## Related questions with answers

The deflection temperature under load for two different types of plastic pipe is being investigated. Two random samples of 15 pipe specimens are tested, and the deflection temperatures observed are as follows (in $^{\circ} \mathrm{F}$): Type 1: 206, 188, 205, 187, 194, 193, 207, 185, 189, 213, 192, 210, 194, 178, 205. Type 2: 177, 197, 206, 201, 180, 176, 185, 200, 197, 192, 198, 188, 189, 203, 192. (a) Construct box plots and normal probability plots for the two samples. Do these plots provide support of the assumptions of normality and equal variances? Write a practical interpretation for these plots. (b) Do the data support the claim that the deflection temperature under load for type 2 pipe exceeds that of type 1? In reaching your conclusions, use $\alpha=0.05$. (c) Calculate a P-value for the test in part (b). (d) Suppose that if the mean deflection temperature for type 2 pipe exceeds that of type 1 by as much as $5^{\circ} \mathrm{F}$, it is important to detect this difference with probability at least 0.90. Is the choice of $n_{1}=n_{2}=15$ in part (a) of this problem adequate?

Solution

VerifiedGiven:

$\begin{align*} n_1&=\text{Sample size}=15 \\ n_2&=\text{Sample size}=15 \\ \alpha&=\text{Significance level}=0.05 \end{align*}$

The mean is the sum of all values divided by the number of values:

$\overline{x}_1=\dfrac{206+188+205+...+194+178+205}{15}\approx 196.4$

$\overline{x}_2=\dfrac{177+197+206+...+189+203+192}{15}\approx 192.0667$

The variance is the sum of squared deviations from the mean divided by $n-1$. The standard deviation is the square root of the variance:

$s_1=\sqrt{\dfrac{(206-196.4)^2+....+(205-196.4)^2}{15-1}}\approx 10.4799$

$s_2=\sqrt{\dfrac{(177-192.0667)^2+....+(192-192.0667)^2}{15-1}}\approx 9.4375$

(a) NORMAL PROBABILITY PLOT

The data values are on the horizontal axis and the standardized normal scores are on the vertical axis.

If the data contains $n$ data values, then the standardized normal scores are the z-scores in table III corresponding to an area of $\dfrac{j-0.5}{n}$ (or the closest area) with $j\in \{1,2,3,....,n\}$.

The smallest standardized score corresponds with the smallest data value, the second smallest standardized score corresponds with the second smallest data value, and so on.

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