Question

2. A postmix beverage machine is adjusted to release a certain amount of syrup into a chamber where it is mixed with carbonated water. A random sample of 25 beverages was found to have a mean syrup content of 1.05 fluid ounces and a standard devia- tion of 0.15 fluid Ounces. Assume that the syrup content of a beverage is normally distributed (a) [1] Give a point estimate for the mean syrup content in a beverage. Give also the standard error of the estimate. (b) (2) Construct a 95% confidence interval for the mean syrup content, (C) (3) The goal is to have about one ounce of syrup per beverage. Use a critical region to test Ho: p= 1 against H: 71 at a significance level of a = 5%. (a) 2) Compute the p-value for the test in part (e).

Answer 1

a) a point estimate for the mean syrup content is

T = 1.05

The standard error of the estimate

50 = = =

b)

Letī denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, ī = 1.050, s2 = 0.0225, s = 0.1500, n = 25 The margin of error is: ME -0.950 24 = 0.150 5 o 0.150 * 0.025,24 = - * * 2.063899 = 0.061917 ~ 0.06 V25 * 125 * 2.0 a 95 percent confidence interval population mean u is S =ī tn* t 1-0,95024 vn 0.150 = 1.050 £ *10.025,24 = 1.050 +4. * 2.063899 V25 = 1.050 + 0.061917 = (0.988083, 1.111917) ~ (0.99, 1.11)

c)

To test null hypothesis Ho :f =1.00 against alternative hypothesis H1 : # 1.00 Let ī denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, = 1.05000, sa = 0.0225, 8 = 0.15000, n = 25 The test statistic can be written as : Vnī – 1.00) which under Ho follows a t distribution with n - 1 df. We reject Ho at 0.05 level of significance if P-value < 0.050 or if tobs | > t0.025.n-1 Now, The value of the test statistic = 1.666667 1.667 associated degrees of freedom = 24 The critical value =+2.063899 ~ +2.064 P-value = P(|tn–1] > tobs) = 2 * P(t24 > 1.666667) = 2 * (1 – 0.945710) = 0.108580 – 0.1086 Since p-value > 0.05 and tobs * tcritical = 2.063899, so We fail to reject Ho at 0.05 level of significance Hence, we can conclude that the population mean is not significantly different from 1.00.