Respuesta :
Answer:
To find the p-value for this hypothesis test, we first need to compute the t-statistic and then determine the probability associated with that t-statistic from the t-distribution table.
Given:
- Sample size for each group (BMW and Chevy Tahoe) is 50.
- Mean cost of an oil change for BMWs (\( \mu_1 \)) is $79 with a standard deviation (\( \sigma_1 \)) of $18.
- Mean cost of an oil change for Tahoes (\( \mu_2 \)) is $49 with a standard deviation (\( \sigma_2 \)) of $12.
- Hypotheses: \( H_0: \mu_1 = \mu_2 \) (mean cost of oil change is the same for BMWs and Tahoes) vs \( H_a: \mu_1 \neq \mu_2 \) (mean cost of oil change is different for BMWs and Tahoes).
First, we calculate the t-statistic:
\[ t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
Where:
- \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means for BMWs and Tahoes, respectively.
- \( s_1 \) and \( s_2 \) are the sample standard deviations for BMWs and Tahoes, respectively.
- \( n_1 \) and \( n_2 \) are the sample sizes for BMWs and Tahoes, respectively.
Plugging in the values, we get:
\[ t = \frac{(79 - 49) - (0)}{\sqrt{\frac{18^2}{50} + \frac{12^2}{50}}} \]
\[ t = \frac{30}{\sqrt{\frac{324}{50} + \frac{144}{50}}} \]
\[ t = \frac{30}{\sqrt{6.48 + 2.88}} \]
\[ t = \frac{30}{\sqrt{9.36}} \]
\[ t ≈ \frac{30}{3.06} \]
\[ t ≈ 9.80 \]
Now, using the t-distribution table with degrees of freedom (df) equal to \( n_1 + n_2 - 2 = 50 + 50 - 2 = 98 \) (assuming equal sample sizes for both groups), we can find the p-value associated with a t-statistic of approximately 9.80.
The p-value associated with a t-statistic of 9.80 is significantly smaller than the smallest value listed in most t-distribution tables. It's essentially zero.
Therefore, we can conclude that the p-value of this test is less than 0.001.
