Question

Question 6: A research team wants to get a quick picture of the job market in the construction industry. They sample 25 companies randomly. Based on the sample shown in the Google sheet, can we say that there was an overall hiring in the last year in the entire construction industry? (The null hypothesis is that there was no hiring, thus the mean of the differences is zero.) Use a 99% confidence level. Assume that the standard deviation of the sample is the same as that of the population. The Google Sheet can be accessed at this link You can also copy &paste this in your browser: https://docs.google.com/spreadsheets/d/u-43h9GYOFieQZPf49RSgsUYoxn682F1apTiDa c/copy th Coypte g s tha foucione Company Number of employees One year ag Now Sample mean Sample Standard Deviation Letter D ference Use "averagel Use "stdew 480 15 129 138 212 152 163 189 Fit to 141 10 40 Nar? 121 178 14 123 61 -28 16 17 95 108 14 74 249 156 164 271 188

Answer 1

Answer Date: 08/05/2018 To test the hypothesis is that the mean difference is different from 0 at 1% level of significance First, give hypothesis, compute test-statistics, p-value then give conclusion based on p value. The null and alternative hypothesis is The t-test statistics is, with the help of following steps: 1) Import the data. 2) Select the Stat and choose the Basic Statistics option. 3) Select the Paired t and choose variable option and put First sample and Second sample. 4) Click option button choose Confidence level, Test difference and alternative hypothesis. 5) Click Ok.

Paired T-Test and Cl: After, Before Paired T for After Before After Before Difference 25 11.72 22.46 4.49 N Mean StDev SE Mean 25 190.5 114.3 22.9 25 178.8 107.121.4 99 CI for mean difference: (-0.84, 24.28) T-Test of mean difference = 0 (vs not 0) : T-Value = 2.61 P-value 0.015 From the MINITAB output, the t-test statistics is 2.61 The p-value for this test is, From the MINITAB output, the p-value for this test is 0.015 Decision The conclusion is that p-value in this context is higher than 0.01 which is 0.015, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that the mean difference is different from 0. The result is not statistically significant