The homogeneity of a standard chloride solution was tested by analyzing portions

Question

The homogeneity of a standard chloride solution was tested by analyzing portions of the matieral at the top and botttom with the following found. Top 26.32, 26.33, 26.38, 26.25 Bottom 26.28, 26.25, 26.38 Is homogeneity indicated at the 95% confidence level, use both t-test and f-test
The homogeneity of a standard chloride solution was tested by analyzing portions of the matieral at the top and botttom with the following found. Top 26.32, 26.33, 26.38, 26.25 Bottom 26.28, 26.25, 26.38 Is homogeneity indicated at the 95% confidence level, use both t-test and f-test
Top 26.32, 26.33, 26.38, 26.25 Bottom 26.28, 26.25, 26.38 Is homogeneity indicated at the 95% confidence level, use both t-test and f-test

Explanation / Answer

Sample 1

n = 4
m = 26.355
s = 0.035118846


Sample 2
n = 3
m = 26.29
s = 0.045825757

"Unequal sample sizes, equal variance."

effective S for the two samples. That value will be

S = 0.00158

The S value is than plugged into the first formula to find the test value called t*. That value is

t* = 2.141048061 with 5 degrees of freedom

. A t table shows that the critical value for 95% confidence and 5 degrees of freedom is 2.5706.

Calculated t* value is less than the critical value,

Hence

there is not a significant difference in the means .

it can be concluded that the sample is homogeneous).

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