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A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. The group chooses 13 cars at random and installs both brands on each car. After all cars are driven over the standard test course for 20,000 miles, the 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.

| Car | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
|------|------|------|------|------|------|------|------|------|------|------|------|------|------|
| Brand 1 | 0.30 | 0.38 | 0.30 | 0.37 | 0.24 | 0.33 | 0.27 | 0.40 | 0.40 | 0.26 | 0.40 | 0.21 | 0.25 |
| Brand 2 | 0.60 | 0.04 | 1.20 | 0.34 | 1.27 | 0.21 | 0.39 | 0.55 | 0.68 | 0.28 | 0.65 | 0.95 | 0.57 |
| Difference (Brand 1 - Brand 2) | -0.30 | 0.34 | -0.90 | 0.03 | -1.03 | 0.12 | -0.12 | -0.15 | -0.28 | -0.02 | -0.25 | -0.74 | -0.32 |

Assume that the population of these differences in tread wear (Brand 1 minus Brand 2) is approximately normally distributed.

Construct a 95% confidence interval for μ_d, the population mean difference in tread wear between the two brands.

Lower limit: __________

Upper limit: __________

Sagot :

To construct a 95% confidence interval for [tex]\( \mu_d \)[/tex], the population mean difference in tread wear between the two brands, we will follow these steps:

1. Calculate the mean of the sample differences:
[tex]\[ \bar{d} = \text{mean of the differences} \][/tex]
Given the sample differences are:
[tex]\[ [-0.3, 0.34, -0.9, 0.03, -1.03, 0.12, -0.12, -0.15, -0.28, -0.02, -0.25, -0.74, -0.32] \][/tex]
We obtain:
[tex]\[ \bar{d} = -0.278 \][/tex]

2. Calculate the standard deviation of the sample differences:
[tex]\[ s_d = \text{standard deviation of the differences} \][/tex]
We obtain:
[tex]\[ s_d = 0.399 \][/tex]

3. Calculate the standard error of the mean difference:
[tex]\[ SE = \frac{s_d}{\sqrt{n}} \][/tex]
where [tex]\( n \)[/tex] is the number of differences. For our sample:
[tex]\[ n = 13 \][/tex]
We obtain:
[tex]\[ SE = \frac{0.399}{\sqrt{13}} \approx 0.111 \][/tex]

4. Determine the critical value:
For a 95% confidence interval and [tex]\( n - 1 = 12 \)[/tex] degrees of freedom, the critical value ([tex]\( t \)[/tex] value) from the t-distribution is:
[tex]\[ t_{\alpha/2} \approx 2.179 \][/tex]

5. Calculate the margin of error:
[tex]\[ ME = t_{\alpha/2} \times SE \][/tex]
We obtain:
[tex]\[ ME = 2.179 \times 0.111 \approx 0.241 \][/tex]

6. Calculate the confidence interval:
[tex]\[ \text{Lower limit} = \bar{d} - ME \][/tex]
[tex]\[ \text{Upper limit} = \bar{d} + ME \][/tex]
We obtain:
[tex]\[ \text{Lower limit} = -0.278 - 0.241 \approx -0.520 \][/tex]
[tex]\[ \text{Upper limit} = -0.278 + 0.241 \approx -0.037 \][/tex]

Therefore, the 95% confidence interval for the population mean difference in tread wear between the two brands is approximately:

[tex]\[ \text{Lower limit}: -0.52 \][/tex]
[tex]\[ \text{Upper limit}: -0.04 \][/tex]
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