Question

The calibration of a scale is to be checked by weighing a 12 kg test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with σ-0.200 kg. Let μ denote the true average weight reading on the scie (a) What hypotheses should be tested? Ha: μ#12 Ha: μ > 12 Ha: μ < 12 Ha: μ < 12 11.847 (Round your answer to four decimal places.) (b) With the sample mean Itself as the test statistic, what is the P-value when 0001 What would you conclude at significance level 0.01? Conclude that the true mean measured weight is the same as 12 kg. Conclude that the true mean measured weight differs from 12 kg. al places. For a test with α"001, what is the probability that recalibration is udged unnecessary when in fact μ-12.17 Round your answer to forded c)

Answer 1

(a) Let udenote the true average weight reading on the scale The null and alternative hypotheses are, Но: и-12-kg H 12-kg Hence, the correct option is (A) (b) The sample size is, n = 25 Let o be the population standard deviation 0.2000. Let the sample mean itself as the test statistic be, =11.84 The z-score is, Z= 11.84 12 0.2000 25 -4 Find the P-value. [1-0)] -2x 1-(4.00 P- value 2: X _ (Since from the standard normal tables) 2x (10.999968) -2 x (0.000032) =0.000064 0.0001

Since, the P-value is less than 0.01, reject the null hypothesis at a 0.01 Hence, conclude that the true mean measured weight differs from 12 kg. (c) Find the probability of recalibration is judged unnecessary when in fact u 12.1 That is, the probability of type II error B given / 12.1. critical value First, find the σ/νn al 2 = 12 2.58 0.200 /V25 (11.8968,12.1032) The probability of recalibration is judged unnecessary when in fact u=12.1 is, B(12.1) P(11.8968 X12.1032, when /u=12.1) X-12.1 11.8968 12.1 P 0.200 25 12.1032 12.1 0.200 25 0.200 25 = P(-5.08Z 0.08) = D(0.08)- D(-5.08) (Since from the standard normal tables) = 0.5319 - 0.0000 0.5319 Therefore, the required probability value is 0.5319. Find the probability of recalibration is judged unnecessary when in fact /u=11.8 That is, the probability of type II errorB given/i =11.8

B(11.8) P(11.8968 T12.1032, when u 11.8) X 11.8 11.8968 11.8 = P 12.1032 11.8 0.200 25 0.200 /25 0.200 /25 = P(2.42 Z S7.58) D(7.58)- D(2.42) = Since from the standard 1.0000 0.9922 normal tables 0.0078 Therefore, the required probability value is 0.0078.