Question

are these two questions correct?

Preliminary data analyses indicate that it is reasonable to use a test to carry out the specified hypothesis test. Perform the best using the critical value approach. Use a significance level of 0.01 to test the claim that > 265. The sample data consists of cows for which 312 and 300 For the toolbar, press ALT+F10 ( Por ALTEN F10/Mac) TT TT Paragraph Arial 312001 3. E. T. %DOQ TT. of hups -1.56 critical value 2.196 cannot reject hypothesis due to insufficient evidence about the mean being more than 2.5 QUESTION 10 Preliminary data analyses indicate that it is reasonable to use are to carry out the specified hypothesis test. Perform the best using the critical approach Use a cance level of - OS to test the claim that the same code for which For the toolbar.press ALT:10POor ALTFN+F10 M) TTTT Paragraph Anda 312 ET %D0Q TTL 2.87 > Critical value.14.48 the data genges evidence against the hypothesis.

Answer 1

1.

Let u denotes the population mean. Hypotheses : To test null hypothesis H = 2.85 against alternative hypothesis H1 :u > 2.85 Let e denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, ã = 3.120, 32 = 0.270400, 8 = 0.52, n=9 The test statistic can be written as: Vn (3 – 2.85) t= s which under H, follows a t distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.01 level of significance if P-value < 0.01 or if tobs > t0.01,n-1 = 2.896459 0.4 0.3 0.2 0.1 0.0- 2 0 t Now, Value of the test statistic: The value of the test statistic tobs V9(3.1200 – 2.85) 0.5200 = 1.557692 ~ 1.56 associated degrees of freedom = 8 The critical value = 2.896459 2.90 P-value = P(tz > 1.557692) = 1- P(t: < 1.557692) = 1-0.921039 = 0.078961 0.0790 Conclusion : Since p-value > 0.01 and tobs > tcritical = 2.896459, so we fail to reject H, at 0.01 level of significance Hence, we can conclude that population mean is not significantly greater than 2.85

2.

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho :p =32.6 against alternative hypothesis H :p # 32.6 Let I denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, S 5.80 = 38.4000, sº = 33.6400, = 5.8000, n = 15, standard error of mean 1.4976 n 15 The test statistic can be written as: (3 – 32.6) te S which under H, follows at distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs | > to.025,1-1 = 2.144787 0.4 0.3 0.2 0.1 0.0 - t Now, Value of the test statistic: The value of the test statistic tobs = V15 (38.4000 – 32.6) 5.8000 3.872983 3.87 associated degrees of freedom = 14 The critical value =Łt 0.05 0,05, n–1 = t0.025,14 = 22.144787 +2.1448 P-value = P(|tn-1| > tobs) = 2 * P(t14 > 3.872983) = 2 * (1 - 0.999155) = 0.001689 0.0017 Conclusion : Since p-value < 0.05 and tobs > tcritical = 2.144787, so we reject H, at 0.05 level of significance Hence, we can conclude that the the population mean is significantly different from 32.6.