Question

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 (in °F) are Type 1 206 188 205 187 194 193 207 185 189 213 192 210 194 178 205 and 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? What is the practical interpretation for these plots?

(b) Do the data support the claim that the deflection temperature under load for type 1 pipe exceeds that of type 2? In reaching your conclusions, use α = 0.05. Calculate a P-value.

(c) If the mean deflection temperature for type 1 pipe exceeds that of type 2 by as much as 5°F, it is important to detect this difference with probability at least 0.90. Is the choice of n1 = n2 = 15 adequate? Use α = 0.05.

Answer 1

Answer:-

Given That:-

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 (in °F) are Type 1 206 188 205 187 194 193 207 185 189 213 192 210 194 178 205 and Type 2 177 197 206 201 180 176 185 200 197 192 198 188 189 203 192 .

(a) The figure below shows the normal probability plot of the two samples of yield data from Mintab. The normal probability plots indicate that there is no problem with the normality assumption . Function, both straight lines have similar slopes , providing some verification of the assumption of equal variances.

The figure below shows comparative box plots from Minitab. The comparative box plots indicate that there is no obvious difference in the two types of pipe. although type 1 has slightly greater sample variability.

(b)the data support the claim that the deflection temperature under load for type 1 pipe exceeds that of type 2  In reaching your conclusions, use α = 0.05. Calculate a P-value.

Find the sample means.

Find the standard deviations.

First we compute the pooled estimator of :

The pooled estimator has degrees of freedom

The test statistic is

Test against the alternative . with The test is one - side That , we reject at the level confidence if

Since from the table of percentage points of the t distribution , we find that for DF = 28, and Therefore , since 0.683<1.1891<1.313. we conclude that lower and upper bounds on the P - value are 0.1<P<0.25

Since the P - value exceeds the null hypothesis cannot be rejected .At the 0.05 level of significance . we do not have strong evidence to conclude that the deflection temperature under load type 1 pipe exceeds that of type 2

(c)If the mean deflection temperature for type 1 pipe exceeds that of type 2 by as much as 5°F, it is important to detect this difference with probability at least 0.90. Is the choice of n1 = n2 = 15

Use α = 0.05.

Using as a rough estimate of the common standard deviation the computer software output is

Power and sample Size

2 - Sample t Test

testing mean 1 = mean 2 (versus not=)

Calculating power for mean 1 = mean 2 + difference

Alpha = 0.05 Assume standard deviation = 9.9723

Difference Sample Size Target Power Actual Power

5 85 0.9 .901392

Therefore , a sample of at least 85 pipe specimens should be used Thus. the choice of is not adeuate.