Showdoing the data provided that SST is 184 ii) The larger the sample size, the closer the t distribution approximates the standard timesow wie compressive strength of these three different lude ing themorning normal distribution. iv) A two-tailed test of hypothesis contains one rejection region and two nonrejection regions. v) A critical point is a value that is used for comparison to determine whether or not to reject the null hypothesis. sintering times and metal types? Justify your answer. (5 marks) QUESTION 4 QUESTION 3 a) Two portfolios were randomly assembled from some stock exchange, and the daily stock a) The repeated weighting of the same object on a particular chemical balance produces prices are shown in the table below. normally distributed results. Past evidence, using experienced operators, reveals that the mean is equal to the object's mass, and the standard deviation is 0.25 mg. A novice Portfolio 36.44 A 44.2 12.21 59.60 55.44 39.42 51.29 48.68 41.59 operator performed seven repeated weights of the same object and obtained the mean and standard deviation of the mass as 19.3 mg and 0.37 mg, respectively. Portfolio B 32.69 47.2 49.35 36.17 63.04 17.74 4.23 34.98 37.02 i) Which distribution should a novice operator use to infer the population mean mass of The Minitab output for the Two-Sample T-Test and Cl is displayed below. the object? Justify the answer. (2 marks) Two-Sample T-Test and Cl: A, B ii) At a 98% confidence level, estimate the interval for the mean mass obtained by the Two-sample T for A vs E novice operator. Hence, interpret the finding. (5 marks) Variable N Mean SE Mean StDev A 9 43.21 4 . 62 13.87 B 9 35.83 5.76 17.29 ii) Does the sample provide sufficient evidence that the standard deviation of the mass obtained by the novice operator differ from 0.25mg? Test at 5% significance level. Difference - p (A) - p (B) (5 marks) Estimate for difference: 7.38 954 CI for difference: (-8. 28, U) T-Value - 1.00 P-Value - 0.333 DF - 16 iv) State if the test conducted in (iii) is a Type | OR Type II error. Explain your answer. Both use Pooled StDev - (2 marks) Assume the two populations are normally distributed. b) Determine whether each statement below is TRUE or FALSE. i) Based on the Minitab output, state the assumption of variances that will be used to construct the 95% confidence interval? i) An estimate is the value assigned to a population parameter based on the value of a (1 mark) sample statistic. ii) Determine the missing values of U and so in the Minitab output. ii) Width of the confidence interval may increase by increasing the sample size and by (4 marks) decreasing the level of confidence.iii) Construct a 95% confidence interval for the ratio of two population variances for prices of the two portfolios. Explain if the interval constructed is consistent with the assumption of variances that was stated in (i) above. (6 marks) iv) Using the 95% confidence interval for the mean difference in the output above, can you conclude a difference in prices between the two portfolios? Justify your answer. (2 marks) b) The manufacturer of a fuel additive claims that the use of this additive increased mileage. A random sample of six cars was selected and these cars were driven for one week without the fuel additive and then for one week with the fuel additive. The Minitab output below gives the paired T-Test and CI for these cars without and with the additive. Assume the distance travelled with or without the additive are normally distributed. Paired T-Test and Cl: Without, With Paired T for Without - With N Mean St.Dev SE Mean Without 6 23.40 5.42 2.21 With 6 25.22 5.79 2.36 Difference 6 D 1. 423 0. 581 T-Test of mean difference - 0: T-Value - T Assume that the population distributions are approximately normally distributed, answer the questions below. i) Find the missing values of D and T in the output. (3 marks) i) Is there enough evidence at 5% significance level to conclude whether the use of fuel additive increases the mean mileage of cars? (5 marks)KEY FORMULAS Binomial probability formula P(X = x) =(1-pyx; x=0,12 ....n Poisson probability formula P(X = x)_ex -; x = 0,1, 2, ... x! CONFIDENCE INTERVALS Parameter & description Two-sided (1 - @)100% confidence interval Mean, u, variance, 6 unknown, small samples of =n-1 Difference in means, #1 -#2 , 1+1 variances of =02" and unknown of = 1 + my -2, 5p = (n, - 1)s,+ (12 - 1)s 2 My + 12 - 2 Difference in means, Hy - H2. variances of #62 and unknown di = - (s,2, + 5 2 2 P m -1 12 - 1 Mean difference for paired samples, My of =n-1 where n is no. of pairs Variance, 62 (n -1)s' (n-1)s' of =n-1 Ratio of the variances ,?/62 1 $7 Fa/2: VIV2 V1 = 0 -1, v2 = 02 -1