Question

An investigation was conducted to determine the source of reduction in yield of a certain chemical product. It was known that the loss in yield occurred in the mother liquor, that is, the material removed at the filtration stage. It was thought that different blends of the original material might result in different yield reductions at the mother liquor stage. The following are the percent reductions for 3 batches at each of 4 preselected blends: Blend 1234 25.6 25.2 20.8 31.6 24.3 28.6 26.7 29.8 27.9 24.7 22.2 34.3 (a) Perform the analysis of variance at the α = 0.05 level of significance. (b) Use Duncan’s multiple-range test to determine which blends differ. (c) Do part (b) using Tukey’s test

Answer 1

Answer:-

Given That:-

An investigation was conducted to determine the source of reduction in yield of a certain chemical product. It was known that the loss in yield occurred in the mother liquor, that is, the material removed at the filtration stage.

Given,

From the information, observe that an investigation was conducted to determine the source of reduction in yeild of a certain chemical product.

(a) Perform the analysis of variance at the α = 0.05 level of significance.

The null and alternative hypothesis are as follows:

Null Hypothesis :

There is no significant direction in mean yield reduction for the 4 preselected blends.

Alternative hypothesis :

There is significant direction in mean yield reduction for thr 4 preselected blends.

The level of significance is, α = 0.05

Use megaStat to perform ANOVA one factor.

Step 1:

Import the data into the spreadsheet.

Step 2:

Select Add-Ins MegaStat Analysis of Variance One factor then a dialog box will open.

Step 3:

Select the data range in the respective field and select as level of significance.

Step 4:

Click on OK button.

Therefore, the MegaStat output is,

From the output, Observe that the p-value is .0121.

Compare the p-value with the level of significance.

Here, the p-value less than the level of significance so reject the Null hypothesis.

Hence, conclude that there is significant direction in mean yield reduction for the 4 preselected blends.

(b) Use Duncan’s multiple-range test to determine which blends differ.

The formula for thr Duncan's least significant range is,

The quantity is the least significant studentized range given in table A.13

The error degrees of freedom is 8.

The level of significance is α = 0.05

From the studentized table values, observe that the quantity values at the 5% level of significance and at 8 error of freedom for p = 2, p = 3 and p = 4 are 3.261, 3.339 and 3.475 respectively.

That is,

Calculate

= 3.261 * 1.368

= 4.461

Calculate

= 3.399 * 1.368

= 4.650

Calculate

  

= 3.475 * 1.368

= 4.754

Therefore, it can be tabulated as follows:

P	2	3	4
	3.261	3.399	4.475
	4.46	4.65	4.75

Now, order the means from smallest to largest.

The mean differences are as follows:

Pair	Mean difference
	31.9-23.233 = 8.663
	31.9-25.933 = 5.967
	26.167-23.233 = 2.933
	31.9-26.167 = 5.733
	26.167-25.933 = 0.2334
	25.933-23.233 = 2.699

Comparisons:

Since (8.666) > (4.75), so the means   and are significantly different.

Since (5.967) > (4.65), so the means   and are significantly different.

Since (2.933) < (4.65), so the means   and are not significantly different.

Since (5.733) > (4.46), so, the means   and are significantly different.

Since (0.2334) > (4.46), so, the means   and are not significantly different.

Since (2.699) > (4.46), so, the means   and are not significantly different.

Therefore, it can be expressed in diagrammatically as,

  

  

__________________ ____

(c) Do part (b) using Tukey’s test

The error degrees of freedom are 8.

The level of significance is,

The number of groups is, k = 4

From the studentized critical values q, observe that the critical value of q 5% level of significance and at 4 for column wise and 8 for row wise is 4.53

calculate Tukey's HSD

HSD =

     

= 6.197

The mean differences are as follows:

Pair	Mean difference
	31.9-23.233 = 8.663
	31.9-25.933 = 5.967
	26.167-23.233 = 2.933
	31.9-26.167 = 5.733
	26.167-25.933 = 0.2334
	25.933-23.233 = 2.699

Since (8.666) > HSD(6.197), so the means   and are significantly different.

Since (5.967) < HSD(6.197), so the means   and are not significantly different.

Since (2.933) < HSD(6.197), so the means   and are not significantly different.

Since (5.733) <  HSD(6.197), so, the means   and are not significantly different.

Since (0.2334) < HSD(6.197), so, the means   and are not significantly different.

Since (2.699) < HSD(6.197), so, the means   and are not significantly different.

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