Question

To estimate efficiency of a drug for weight loss, the clinical trial was performed. The results are presented in the table below. Patient number Weight before trial, kg Weight after trial, kg 1 85.0 83.5 2 79.6 78.1 3 75.0 73.2 4 76.4 74.0 5 91 90.2 6 89.3 87.0 7 82.4 79.9 8 81.5 78.5 9 67.4 64.0 10 68.1 65.1 11 70 67.8 12 741 70.0 13 66.6 64.6 14 60 58.6 15 94 92.9 16 88.2 88.0 (a) Use the paired t-test to investigate the claim that the drug affects the weight. Test the null hypothesis HO: udp before- after = 0 against H 1: udio using a =0.01. The parameter of interest is the difference in mean weight, ud, where di = weight before - weight after. HO:ud = 0 H1: udio Round your answer to three decimal places (e.g. 98.765). Reject the null hypothesis if|t0|> i for a = 0.01 Find the test statistic Round your answer to two decimal places (e.g. 98.76). tori (b) Construct a 99% two-sided confidence interval for the difference in weight. Round your answer to two decimal places (eg. 98.76). suds i

Use minitab if possible. If it is, please show step by step process of how you did it.

Answer 1

Minitab:

Stat --> Basic Statisitics --> Paired t test

Session Paired t for the Mean ci C2 Before After Each sample is in a column Sample 1: Before Sample 2: After Select Options... Graphs... Paired t: Options X | Worksheet 1 *** Difference = mean of (sample 1 - sample 2) + C1 C2 C9 After Confidence level: 991 Before 85.0 1 83.5 Hypothesized difference: 0.0 79.6 78.1 2 3 75.0 Alternative hypothesis: Difference # hypothesized difference 73.2 4 76.4 74.0 Help OK Cancel 5 91.0 90.2 6 89.3 87.0 7 82.4 79.9 8 815 785

Output:

Paired T-Test and Cl: Before, After Descriptive Statistics Sample N Mean StDev SE Mean Before 16 78.04 9.98 After 16 75.96 10.47 2.62 2.50 Estimation for Paired Difference 99% Cl for Mean StDev SE Mean u_difference 2.075 1.007 0.252 (1.33, 2.82) u_difference: mean of (Before - After) Test Null hypothesis Ho: u_difference = 0 Alternative hypothesis Hipi_difference 0 T-Value P-value 8.24 0.000

Answers:

2.977 (using t table with alpha = 0.01 and df=n-2 = 16-2=14)

8.24

1.33 and 2.82