Question

A reliable assay shows that the ATP (adenosine triphosphate) content of a certain cell type is. You developed a new assay, which gave the values117,119,111, 115, 120$\mathit{\mu }\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathbf{100}\mathit{m}\mathit{L}$(average = 116.4) for replicate analyses. Do your results agree with the known value at the 95%confidence level?

Short answer

The 95% confidence interval does not include the accepted value of$111\mu mol/100mL$

Step-by-step solution

Definition of standard deviation and mean.

• Arithmetic mean: By dividing the amount of measured values by the number of measurements , the arithmetic mean $\overline{x}$is determined.

Mean:

• Standard deviation: The standard deviation is a measurement of how tightly the data cluster around the mean.

The standard deviation (s) is given by the formula:

$s=\sqrt{\frac{{\displaystyle \sum _i}{\left(x_i-\overline{x}\right)}^2}{n-1}}$

Standard deviation of the mean of sets of n values is:

${\sigma }_n=\frac{\theta }{\sqrt{n}}$

Explanation.

The mean:

$$\begin{gathered} \overline{x}=\frac{{\displaystyle \sum _i}{x}_i}{n} \\ =\frac{117+119+111+115+120}{5} \\ =116.4 \\ =116\left(roundedoff\right) \end{gathered}$$

Standard deviation:

$$\begin{gathered} s=\sqrt{\frac{\sum _{}{\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(116-117\right)}^2+.....+{\left(116-120\right)}^2}{\left(5-1\right)}} \\ =3.58 \\ =4\left(roundedoff\right) \end{gathered}$$

95% Confidence interval:

The 95% confidence interval for four degrees of freedom is calculated as follows,

Confidence interval$=\overline{x}\pm \frac{ts}{\sqrt{n}}$

$$\begin{gathered} =116.4\pm \frac{\left(2.776\right)\left(3.58\right)}{\sqrt{5}} \\ =116.4\pm 4.4 \\ =116\pm 4\left(roundedoff\right) \\ =112.0 \end{gathered}$$

or 120.8 (digits in subscript denotes insignificant figures)

The 95% confidence interval does not include the accepted value of$111\mu mol/100mL$ Hence, the difference is significant.