A personal computer manufacturer is interested in comparing assembly times for two keyboard assembly processes. The company selects 10 workers at random and asks each to use both assembly processes. The assembly times (in minutes) for Process 1 and Process 2 are recorded, and the differences (Process 1 minus Process 2) are shown in the table below.

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Worker & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
Process 1 & 73 & 52 & 57 & 89 & 39 & 82 & 83 & 31 & 76 & 43 \\
\hline
Process 2 & 59 & 54 & 67 & 71 & 46 & 68 & 73 & 43 & 67 & 52 \\
\hline
Difference & 14 & -2 & -10 & 18 & -7 & 14 & 10 & -12 & 9 & -9 \\
\hline
\end{tabular}
\][/tex]

Assume that the population of these differences in assembly times (Process 1 minus Process 2) is approximately normally distributed.

Construct a 90% confidence interval for [tex]\(\mu_d\)[/tex], the population mean difference in assembly times for the two processes. Find the lower and upper limits of the 90% confidence interval. Carry your intermediate computations to three or more decimal places. Round your answers to two or more decimal places.

- Lower limit: [tex]\(\square\)[/tex]
- Upper limit: [tex]\(\square\)[/tex]



Answer :

To construct a 90% confidence interval for [tex]\(\mu_d\)[/tex], the population mean difference in assembly times for the two processes, we proceed with the following steps:

1. Determine the differences:
From the given data:
Differences (Process 1 - Process 2): [tex]\( [14, -2, -10, 18, -7, 14, 10, -12, 9, -9] \)[/tex]

2. Calculate the sample mean ([tex]\(\bar{x}\)[/tex]) of the differences:
The mean difference is given as:
[tex]\[ \bar{x} = 2.5 \][/tex]

3. Calculate the sample standard deviation ([tex]\(s\)[/tex]) of the differences:
The standard deviation of the differences is:
[tex]\[ s = 11.607 \][/tex]

4. Determine the number of samples (n):
The number of differences is:
[tex]\[ n = 10 \][/tex]

5. Determine the degrees of freedom (df):
[tex]\[ \text{df} = n - 1 = 10 - 1 = 9 \][/tex]

6. Determine the critical t-value for the 90% confidence level:
Using the t-distribution table or an appropriate statistical tool, we find the critical t-value for a 90% confidence level and 9 degrees of freedom:
[tex]\[ t_{\text{critical}} = 1.833 \][/tex]

7. Calculate the margin of error (ME):
The margin of error is given by:
[tex]\[ \text{ME} = t_{\text{critical}} \times \left(\frac{s}{\sqrt{n}}\right) \][/tex]
Substituting the values, we get:
[tex]\[ \text{ME} = 1.833 \times \left(\frac{11.607}{\sqrt{10}}\right) = 6.728 \][/tex]

8. Calculate the lower and upper limits of the confidence interval:
The lower limit is:
[tex]\[ \text{Lower limit} = \bar{x} - \text{ME} = 2.5 - 6.728 = -4.23 \][/tex]
The upper limit is:
[tex]\[ \text{Upper limit} = \bar{x} + \text{ME} = 2.5 + 6.728 = 9.23 \][/tex]

Therefore, the 90% confidence interval for the population mean difference in assembly times for the two processes is:

- Lower limit: [tex]\(-4.23\)[/tex]
- Upper limit: [tex]\(9.23\)[/tex]