Showing pages 1 to 4 of 13 pages

Assignment 3Question 1Let1and2be the population means of useful life times (in months) of Cooper paint and King paint,respectively. Then, the null and alternative hypotheses are:H0:1=2H1:1≠2This is a two-tailed test.We have chosen= 0.01.The samples are chosen independently, the population variances are unknown and they are not knownto be equal. Hence, we shall use as test statistic,12221212()0XXSSnn.The population distributions are approximately normal and the sample sizes are large enough. Hence,this is a two-tailedt-testdf =22211222222112212////11snsnsnsnnn=( 212/ 54 + 282/ 60)2/ ((212/ 54)2/ 53)) + ((282/60)2/ 59))= 108.59For df ≈ 109,tα=t0.025= 1.960.Decision rule : rejectH0in favor ofH1if the computedt-value is less than -1.960 or if it is greater than1.960.t= (345 – 351) / (212/54 + 282/60)= - 0.2826

Since –0.2826 is more than –1.960, there is sufficient evidence, at= 0.05, that we cannot rejectH0,that is, to infer that the population mean of the number of units produced on the afternoon shift issmaller.α = 1 – 0.99 = 0.01. For df = 109,tα/2=t0.005= 2.576.A 995 percent confidence interval estimate for the difference12()is(345 – 351)+(2.576) √ ((21)2/ 54)+ ((28)2/ 60)= (-6+11.87) =(-17.87, 5.87)Question 2Let the average weight (in g) be. Then, the null and the alternative hypotheses are:H0:≤250H1:> 250We have selected= 0.05.We shall use as test statistic t = (x̅-μ)/ s /√n.Since the population distribution is approximately normal and the sample size (n = 50) is large enough,the distribution of the test statistic,when= 250, is approximately student’st-distribution with df = (n–1) = 49. Thus, this is a lower-tailedt-test.For df = 49,t=t0.05= approximately 1.678.Our decision rule is: rejectH0in favor ofH1if the computedt-value is less than -t0.05-1.678.The computedt-value is:(252 – 250) / 6/√50= 2.357Since the computedt-value (= -2.357) is less than -1.678, we conclude that there is sufficient evidence at= 0.05, to rejectH0in favor ofH1, that is to infer that the mean weight is greater than 250g.

Question 3Let us formulate the null and alternative hypotheses.H0: μ = $50000 (The mean income does not defer)H1: μ ≠ $50000 (The mean income defers)At the centre of the hypothesized distribution will be the highest possible value for which H0could betrue, μ0= $50000.The significance level as given will be α = 0.1The population standard deviation (σ) is known and the sample size is large, so the normal distribution isappropriate and the test statistics (ᴢ) is calculated as:z = (x̅ -μ0)/σ= (50500 – 50000)/ (3000/√120)= 1.825For a two-tail z-test in which α = 0.1, z = -2.326 and z = +2.326 will be the respective boundaries forlower and upper tails of 0.01 each. These are the critical values for the test and they identify the nonrejection and rejection regions i.e. reject H0if calculated z < -2.326 or z > +2.326, otherwise do notreject.The calculated value, z = 1.825 falls within the non rejection region of the diagram. So at the 0.1 level ofsignificance, the null hypothesis is not rejected.The results suggest that the mean income defers from $50000P-value = (-1.825≤Z≤ 1.825) = approximately 2*(1 – 0.96562) = 0.06876.

Question 4a) True.b) True.c) False.d) True.e) True.