Vehicle Speeds At A Highway Location Are Believed To

Vehicle Speeds At A Certain Highway Location Are Believed To Be Approx

Vehicle speeds at a certain highway location are believed to be approximately normally distributed with a mean of 60 mph and standard deviation of 6 mph. Using the Empirical Rule, fill in the blanks for each scenario:

  • a. There is about a 68% probability that a randomly selected vehicle's speed will be between ____ and ____ mph.
  • b. There is about a 95% chance that a randomly selected vehicle's speed will be between ____ and ____ mph.
  • c. For samples of 16 vehicles, there is about a 68% chance that the sample mean speed will be between ____ and ____ mph.
  • d. For samples of 36 vehicles, there is about a 95% chance that the sample mean speed will be between ____ and ____ mph.
  • e. Given a sample of 36 vehicles with a mean speed of 57 mph, find the probability that a sample would have a mean of 57 mph or less.
  • f. Based on your calculation in (e), do you think it is likely that this sample of 36 vehicles came from a population with a mean of 60 mph? Why or why not?
  • g. How do you calculate the critical t-value and degrees of freedom for data: 3-a) N=10, p=.10; 3-b) N=47, p=.05; 3-c) N=80, p=.01?

Paper For Above instruction

The analysis of vehicle speeds at a specific highway location relies on understanding the properties of the normal distribution and the application of the Empirical Rule. The problem provides a mean speed of 60 mph and a standard deviation of 6 mph and asks to estimate intervals capturing specific probabilities, as well as performing inferential statistics to assess the likelihood of observed data under different assumptions.

Using the Empirical Rule, which states that approximately 68% of data falls within one standard deviation of the mean, about 95% within two standard deviations, and about 99.7% within three standard deviations, we can find the required speed intervals in parts (a) through (d). For part (a), the interval for about 68% of vehicles is one standard deviation from the mean: from 60 - 6 = 54 mph to 60 + 6 = 66 mph. For part (b), about 95% of vehicles fall within two standard deviations: from 60 - (2×6) = 48 mph to 60 + (2×6) = 72 mph.

Similarly, for sample means, the standard error (SE) is essential. The standard error is calculated as the population standard deviation divided by the square root of the sample size. For part (c), with n=16, SE = 6 / √16 = 6 / 4 = 1.5 mph. Since sample means tend to be normally distributed (by the Central Limit Theorem), the interval that captures 68% of sample means is approximately one standard error around the population mean: from 60 - 1.5 = 58.5 mph to 60 + 1.5 = 61.5 mph.

For part (d), with n=36, SE = 6 / √36 = 6 / 6 = 1 mph. The 95% confidence interval for sample mean is then from 60 - (2×1) = 58 mph to 60 + (2×1) = 62 mph, applying the Empirical Rule for two standard errors.

The probability of observing a sample mean of 57 mph or less from a sample of 36 vehicles (part (e)) assumes the sampling distribution of the mean is normal with mean 60 and standard deviation 1 mph. Standardized z = (57 - 60) / 1 = -3. Consulting standard normal tables, the probability that the sample mean is less than or equal to 57 mph is approximately 0.0013, indicating a very low likelihood of such an observation if the true population mean is 60 mph.

Considering part (f), since the probability (p-value) is very low, it suggests that observing such a low sample mean under the assumption that the population mean is 60 mph is unlikely. Therefore, the data cast doubt on the hypothesis that the true mean speed is 60 mph, unless there are reasons to suspect sampling variability or measurement errors.

Finally, to determine the critical t-values for the specified data (parts (a), (b), and (c)), the degrees of freedom (df) and significance level (p) are essential. For a two-tailed test with N=10 and p=0.10, df = N - 1 = 9, and the critical t-value can be found using a t-distribution table or software. For N=47 with p=0.05, df=46; and for N=80 with p=0.01, df=79. The critical t-value depends on the chosen significance level and degrees of freedom, which can be obtained from statistical tables or software like R or SPSS.

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