Question

The answers are provided in red but I would like to see the calculations and steps about how to get the answers, Thanks:)

Q5. A machine pours juice in bottles marked 1000ml. A quality control supervisor wants to verify that the machine is well calibrated and pours the right amount on average. He selects a random sample of 9 filled bottles and observes a sample mean of 998ml with a sample standard deviation of 5ml. a) Write the relevant hypotheses (Ho and H) to be tested. Hou = 1000 ml Haru # 1000 ml b) Compute the relevant test statistic t=-1.2 c) What is the distribution of this statistic under Ho? t(8) d) Compute the 95% confidence interval for the true mean amount filled. (994.16,1001.84) e) Should Ho be rejected? Why? Don't reject at a 5% significance level.

Answer 1

Answer Date: 03/12/2019 a) To test the hypothesis is that the mean amount filled is different from 1000 ml at 5% significance level. The null and alternative hypothesis is, Ho: 4 = 1000 H: #1000 b) The t-test statistics is, t=x-Mo 998-1000 5/19 =-1.200

The t-test statistics is - 1.200. The distribution this statistics is t due to the sample size is small and the population standard deviation is unknown. The degree of freedom is, d.f.=n-1 = 9-1 The 95% confidence interval for the true mean amount filled is, First, compute t-critical value then find confidence interval. The t critical value for the 95% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed a = 0.05 and the row headed in the t distribution table by using degree of freedom is, d.f.=n-1 = 9-1 = 8 The t critical value for the 95% confidence interval is 2.306. 95%C.I. = ă ft. xs/In = 998 – 2.306x5/19 =998 +3.84 = 994.16 to 1001.84 The 95% confidence interval for the true mean amount filled is 994.16 to 1001.84). e) The t critical value is,

The sample size is small and two-tailed test. Look in the column headed a = 0.05 and the row headed in the t distribution table by using degree of freedom is, d.f.=n-1 =9-1 The t critical value is - 2.306 and 2.306. Decision The conclusion is that the t value corresponds to sample statistics is not fall in the critical region, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that the mean amount filled is different from 1000 ml. The result is not statistically significant.