The Situation The manager of a large manufacturing company claims that the mean

Question

The Situation The manager of a large manufacturing company claims that the mean income for all his senior assembly-line workers is $500 per week. The workers believe that the mean wage is actually less than $500 and decide to test their manager's claim. A random sample of 16 workers had the following weekly incomes: 495 470 475 480 485 500 465 475 490 495 505 510 485 490 460 490

You may use any statistical software to answer the following questions. Part A. Test of Significance STATE: At the = 0.05 level, is there sufficient evidence to conclude that the mean weekly salary of all senior assembly-line workers is less than $500 per week?

STATE: What is a 95% confidence interval estimate of the mean weekly salary of all senior assembly-line workers in this manufacturing company?

PLAN: 1. State the name of the appropriate estimation procedure (e.g. matched-pairs t-confidence interval for estimating the mean difference).

(Note: You have already described the parameter being estimated in part A, so you don't have to repeat it here.)

2. State the confidence level. SOLVE: (Note: You have already plotted the data and checked the conditions in part A, so you don't have to redo it here.)

3. Restate the values of the sample mean and sample standard deviation.

4. Write down the t* critical value. Calculate and record the confidence interval in interval form. Be sure to show your work.

CONCLUDE: 5. Interpret your confidence interval in context.

Do this by including these three parts in your conclusion (3 pts): • Level of confidence • Parameter of interest in context • The interval estimate

Explanation / Answer

1) HERE we use one sample t test for mean as sample size is less then 30 and stdd deviation of population is not given.

for 0.05 level, and one sided t test , critical value of t =-1.753

as we know std error =std deviation/(n)1/2 =3.532

therefore tstat =(X-mean)/std error =(500-485.625)/3.5318 =-4.07

as t stat is less then critical value , we reject that mean is equal to 500 .

for 95% CI , value of t at 15 degrees of freedom =+/- 2.13145

hence margin of error =std error*t =7.528

therefore confidence interval =mean+/- margin of error =485.625 +/- 7.528

X 495 470 475 480 485 500 465 475 490 495 505 510 485 490 460 490 mean 485.625 std deviation 14.127

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