Question

A variable has a mean of 100 and a standard deviation of 16. Sixteen observations of this variable have a mean of 113 and a sample standard deviation of 36. Determine the observed value of the a. standardized version of x. b. studentized version of x. a. Z= (Round to three decimal places as needed.) b.t- (Round to three decimal places as needed.)

Answer 1

1. a)

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho :f =100 against alternative hypothesis H1 :pi € 100 Let denotes the sample mean, o denotes the population standard deviation and n denotes the sample size. Here, ū = 113.0000, o2 = 256.0000,0 = 16.00, n = 16 The test statistic can be written as: m (– 100) which under H, follows a standard normal distribution. Decision rule / Rejection region : We reject Ho at 0.05 level of significance if P-value < 0.05 or if |zobs | > 20.025 = 1.959964 0.4- 0.3- density 0.2 0.1- 0.0- -4 -2 0 Z Now, Value of test statistic: The value of the test statistic is /16 (113.0000 – 100) 16.00 Zobs = 3.250000 3.25

b)

2. a) and c)

Let u denotes the population mean. Hypotheses : To test null hypothesis H. :f =100 against alternative hypothesis H1 :pi € 100 Let denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, S ã = 113.0000, s2 = 1296.0000, 8 = 36.0000, n = 16, standard error of mean 36.00 ✓16 9.0000 n The test statistic can be written as: Vn(– 100) S which under H, follows a t 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 = 2.131450 0.4 0.3 - 0.2- 0.1 0.0- Now, Value of the test statistic : The value of the test statistic tobs 16(113.0000 – 100) 36.0000 = 1.444444 1.44

b) Length of confidence interval = Upper bound - Lower bound = 32.62 - 29.38

Letī denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, c= 31.000, sa = 16.000000, s = 4.000, n = 36, standard error = 4.000 V36 0.666667 ~ 0.667 The critical value is tcritical = t 1-0.98 t0.01,35 2.437723 ~ 2.438 ,35 At 98 percent confidence level, The margin of error is: 2 S 4.000 4.000 ME *t 1-0.98 = * 10.01,35 * 2.438 = 1.624000 = 1.624 ,35 2 V36 136 a 98 percent confidence interval of population mean р is S = i + *t 1-0.98 vn ,35 2 4.000 = 31.000 + * +0.01,35 36 = 31.000 + 4.000 136 * 2.438 31.000 + 1.624 = = (29.376000, 32.624000) ~ (29.38,32.62)

Margin of error $= \frac{32.62 - 29.38}{2} = 1.62$