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rjkz [21]
3 years ago
14

Five students performed a Kjeldahl nitrogen analysis of a protein sample. The following weight % nitrogen values were determined

: 15.5, 13.8, 15.7, 15.4, 15.8. What is the value of G_calculated for the Grubbs test? Should the outlier be rejected with 95% confidence?
Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

G_calculated = 1.756

The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.

Explanation:

The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,

G_calculated = | suspect value - mean| / s

Here,  suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.

s is calculated from the following formula:

s = (Σ(xi - x)²/(N-1))^1/2

Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).

From the above formula, s is found to be

Standard Deviation, s = 0.820

Now for G value,

G_calculated = | 13.8 - 15.24| / (0.820)

G_ calculated = 1.756

The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.

As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.  

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\bold{\huge{\underline{ Solution }}}

<h3><u>Basic </u><u>Characteristic </u><u>of </u><u>acids </u></h3>

  • Acids are sour in taste
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According to Arrehinus,

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<u>Second </u><u>definition </u><u>of </u><u>Acids </u><u>was </u><u>given </u><u>by </u><u>Bonsted </u><u>Lowry </u><u>:</u><u>-</u>

According to Bonsted Lowry

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<u>3rd </u><u>definition </u><u>was </u><u>given </u><u>by </u><u>Lewis </u>

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  • Bases are bitter in taste
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<u>Arrehinus definition of bases :-</u>

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<u>Bonsted Lowry definition </u>

According to Bonsted Lowry

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4) </span><span>An increase in the activation energy of the forward reaction:
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Result:
          Hence, the correct answer is,"</span>An increase in the temperature of the system".
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