Statistical inspection of food physical and chemical testing results (2)

[Example 1-5] Two different methods are used to determine the fat content (%) in milk powder.

Try to compare the detection results of the two methods for significant differences

Solution: According to the two sets of data, calculate the average and standard deviation of the two methods respectively, and get:

=1.

=1.

Calculate the combined standard deviation:

Calculate statistics t _meter :

Check Table 1-17, degree of freedom f=n1+n2-2, when the significance level a=0.

t _meter = 1.

For the significance test of the above two measured means, the primary condition is that the variance of the two sets of data is not significantly different, so the F-test method must be used to judge first

3.

Different analysts, different laboratories or using different analytical methods to measure the same sample have different standard deviations

Inspection steps:

(1) Calculate the statistic variance ratio F _meter

(2) Check the F distribution table (see Table 1-18 ) according to the confidence and degree of freedom f _large and f _small to obtain the critical value F _table

(3) Judgment: F _meter ≤ F _table , there is no significant difference in the variance of the two groups of data

F _meter >F _table , there is a significant difference in the variance of the two groups of data

It shows that there is a significant difference in the precision of the two groups of data

Table 1-18 Critical F value table (95% confidence level)

[Example 1-6] Still taking the experimental data in Example 1-5 as an example, compare the precision of the two methods by F test to see if there is a significant difference

Solution: Calculate the variance of the two methods separately

Check the F distribution table (table 1-18): α=0.
05, f _large =6-1=5, f _small =5-1=4, F _table =6.
26

Judgment: 5.
75<6.
26, that is, F _meter <F _table , indicating that the variance of the two methods is not significantly different, that is, the precision of the two measurement methods is the same
.
Both methods are available
.

2.
Confidence degree and confidence interval

Any sample can only be measured a limited number of times to get the measured mean and standard deviation s of the sample.
What we want to know is the true value of the sample, that is, the overall mean u.
Only when the number of measurements n tends to infinity will it tend to u, and the actual measurement The number n cannot tend to infinity
.
Therefore, it can only be an estimate of u
.
According to statistical derivation, the relationship between the population mean u and the sample mean is:

Called the confidence interval of the mean
.
It is the error limit, also called the estimation accuracy.
The smaller the value, the higher the measurement accuracy
.
t is the confidence coefficient under a certain degree of confidence (the probability that the measured value appears in a certain range, such as 95% or 90%) and the degree of freedom f (f=n-1), which can be obtained by referring to Table 1-17
.

The above formula has a clear probability meaning, it shows that the true value u falls in the confidence interval:

The confidence probability of is p(p=1-a, a is the confidence level)
.
For example: after 4 measurements, the confidence interval of the mass fraction of iron in a sample is 0.
1915±0.
0009 (the confidence is 95%), which means: the average value of these 4 measurements is 0.
1915, and there is 95% confidence that the quality of iron is considered The true value of the score falls between 0.
1906 and 0.
1924
.

[Example 1-7] Analyze the starch content in the cake to obtain the following data (%): 37.
45, 37.
20, 37.
50, 37.
30, 37.
25
.
Find the confidence interval when the confidence level is 90%
.

solution:

Looking up Table 1-17, f=n-1=5-1=4, when the confidence level is 90%, t=2.
13, then