.8 6.5 7.7 6.1

2 7.1 9.9 11.2 10.5 8.8

3 5.7 5.9 8.2 6.6 5.1

4 9.5 12.1 10.3 12.4 11.7

Test at level .10 to see whether true average strontium-90 concentration differs for at least two of the regions.

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: the X and Y distributions are identical

versus

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distribution

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Consider modifying the procedure for assigning ranks as follows: After the combined sample of m + n observations is ordered, the smallest observation is given rank 1, the largest observation is given rank 2, the second smallest is

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