Answer to Question #116632 in Statistics and Probability for Muzzamil

Question

The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint:
3.4 2.5 4.8 2.9 3.6
2.8 3.3 5.6 3.7 2.8
4.4 4.0 5.2 3.0 4.8
Assuming that the measurements represent a random sample from a normal population, find a 95% prediction interval for the drying time for the next trial of
the paint

Answer 1

$2.5, 2.8, 2.8, 2.9, 3.0,$

$3.3, 3.4, 3.6, 3.7, 4.0,$

$4.4, 4.8, 4.8, 5.2, 5.6$

$$\sum_ix_i=2.5+ 2.8+ 2.8+2.9+3.0+$$ $$+3.3+3.4+3.6+ 3.7+ 4.0+4.4+$$ $$+4.8+4.8+5.2+5.6=56.8$$

$$mean=\bar{x}={\sum_ix_i\over n}={56.8\over 15}\approx3.7867$$

$$\sum_i(x_i-\bar{x})^2={2969.4\over 225}\approx13.1973$$

$$s^2={\sum_i(x_i-\bar{x})^2\over n-1}={2969.4\over 225(15-1)}={212.1\over 225}$$

$$s=\sqrt{s^2}=\sqrt{{212.1\over 225}}\approx0.9709$$

The provided sample mean is $\bar{x}=3.7867$ and the sample standard deviation is $s=0.9709.$

The size of the sample is $n=15$ and the required confidence level is 95%.

The number of degrees of freedom are $df=n-1=15-1=14,$ and the significance level is 

$\alpha=0.05.$

Based on the provided information, the critical t-value for $\alpha=0.05$ and $df=14$ degrees of freedom is $t_c=2.144787$

The 95% confidence for the population $\mu$ is computed using the following expression

$$CI=(\bar{x}-t_c\times s \sqrt{1+1/n},\bar{x}+t_c\times s \sqrt{1+1/n})$$

$$=(3.7867-2.144787\times{0.9709\sqrt{14}\over \sqrt{15}},3.7867+2.144787\times{0.9709\sqrt{14}\over \sqrt{15}})$$

$$=(1.7749, 5.7985)$$