A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2 . To do this, the group chooses 13 ars at random and installs both brands on each car. After all of the cars are driven over the standard test course for 20,000 miles, the amount of tread wear (in inches) on each brand of tire on each car is recorded. The data and the differences (Brand 1 minus Brand 2) are shown in the table below.
$$\text{Unknown environment 'tabular'}$$ & 0.91 & -0.32 & 0.01 & 0.95 & 0.26 & 0.85 & 0.80 & 0.37 & 0.94 & 0.21 & 0.10 & 0 & -0.39 \
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\end{tabular}

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Assume that the population of these differences in tread wear (Brand 1 minus Brand 2) is approximately normally distributed.
Construct a $90\mathrm{\%}$ confidence interval for ${\mu }_d$, the population mean difference in tread wear between the two brands. Then find the lower and upper limits of the $90\mathrm{\%}$ confidence interval. Camy your intermediate computations to three or more decimal places. Round your answers to two or more decimal places. (If necessary, consult a list of formulas.)
Lower limit: [
Upper limit: $◻$

Step 1 ：The problem is asking for a 90% confidence interval for the population mean difference in tread wear between the two brands. To calculate this, we need to use the formula for a confidence interval which is: $\overline{x}\pm z\frac{s}{\sqrt{n}}$, where $\overline{x}$ is the sample mean, $z$ is the z-score corresponding to the desired confidence level (for a 90% confidence level, the z-score is approximately 1.645), $s$ is the sample standard deviation, and $n$ is the sample size.

Step 2 ：First, we need to calculate the sample mean and the sample standard deviation of the differences. The differences are [ 0.91, -0.32, 0.01, 0.95, 0.26, 0.85, 0.8, 0.37, 0.94, 0.21, 0.1, 0, -0.39]. The sample size $n$ is 13.

Step 3 ：The sample mean $\overline{x}$ is calculated as the sum of the differences divided by the sample size, which gives $\overline{x}=0.36076923076923073$.

Step 4 ：The sample standard deviation $s$ is calculated using the formula for the sample standard deviation, which gives $s=0.4828640515794195$.

Step 5 ：Substitute these values into the formula for the confidence interval to find the lower and upper limits. The z-score $z$ for a 90% confidence level is approximately 1.645. This gives the lower limit as $0.14048649842542787$ and the upper limit as $0.5810519631130335$.

Step 6 ：Round these values to two decimal places to get the final answer. The lower limit of the 90% confidence interval for the population mean difference in tread wear between the two brands is $\boxed{{\displaystyle 0.14}}$ and the upper limit is $\boxed{{\displaystyle 0.58}}$.