Math Problem Statement

Please solve

Two models of a popular pick-up truck are tested for miles per gallon (mpg) gasoline consumption. The Pacer model was tested using a random sample of n₁ = 9 trucks and the sample mean was x̄₁ = 27.3 mpg with sample standard deviation s₁ = 6.2 mpg. The Road Runner model was tested using a random sample of n₂ = 14 trucks. The sample mean was x̄₂ = 22.5 mpg with sample standard deviation s₂ = 6.8 mpg. Does this indicate that the population mean gasoline consumption for the Pacer is higher than that of the Road Runner? Use α = 0.01.

Solution

Let's work through the problem step by step.

Given:

  • Pacer model:
    xˉ1=27.3\bar{x}_1 = 27.3, s1=6.2s_1 = 6.2, n1=9n_1 = 9
  • Road Runner model:
    xˉ2=22.5\bar{x}_2 = 22.5, s2=6.8s_2 = 6.8, n2=14n_2 = 14
  • Level of significance α=0.01\alpha = 0.01
  • Hypothesis: Test if the Pacer’s mean gasoline consumption is higher than the Road Runner's (right-tailed test).

(a) State the null and alternate hypotheses:

Null Hypothesis (H0H_0):
H0:μ1μ2H_0: \mu_1 \leq \mu_2
(Pacer's mean gasoline consumption is less than or equal to the Road Runner's)

Alternate Hypothesis (HaH_a):
Ha:μ1>μ2H_a: \mu_1 > \mu_2
(Pacer's mean gasoline consumption is higher than the Road Runner's)

This is a right-tailed test.


(b) Identify the appropriate sampling distribution and calculate the test statistic:

The sample sizes are small (n1=9n_1 = 9, n2=14n_2 = 14), so we use the Student's tt-distribution.

The formula for the test statistic is:
t=xˉ1xˉ2sp2(1n1+1n2)t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2 \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}}

Where sp2s_p^2 is the pooled variance:
sp2=(n11)s12+(n21)s22n1+n22s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}

Step 1: Calculate sp2s_p^2

sp2=(91)(6.2)2+(141)(6.8)29+142s_p^2 = \frac{(9 - 1)(6.2)^2 + (14 - 1)(6.8)^2}{9 + 14 - 2}
sp2=8(38.44)+13(46.24)21=307.52+601.1221=908.642143.27s_p^2 = \frac{8(38.44) + 13(46.24)}{21} = \frac{307.52 + 601.12}{21} = \frac{908.64}{21} \approx 43.27

Step 2: Calculate the test statistic tt

t=27.322.543.27(19+114)t = \frac{27.3 - 22.5}{\sqrt{43.27 \left( \frac{1}{9} + \frac{1}{14} \right)}} t=4.843.27(0.1111+0.0714)=4.843.270.1825=4.87.894=4.82.811.71t = \frac{4.8}{\sqrt{43.27 \left( 0.1111 + 0.0714 \right)}} = \frac{4.8}{\sqrt{43.27 \cdot 0.1825}} = \frac{4.8}{\sqrt{7.894}} = \frac{4.8}{2.81} \approx 1.71

The test statistic is t1.71t \approx 1.71.


(c) Find or estimate the PP-value:

Degrees of freedom (dfdf):
df=n1+n22=9+142=21df = n_1 + n_2 - 2 = 9 + 14 - 2 = 21

Using a tt-distribution table or software for t=1.71t = 1.71 with df=21df = 21, we find:
P-value0.05P \text{-value} \approx 0.05


(d) Conclusion:

Compare the PP-value to the significance level (α=0.01\alpha = 0.01):
P-value(0.05)>α(0.01)P \text{-value} (0.05) > \alpha (0.01)

Since the PP-value is greater than the significance level, we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to conclude that the Pacer’s mean gasoline consumption is higher than the Road Runner’s.


Would you like more explanation or calculations?

5 Related Questions:

  1. How is the pooled variance calculated, and why is it necessary?
  2. What are the assumptions of a tt-test, and are they satisfied here?
  3. What would happen if the test were two-tailed instead of one-tailed?
  4. How does increasing the sample size impact the results?
  5. How can software tools like Python or R be used to calculate the PP-value?

Tip:

Always check assumptions of normality and equal variances before performing a pooled tt-test.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Student's t-Test
P-Value Calculation
Pooled Variance

Formulas

Pooled Variance: sp² = [(n₁ - 1)s₁² + (n₂ - 1)s₂²] / (n₁ + n₂ - 2)
Test Statistic: t = (x̄₁ - x̄₂) / sqrt(sp²(1/n₁ + 1/n₂))

Theorems

Student's t-Distribution
Hypothesis Testing Framework

Suitable Grade Level

College Level (Introductory Statistics)