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Chapter 9: Problem 55

A new rocket-launching system is being considered for deployment of small, short-range rockets. The existing system has \(p=0.8\) as the probability of a successful launch. A sample of 40 experimental launches is made with the new system and 34 are successful. (a) Construct a \(95 \%\) confidence interval for \(p\). (b) Would you conclude that the new system is better?

Short Answer

Expert verified
The 95% confidence interval for the success rate of the new rocket-launching system is (0.734, 0.966). There is not enough evidence from the sample data to conclude that the new system is better than the old one, as the confidence interval includes the success rate of the old system, 0.8.

Step by step solution

01

Calculate Sample Proportion

The sample proportion success rate \(\hat{p}\) is calculated as the number of successful launches divided by the total number of launches, in this case, 34 out of 40: \(\hat{p} = \frac{34}{40} = 0.85\). This is the observed probability of success in the sample.
02

Calculate Standard Error

The standard error (SE) for the proportion is calculated using the formula: \(SE = \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} } = \sqrt{ \frac{0.85 \cdot (1 - 0.85)}{40} } = 0.059\), where \(n\) is the sample size, which equals 40.
03

Calculate Confidence Interval

A 95% confidence interval for the population proportion is given by \(\hat{p} \pm Z \cdot SE\), where \(Z\) is the z-score that corresponds to the desired level of confidence. For a 95% confidence level, \(Z = 1.96\). The interval is then \(0.85 \pm 1.96 \cdot 0.059 = (0.734, 0.966)\). This means that we can be 95% confident that the true success rate of the new system lies within this interval.
04

Compare With Old System

The success rate of the old system is 0.8. This value is within the calculated confidence interval for the success rate of the new system (0.734, 0.966). Therefore, based on this sample, there is not enough evidence to conclude that the new system has a higher success rate than the old one.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
When conducting experiments or surveys, the sample proportion is a key statistic that helps us understand the behavior of the sample data. In this context, the sample proportion, denoted as \( \hat{p} \), is the ratio of favorable outcomes to the total number of observations in the sample.
To calculate the sample proportion, you divide the number of successful outcomes by the total trials. In our example, there were 34 successful launches out of 40 trials. Thus, \( \hat{p} = \frac{34}{40} = 0.85 \).
This means that 85% of the trials in the sample were successful.

Understanding the sample proportion is crucial because it provides a snapshot of how often the event of interest occurs in the sample. This value will help us estimate the likelihood of success in the entire population.
Standard Error
The standard error (SE) of a sample proportion is a measure of how much the sample proportion might vary from the actual population proportion. It helps indicate the precision of our sample estimates.
To find the standard error, use the formula:
  • \( SE = \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} } \)
In this formula, \( \hat{p} \) represents the sample proportion, and \( n \) is the sample size.
For the given problem, \( SE = \sqrt{ \frac{0.85 \cdot (1 - 0.85)}{40} } \approx 0.059 \).
This tells us that the sample proportion of 0.85 could vary by about 0.059 from the real population proportion about the rockets' success.
Hence, a smaller standard error means a more reliable estimate of the true population proportion.
Z-score
The Z-score is a statistical measure that tells us how many standard deviations a data point is from the mean. In confidence interval calculations, it helps determine the spread of the interval at a specified confidence level.
For a 95% confidence level, the commonly used Z-score is 1.96. This score is derived from the standard normal distribution table and reflects how confident we can be that the population parameter falls within the calculated range.
In our scenario, the Z-score of 1.96 was used to calculate the confidence interval around the sample proportion. This means that, approximately 95% of other similar samples should have a sample proportion lying within this range.

Identifying the correct Z-score for your confidence level is important for constructing accurate intervals.
Population Proportion
The population proportion is the actual value of the proportion of a characteristic or outcome in the entire population. While we might not always know this value, we use sample statistics to estimate or infer it.
In the context of the rocket launches, the population proportion \( p \) represents the true success rate of the rocket-launching system in all potential launches.
By calculating a confidence interval, we make an educated guess about where this true proportion might lie, based on the sample data.

In our example, the old system had a known population proportion of 0.8. We are trying to see if the new sample-derived proportion indicates an improvement, but because 0.8 falls within our confidence interval for the new system, we can't confidently say the new system is better.
Therefore, understanding the population proportion is integral in evaluating new methods or systems against established benchmarks.

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Most popular questions from this chapter

A consumer group is interested in comparing operating costs for two different types of automobile engines. The group is able to find 15 owners whose cars have engine type \(A\) and 15 who have engine type B. All 30 owners bought their cars at roughly the same time and all have kept good records for a certain 12 month period. In addition, owners were found that drove roughly the same mileage. The cost statistics are \(y_{A}=\$ 87.00 / 1,000\) miles, \(y_{B}=\$ 75.00 / 1,000\) miles, \(s_{A}=\$ 5.99,\) and \(S B-\$ 4.85 .\) Compute a \(95 \%\) confidence interval to estimate \(\mu_{A}-\mu_{B_{1}}\) the difference in the mean operating costs. Assume normality and equal variance.

Students may choose between a 3-semester-hour course in physics without labs and a 4-semester-hour course with labs. The final written examination is the same for each section. If 12 students in the section with labs made an average examination grade of 84 with a standard deviation of \(4,\) and 18 students in the section without labs made an average: grade of 77 with a standard deviation of \(6,\) find a \(99 \%\). confidence interval for the difference between the average grades for the two courses. Assume the populations to be approximately normally distributed with equal variances.

A random sample of 20 students obtained a mean of \(\bar{x}=72\) and a variance of \(s \sim^{2}=16\) on a college placement test in mathematics. Assuming the scores to be normally distributed, construct a \(98 \%\) confidence interval for \(\sigma^{2}\).

A manufacturer of car batteries claims that his batteries will last, on average, 3 years with a variance of 1 year. If 5 of these batteries have lifetimes of 1.9 \(2.4,3.0,3.5,\) and 4.2 years, construct a \(95 \%\) confidence interval for \(\sigma^{2}\) and decide if the manufacturer's claim that \(\sigma^{2}=1\) is valid. Assume the population of battery lives to be approximately normally distributed.

The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. (a) Construct a \(98 \%\) confidence interval for the mean height of all college students. (b) What can we assert with \(98 \%\) confidence about the possible size of our error it we estimate the mean height of all college students to be 174.5 centimeters?
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