# Math 370 HW 6 - Homework 6 Solution 4-43 (a, c), 4-46, 4-54...

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1Homework 6 Solution4-43 (a, c), 4-46, 4-54 (a, c), 4-55 (a, c, d), 4-56, 4-574-43.a) 1) The parameter of interest is the true mean compressive strength,.2) H0:= 35003) H1:35004)zxn0 /5) Reject H0if z0<z/2wherez0.005=2.58 or z0> z/2where z0.005= 2.586)42.3255x,= 31.6279.2612/62.31350042.3255z07) Because26.79 <2.58, reject the null hypothesis and conclude the meancompressive strength is significantly different from 3500 at= 0.01.b) Smallest level of significance = P-value = 2[1(26.84) ]= 2[11] = 0The smallest level of significance at which we are willing to reject the null hypothesisis approximately 0.c) z/2= z0.025= 1.96xznxzn 0 0250 025..1262.3196.142.32551262.3196.142.32553237.533273.31With 95% confidence, the true mean compressive strength is between 3237.53 psi and3273.31psi.
2d) z/2= z0.005= 2.58xznxzn 0 0050 005..1262.3158.242.32551262.3158.242.32553231.963278.88