5. Your workers in one store appear to be disgruntled and unhappy. You know this

Question

5. Your workers in one store appear to be disgruntled and unhappy. You know this because you designed a job-satisfaction questionnaire and gave it to 16 of your employees. This sample of 16 workers scored an average of 89 on the questionnaire, with a sample standard deviation of 6. You do not know the standard deviation of the population a) Would you use a z-score or a t-score for this confidence interval? Hint: do you know the population variability?) b) Please identify the degrees of freedom: c) What would your critical t-score be? d) What would your standard error of the mean be? e) Given that the job-satisfaction score is 89 and the standard error of the mean is 1.50 Calculate the lower boundary of your confidence interval Calculate the upper boundary of your confidence interval f) How would you summarize this? g) What is the point estimate for of the population mean? h) You would use a t score of to construct a 95% confidence interval, and a t-score of to construct a 99% confidence interval t-table: Confidence Level 90 Confidence Level 95 98 95 98 80 3.078 4 1.533 2.719 63.656 9.925 31.821 36 1306 37 1.305 8 1.304 39 1.304 40 1.303 12.706 2.026 2.024 2.920 6.965 1.476 3.747 3.365 4.032 2.423 2 421 2.571 41 1.303 42 1.302 3 1.302 2.020 2.416 2.695 .355 169 3.055 2.977 8 1.397 1833 2.228 2.201 10 1.372 45 1.301 2.690 6 1.300 47 1.300 48 1.299 49 1299 50 1299 2.408 407 2.403 2.396 2.385 2.654 13 1.350 1.677 2.602 2.009 2.921 2.898 2.878 2.861 55 1.297 60 1.296 65 1295 70 1294 5 1.293 .673 6 1.337 7 1.333 2.004 2.000 2.110 2.101 2.093 2.086 2.381 19 1.328 1.325 1.725 2.074 2.069 2.064 2.060 80 1.292 85 1.292 1291 95 1.291 00 1.290 2.374 2.371 2.366 2.629 2.364 22 .321 2.635 2.508 2.500 2.492 .485 25 1.316 2.626 10 1289 20 1.289 30 1.288 140 1288 150 1287 2.056 2.052 26 1.315 2.473 2.467 1.703 1.978 1.977 3552.614 2.353 2.351 2.326 2.576 2.045 2.756 2.457 2.453 1.655 2.040 2037 32 .309 33 .308 34 1.307 692 2.0352.4452.733t-scores for confidence intervals 2.728 2.724 2.032 (Student's t-distribution)

Explanation / Answer

a)
Here, we use t test because sample size is less than 30 and has an unknown population std.dev

b)
df = 16 - 1 = 15

c)
Critical t score be for 99% = +/-2.947

d)
SE = s/sqrt(n)
= 6/sqrt(16) = 1.5

e)
mean = 89 , SE = 1.5

Lower boundary = mean - t * SE
= 89 - 2.947 * 1.5
= 84.5795
Upper Boundary = mean + t * SE
= 89 + 2.947 * 1.5
= 93.4205
For 95% use t value = +/- 2.131

f) For 99% confidence interval is between 84.5797 to 93.4205

g)
Point estimate = 89

h)
you would use a t score of +/-2.131 to construct a confidence interval, and t score of +/-2.497 to construct a 99% confidence interval.

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