Question

I just need to know how they got .3635 in D.) and how they got 4.032 in f.)

(15) The following data represent the muzzle velocity (in feet pe gun. For each round, two measurements devices, resulting in the following data. (round to THREE decimal places as needed) Observation 9. r second) of rounds fired from a 155-mm of the velocity were recorded using two different measuring 791 793 791 795 793793 799 790 798 790 794 790 Is there a difTference in the measurement of the muzzle velocity between device A and device B at the alpha equals 0.01 level of significance? Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers (a) Determine dy A-B for each pair of data Observation 5 791 793 791795 793 793 799 790 798 790 794790 (b) Using calculator to compute d and Sa (c) Identify the null and alternative hypotheses. (d) Using calculator to determine the test statistic for this hypothesis test and find the P-value. (e) What is your conclusion regarding Ho? (f) Using calculator to construct a 99% confidence interval about the population mean difference. (Round to two decimal places as needed.) and interpret the confidence interval.

Answer 1

In part D) the value 0.3635 comes from t distribution table. Now a t-table shows the probabilities(areas) under the probability density function of the t distribution for different degrees of freedom.

Here the test statistic is t= -0.369. So P[T<=t]=0.3635 from t-table (this is a inbuilt table) some table only shows positive value of t then we take the value 0.369 ,based on symmetry the area above 0.369 is equals to area under-0.369.

so,we look across the row for df=5 for values near 0.369.

Similarly in part F) using a t-table for df =5 we can get the critical value for t_5,0.005 which is 4.032.