Question

A manager of a children’s toy plant believes that a two-day training class helps reduce the number of products that an employee breaks during the assembly process. She randomly selects seven senior employees to receive the training. She collects data on the number of toys they break per week before the training and then measures the same thing for the week after the training. The standard deviation of the difference is 3.45. Assume that toy breaking is normally distributed. The data collected is included in the table below:

Worker	Toys Broken Before Training	Toys Broken After Training
Frank	16	10
John	20	14
Suzanne	18	17
Allyson	22	23
Whitney	23	20
Gray	25	28
Brent	20	20

Calculate d¯ or the average of the differences using an after-minus-before approach. Round your answers to 4 decimals

Calculate the appropriate test statistic needed to test your claim. Round your final answer to 4 decimals. Hint: Make sure you are using the correct formula based on the information given in the problem and use the minus sign (-) for a negative answer.

Find and enter the correct critical value needed to test your claim at the 5% significance level. Round your final answer to 4 decimals. Hint: Things you will need to know: alpha, df, is this one or two-tailed test, and which table to use! Use the minus sign (-) for a negative answer.

Answer 1

From the information, observe that a manager of a children's toy plant believes that a two- way training class helps reduce the number of products than an employee breaks during the assembly proce Consider a random sample of 7 senior employees for training The researcher collects the data on the number of toys they break per week before the training and after the training The Null and Alternative hypothesis is as follows ss Null hypothesis, Ho: H -0 Alternative hypothesis, H .14 < 0 Calculate d Toys broken Toys brokern Employee before training after training d-x-yd 16 20 18 10 36 36 Frank John Suzanne Alyson Whitne Gra Brent 6 6 23 25 20 Total 17 23 20 28 20 3 9 9 0 92 0 12 The mean of the difference iS, (12) 1.7143 Calculate the test statistic. The standard deviation of the paired test 1S,

The standard deviation of the paired test is, 12 92 11.90476 - 3.4503 Therefore, the test statistic value is, 1.7143 3.4503 1.7143 1.3041 1.3146 The level of significance is, The degrees of freedom is a 0.05 df n-1 The tail of the test is one tail because the alternative hypothesis is defined as not equal to type

Use t table for identifying the t critical value. From the standard t table values, observe that the critical value of t for the one tail test at the 5% level of significance and 6 degrees of freedom is 1.9432 Compare the calculated value with the critical value Here, the calculated value less than the critical value so fails to reject the Null hypothesis Hence, conclude that there is insufficient evidence that the two-way training class helps reduce the number of products that an employee breaks the assembly process.