Question

2 .c) Test the given hypothesis. Assume that the population is normally distributed and that the sample has been randomly selected. A manufacturer uses a new production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standard deviation of 4.7 cm. At the 0.10 significance level, test the claim that the new production method has lengths with a standard deviation different from 3.5 cm, which was the standard deviation for the old method. Test statistic: x = 7.788. Do not support the claim B/ Test statistic: X'= 25.344. Support the claim Test statistic: x* = 16.345. Do not support the claim D/ Test statistic: X = 28.852. Support the claim Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim. A light-bulb manufacturer advertises that the average life for its light bulbs is 900 hours. A random sample of 15 of its light bulbs resulted in the following lives in hours. 995 590 510 539 739 917 571 555 916 728 664 693 708 887 849 At the 10% significance level, test the claim that the sample is from a population with amean life of 900 hours. Use the P-value method of testing hypotheses. A/ B/ Test statistic: t = 4.342. Do not support the claim Test statistic: t = -2.05. Support the claim Test statistic: 1 = 3.111. Do not support the claim Test statistic: 1 = 2.575. Support the claim

question 2-c and 2-d

Answer 1

2.0: a random sample of n=17 steel rods have standard deviation s=4.7cm. A 10% level of significance, we want to test the claim that the new production method has lengths with standard deviation different from 3.5cm. The hypothesis testing problem is: Ho: 6=3.5 vs H,:683.5 The test statistic is X= (n-1) 5? -(17-1)x147)* 16x2209_ 353.44 . (3.52 12. 25 12 .2.5 2 ::? = 24.852. x critical value at 10% level of significance for two tail test is xaga, 16 = 7.962 :.x²> Xogs, 16 , we reject the null hypothesis. Therefore, we have enough evidence to support the claim that the new producho method has lengths with standard deviation different from 3.5cm. :: correet choice is D.

2d). Let 'x' be the life of a light bulbs. The sample mean is 2=724.07 #using Excel function = Average (A2: A16) 5=156.92 #by using Excel function. = stdere A2:A1C) At 10% level of significance we want to dest the claim that the sample is from the population with mean life of goo hours. The hypothesis testing problem is: Ho: M=900 vs Hi! "# 900 The test statistic is t. īre population standard deviation ! to is not known we used test. : 722707-90o 136.92/JTS - - -175.93 40.51GE = -4.342 it=-4.342 The pralue for two tail test is P = 2xP (t? 1-4.3421) = 2P(t> 4.342) PZ 2 x0001 Plt), 41342) <0.001 To<0.002 / Here degrees of freedom are we used t-table, since pe 0.002 is less than 0.10, we reger - the null hypothesis. .. corre et choice is Ai> t = -3,342 . Do not support Claim. 14, e reject