Question

A new concrete mix is being designed to provide adequate compressive strength for concrete blocks. The specification for a particular application calls for the blocks to have a mean compressive strength μ greater than 1350 kPa. A sample of 100 blocks is produced and tested. Their mean compressive strength is 1356 kPa and their standard deviation is 70 kPa. A test is made of H₀ : µ ≤ 1350 versus H₁ : µ > 1350.

a. (True or False) The P-value calculated for testing H₀ : µ ≤ 1350 versus H₁ : µ > 1350 is not a small probability; hence, we can conclude that the mean is greater than 1350 kPa and that the blocks meet the specification.

Answer 1

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho : < 1350 against alternative hypothesis H1 : > 1350 Let i denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, 7 = 1356.000, s2 = 4900.000000, s = 70.00, n = 100 The test statistic can be written as : /n (a - 1350) te S which under H, follows a t distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > t0.05,1–1 = 1.660391 0.4 0.3 0.2 ^ 0.1 0.0 -2 0 t Now, Value of the test statistic : The value of the test statistic tobs V100(1356.0000 - 1350) 70.0000 = 0.857143 0.86 associated degrees of freedom = 99 The critical value = 1.660391 1.66 P-value = P(tgg > 0.857143) =1- P(t99 <0.857143) = 1-0.803282 = 0.196718 0.1967 Conclusion : Since p-value > 0.05 and tobs * tcritical = 1.660391, so we fail to reject H, at 0.05 level of significance Hence, we can conclude that population mean is not significantly greater than 1350

ans-> FALSE