Dave The Jogger Runs The Same Route Every Day About 22 Miles On 18 ✓ Solved

Dave the jogger runs the same route every day (about 2.2 miles). On 18 consecutive days, he recorded the number of steps using a pedometer. The results were 3,,,,,,,,,,,,,,,,,,129 a) Construct a 95% confidence interval for the true mean number of steps Dave takes on his run. (Round your answers to 4 decimal places.) b) What sample size would be needed to obtain an error of ± 20 steps with 95% confidence? (Enter your answer as a whole number (no decimals). Use a z-value taken to three decimal places in your calculations.) SHOW THE WORK IF POSSIBLE. I know the mean is 3278.7222 and the standard deviation is 97.0535.

One of the nation's biggest regional airlines has tracked 4,000 landings and take-offs during the past month. Treating these data as the population of interest, the company found that the average time the planes spent on the ground (called the turn time) was 17.23 minutes with a standard deviation of 3.79 minutes. Further, they determined that the distribution of turn times is normally distributed. Then, the probability that a single turn time selected at random from this population would exceed 20 minutes is approximately 0.2327.

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Assignment Solution

This assignment encompasses two parts: constructing a 95% confidence interval for the true mean number of steps Dave takes during his runs and determining the required sample size to achieve a ±20 steps error margin with 95% confidence. Additionally, we will analyze the turn times for a regional airline to illustrate how we can derive probabilities from a normally distributed dataset.

Part A: Constructing a 95% Confidence Interval

Given Data:
- Sample Mean (\( \bar{x} \)): 3278.7222
- Standard Deviation (\( s \)): 97.0535
- Sample Size (\( n \)): 18
To construct a 95% confidence interval (CI) for the true mean, we will use the formula:
\[
CI = \bar{x} \pm z \frac{s}{\sqrt{n}}
\]
Where:
- \( z \) is the critical z-value for a 95% confidence level.
- The critical value can be found in z-tables or using statistical software. For a two-tailed test at a 95% confidence level, the \( z \) value is approximately 1.960.
Calculations:
1. Calculate the standard error (SE):
\[
SE = \frac{s}{\sqrt{n}} = \frac{97.0535}{\sqrt{18}} \approx \frac{97.0535}{4.2426} \approx 22.87
\]
2. Calculate the margin of error (ME):
\[
ME = z \cdot SE = 1.960 \cdot 22.87 \approx 44.909
\]
3. Construct the confidence interval:
\[
CI = 3278.7222 \pm 44.909
\]
Calculating the lower and upper bounds:
- Lower Bound: \( 3278.7222 - 44.909 \approx 3233.8132 \)
- Upper Bound: \( 3278.7222 + 44.909 \approx 3323.6312 \)
Thus, the 95% confidence interval for the true mean number of steps Dave takes is approximately:
\[
CI: [3233.8132, 3323.6312]
\]

Part B: Calculating Required Sample Size

To determine the required sample size (\( n \)) to achieve a margin of error (ME) of ±20 steps with 95% confidence, we can rearrange the confidence interval formula:
\[
ME = z \cdot \frac{s}{\sqrt{n}}
\]
Solving for \( n \), we have:
\[
n = \left( \frac{z \cdot s}{ME} \right)^2
\]
Plugging in the values:
- \( z = 1.960 \) (for 95% confidence)
- \( s = 97.0535 \)
- \( ME = 20 \)
\[
n = \left( \frac{1.960 \cdot 97.0535}{20} \right)^2
\]
Calculating the numerator:
\[
1.960 \cdot 97.0535 \approx 190.058
\]
Now, substituting this into the equation for \( n \):
\[
n = \left( \frac{190.058}{20} \right)^2 = (9.5029)^2 \approx 90.051
\]
Since \( n \) should be a whole number, we round up to the nearest whole number:
\[
n \approx 91
\]

Analysis of Turn Times of the Airline

The regional airline has already identified that:
- Mean Turn Time (\( \mu \)): 17.23 minutes
- Standard Deviation (\( \sigma \)): 3.79 minutes
Given these statistics and the fact that turn times are normally distributed, we can calculate the probability of a randomly selected turn time exceeding 20 minutes.
To find this probability (\( P(X > 20) \)), we first need to calculate the z-score:
\[
z = \frac{X - \mu}{\sigma} = \frac{20 - 17.23}{3.79} \approx \frac{2.77}{3.79} \approx 0.731
\]
We use the z-table to find the probability of a z-score being less than 0.731. Upon consulting the z-table, we find:
\[
P(Z < 0.731) \approx 0.7673
\]
To find the probability that a randomly selected turn time exceeds 20 minutes:
\[
P(X > 20) = 1 - P(Z < 0.731) = 1 - 0.7673 \approx 0.2327
\]
This confirms that the probability of a single randomly selected turn time exceeding 20 minutes is approximately 0.2327, as concluded by the airline.

Conclusion

We successfully constructed a 95% confidence interval to estimate the mean number of steps Dave takes and calculated the necessary sample size to achieve a desired margin of error. Furthermore, we demonstrated the application of statistical concepts to analyze turn times in the airline industry, utilizing the normal distribution to derive probabilities.

References

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