Question

Hypothesis Testing_03 (two independent samples) m the detection temperature under load for two different vnes of plastie pipe is being investee samples of 15 pipe specimens are tested and the detection temperatures observed are as follows ( 7) Type_01: 193 207 206 210 205187194 178 205 185 189213192 188 194 Type_02: 176 180 185 200 197 192 198 177 188 197 189 206 201 203192 a. Write down the null and alternative hypotheses to test if the deflection temperature under load for Type_02 pipe exceeds that of Type_01. b. What is the type of statistical test that is appropriate? Explain Test the hypotheses at 0.05, using the critical value approach. What is your decision on null hypothesis(HO)? Interpret the test results. d. What is the P Value for the test? Test the hypotheses using the P Value. Do you get the same answer as part (c)? construct 95% confidence interval for the mean difference of deflection temperatures. Test the wypotheses using the confidence interval approach. Do you get the same answer as in parts and 2

Answer 1

For Type 1 :  

Sample mean using excel function AVERAGE(), x̅1 = 196.4

Sample standard deviation using excel function STDEV.S, s1 = 10.4799

Sample size, n1 = 15

For Type 2 :  

Sample mean using excel function AVERAGE(), x̅2 = 192.0667

Sample standard deviation using excel function STDEV.S, s2 = 9.4375

Sample size, n2 = 15

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a) Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 < µ2

b) we will use a two independent sample t test with pooled variance.

c) Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((15-1)*10.4799² + (15-1)*9.4375²) / (15+15-2) = 99.4476

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (196.4 - 192.0667) / √(99.4476*(1/15 + 1/15)) = 1.19

df = n1+n2-2 = 28

Critical value :

Left tailed critical value, t crit = T.INV(0.05, 28) = -1.701

Reject Ho if t < -1.701

Decision:

As t = 1.19 > -1.701, Do not reject the null hypothesis

d) p-value :

Left tailed p-value = T.DIST(1.19, 28, 1) = 0.8780

Decision:

p-value > α, Do not reject the null hypothesis

Yes, we get the same answer as in c).

e)

95% Confidence interval :

At α = 0.05 and df = n1+n2-2 = 28, two tailed critical value, t-crit = T.INV.2T(0.05, 28) = 2.048

Lower Bound = (x̅1 - x̅2) - t-crit*√(S²p*(1/n1 +1/n2)) = (196.4 - 192.0667) - 2.048*√(99.4476*(1/15 + 1/15)) = -3.1257

Upper Bound = (x̅1 - x̅2) + t-crit*√(S²p*(1/n1 +1/n2)) = (196.4 - 192.0667) + 2.048*√(99.4476*(1/15 + 1/15)) = 11.7924

-3.1257 < µ1 - µ2 < 11.7924

As the confidence interval contains so we can not reject the null hypothesis.

Yes, the result is same as in c) and d)