Question

Hypothesis Testing O2 (when is 2. Cloud seeding has been studied for many decades as a weather modification procedure. The rainfall in a from 20 clouds that were selected at random and seeded with silver nitrate follows: 180 30.7 19.8 27.1 22.3 18.8 31.8 23.4 21.2 27.9 31.9 27.1 25.0 24.7 26.9 218 29.2 34.8 26.7 31.6 Write down the null and alternative hypotheses to test the claim that mean rainfall from seeded clouds exceeds 25 acre feet. a b. What is the type of statistical test that that should be used? Explain. Test the hypotheses at 0.05, using the critical value approach. What is your decision on null hypothesis(H.)? Interpret the test results. d. What is the P Value for the test? Test the hypotheses using the P Value. Do you get the same answer as part (c)? e. Construct a 95% confidence interval for the mean rainfall. Test the hypotheses using the confidence interval approach. Do you get the same answer as parts (c) and (d)?

Answer 1

(a) To perform the right-tail hypothesis test, we use the six-step method. Step1: The null and alternative hypotheses are as follows. H,:41= 25 (The population mean is equal to 25) versus 4:4>25 (The population mean is greater than 25) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (b) Step 2: We have selected a sample of n = 20. We decide to use a = 0.05. Step3. Because o is unknown, you use the t - distribution and the t-test statistic. We must assume that the population mean are normally distributed. --------------------------------------- ---- (c) Step 4: For a given sample size, n, the test statistic t follows a t distribution with n-1 degrees of freedom. The critical values of the t distribution with 20-1 =19 degrees of freedom. The table of t test is obtained by using the Excel2010 function, = tinv(probability, degrees of freedom) tit( 0.05,19) = +1.729133 10.05.10 = 41.729133 Step 5: The sample statistics: The sample mean is X = 26. The standard deviation is sz = 4.84. The sample size is n = 20. The formula for calculating the t-test statistic is given by 26-25 = 4.84 120 1 4.84 4.472136 1.082257 = 0.923995 t = 0.923995 The Decision Rule: Since, the test statistic is t > 60.05.19 = 1.729133. We reject the null hypothesis Ho Otherwise, we do not reject the null hypothesis H. Since, the test statistic is f\= 10.9239]< t00:19 = 11.729133] We do not reject the null hypothesis H, -------------------------------------------

The P-value approach: P(t> )=(1-P(1 st)) = (1-P(t <0.923995)) = (1 -0.183538) = 0.816462 P-value = 0.816462) > a = 0.05. We do not reject the null hypothesis 14 Yes, the conclusion is matched with the part () Step 6: Decision: Since, the test statistic is = 0.9239< t005.19 = 1.729133 We do not reject the null hypothesis H. Hence, we conclude that there is insufficient evidence to conclude that the mean rain fall exceeds 25 acre feet at the 5% level of significance.

(e) The 95% confidence interval estimate of the mean rain fall is The Sample Size is n = 20 The Sample Mean is X = 26 The Sample Standard Deviation s = 4.84. df = n-1 = 20-1 =19 df = 19 The level of significance is a =0.05. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The table values are calculated by using the Excel2010 function, = tinv(probability, degrees of freedom) timv(0.05, 193 = 2.093024 40- 12.093024 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The formula for calculating the 95% confidence interval for the true population mean is 8-way to su 58+Laina- =-26-209302 (2) Sus 26+209302 ( 9 ) 4.84 14.84 =26-2.093024 Sus 26+ 2.093024 14.472136) 14.472136) = 26-2.093024(1.082257)<4526 +2.093024 (1.082257) = 26-2.26519 Sus 26+2.26519 = 23.7348134 528.26519 X-41 in SusX+4-2= 23.73485 us 28.2652 - - - - - - - - - - - - - - - - - - - - - ----- -------- ---------- - Interpretation: We can be 95% confident that the true population mean rain fall is between 23.7348 <u<28.2652 Since, the population mean is 25 lies in between the calculated confidence interval. We can say that the null hypothesis is not rejected. Hence, we can conclude that the popopulation mean rain fall is equal to 25 acre feet at the 5% level of significance. Yes, the conclusion of the parts (c), (d) and (e) are same.