Question

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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 °F): Type 1 213 195 212 194 201 200 214 192 196 220 199 217 201 185 212 Type 2 185 205 214 209 188 184 193 208 205 200 206 196 197 211 200 (a) 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 a = 0.05. Suppose that the hypotheses are H0: 1- 2 = 0 versus H1:21-2>0. Calculate a P-value. Use lower and upper bounds for the P-value. <P Select the correct conclusion. There is insufficient evidence to conclude that the mean deflection temperature under load for type 1 exceeds the mean for type 2 at the 0.05 level of significance. There is sufficient evidence to conclude that the mean deflection temperature under load for type 1 exceeds the mean for type 2 at the 0.05 level of significance. (b) If the mean deflection temperature for type 1 pipe exceeds that of type 2 by as much as 4°F, it is important to detect this difference with probability at least 0.90. Is the choice of n 1 = n2 = 15 adequate? Use a = 0.05. O No. There is not enough information to make a conclusion about the sample size. O Yes.

Answer 1

we are given that The deflection temperature under load for two types of plastic pipe samplesizes ny =15 h =15 we have to test the claim that the deflection temperature for type 1 pipe exceeds that of type a (6) Mull and alternative hypotheses Hoi My - Mg = 0 cone-tailed) Ho00, test statistic is given by Hai Mi-Mazo ta so? Chiche] with third-2) dif where, sf = pooled variance [C-5)?+ E62-5o)? bitha-2

typy1 x1 type2 ( x1- mean(x1)) (x2-mean(x2)) ( x1- mean(x1))? |(x2-mean(x2)) ? x2 213 185 195 205 - 15.07 4.93 13.93 92.16 70.56 73.96 212 214 194 209 9.6 -8.4 8.6 -9.4 -2.4 -3.4 10.61 8.93 88.36 201 188 5.76 200 184 -12.07 - 16.07 -7.07 11.56 214 193 112.36 192 208 -11.4 7.93 129.96 227.10 24.30 194.04 79.74 145.68 258.24 49.98 62.88 24.30 0.00 35.16 16.56 9.42 119.46 0.00 1246.93 196 205 54.76 4.93 -0.07 5.93 220 199 217 201 185 212 3051 -4.07 -7.4 16.6 -4.4 13.6 -2.4 -18.4 8.6 200 206 196 197 211 200 -3.07 10.93 275.56 19.36 184.96 5.76 338.56 73.96 1537.6 -0.07 total 3001

= 1537.671446193 15tis - 2 2 sp= 2784:53 $ = 94: -- thus, 203.4 200107 ta 99.45615 ta 3:33 (13.26 t. 3.33 3.64 t = 0.91 we have Alcalculated 0.91 def. nithg-2 dot - 28 using t-table, one tauled is 0.15 < p-value Condo p-value

conclusion we have p-valiee = 0.15 LP20120 05 8 thus, b-value> so, we accept the null hypothesis Ho at a=0005

conclusion of the test is " There is insufficient evidence to conclude that the mean deflection tomberature under load for type I exceeds the mean for type a at the 0.05 level of significance.

(b) Here 0.15 <p-valuce co.ro then by the definition of p-valele, there is maximum 0.80 -0.85 probability of detecting the deflection in temperature of pipes at least 3.33of which means for chaice of nEng=15, deflection may be detected up to 4of beet probability will not increase to 0.85 so, to detect the deflection of - in temperatures with oogo, we have to deal with another set of hi and na