Answer to Question #238444 in Statistics and Probability for baber

Question

b) Repeated tests on the determination of human blood composition during a laboratory analysis are known to be normally distributed. Ten tests on a given sample of blood yielded the values 2.002, 1.958, 2.034, 3.987, 3.987 4.014, 3.009, 1.031, 2.048, 1.024, 1.0886, 1.019, 0.020. Find a 90 per cent confidence interval for true composition of the blood in repeated tests of the sample, and what is the margin of error?

Answer 1

$$mean=\bar{x}=\dfrac{1}{13}(2.002+1.958+2.034+3.987$$

$$+3.987+ 4.014+3.009+1.031+2.048$$

$$+1.024+1.0886+1.019+0.020)=2.094$$

$$s^2=\dfrac{1}{13-1}((2.002-2.094)^2+(1.958-2.094)^2$$

$$+(2.034-2.094)^2+(3.987-2.094)^2$$

$$+(3.987-2.094)^2+(4.014-2.094)^2$$

$$+(3.009-2.094)^2+(1.031-2.094)^2$$

$$+(2.048-2.094)^2+(1.024-2.094)^2$$

$$+(1.0886-2.094)^2+(1.019-2.094)^2$$

$$+(0.020-2.094)^2)=1.7055$$

$$s=\sqrt{s^2}=1.306$$

The critical value for $\alpha=0.1$ and $df=n-1=13-1=12$ degrees of freedom is $t_c=z_{1-\alpha/2, n-1}=1.782288.$

The corresponding confidence interval is computed as shown below:

$$(1.448, 2.740)CI=(\bar{x}-t_c\times \dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times \dfrac{s}{\sqrt{n}})$$

$$=(2.094-1.782\times \dfrac{1.306}{\sqrt{12}}, 2.094+1.782\times \dfrac{1.306}{\sqrt{12}})$$

$$=(1.448, 2.740)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $1.448<\mu<2.740,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(1.448, 2.740)$.

$$margin\ of\ error=t_c\times \dfrac{s}{\sqrt{n}}$$

$$=1.782\times \dfrac{1.306}{\sqrt{12}}=0.646$$