Applied Statistics And Process Management

Mas284 Applied Statistics And Process Managementinternalexternal Ass

Identify the core assignment questions: statistical analysis involving probability distributions, hypothesis testing, probability calculations for binomial and Poisson distributions, process control limits, and reliability calculations based on normal distribution assumptions.

These questions involve graphing probability distributions, calculating specific probabilities using binomial and Poisson models, hypothesis testing concerning a population mean, calculating probabilities related to normal distributions, and estimating process control chart limits.

Paper For Above instruction

The given assignment encompasses several fundamental topics in applied statistics, including probability distributions, hypothesis testing, process control, and reliability assessment. To provide a comprehensive understanding, each section of the problem set is approached systematically, illustrating relevant statistical concepts, calculations, and interpretations.

1. Testing the Success Rate of Hip Joint Replacement Parts

a) Distribution of X

The problem involves testing 8 independent hip joint replacement parts, each with a 0.9 probability of success. The number of successful parts, X, follows a binomial distribution with parameters n=8 and p=0.9. The probability mass function (pmf) is given by:

\[ P(X = k) = \binom{8}{k} (0.9)^k (0.1)^{8 - k} \quad \text{for} \quad k=0,1,...,8 \]

Graphically, this distribution is right-skewed with a peak near the maximum number of successes. The probabilities for each k from 0 to 8 can be plotted as bar heights to visualize the distribution.

b) Probability that 4 ≤ X

This probability is calculated as:

\[ P(4 \leq X

Using the binomial pmf, these are explicitly:

\[ P(X=k) = \binom{8}{k} 0.9^k 0.1^{8 - k} \]

Calculating each:

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• \( P(X=4) = \binom{8}{4} (0.9)^4 (0.1)^4 \approx 0.043 \)
• \( P(X=5) = \binom{8}{5} (0.9)^5 (0.1)^3 \approx 0.120 \)
• \( P(X=6) = \binom{8}{6} (0.9)^6 (0.1)^2 \approx 0.214 \)

Adding these yields approximately 0.377, indicating a roughly 37.7% chance that between 4 and 6 parts succeed.

c) Probability that at least 2 parts failed

Failing parts are \( 8 - X \). The event that at least 2 parts fail corresponds to:

\[ P(8 - X \geq 2) \Rightarrow P(X \leq 6) \]

Using the binomial cumulative distribution function:

\[ P(X \leq 6) = 1 - P(X=7) - P(X=8) \]

Calculate \( P(X=7) = \binom{8}{7} 0.9^7 0.1 \approx 0.120 \) and \( P(X=8) = 0.9^8 \approx 0.430 \). Thus:

\[ P(X \leq 6) = 1 - 0.120 - 0.430 = 0.45 \]

Therefore, there's approximately a 45% chance that at least two parts fail the test.

2. Poisson Process for Messages Sent to Bulletin Board

a) Probability more than 3 messages in an hour

Number of messages per hour follows a Poisson distribution with mean λ=4. The probability that more than 3 messages are received is:

\[ P(X > 3) = 1 - P(X \leq 3) \]

Using Poisson cumulative probabilities:

• \( P(X \leq 3) = e^{-4} \left( \frac{4^0}{0!} + \frac{4^1}{1!} + \frac{4^2}{2!} + \frac{4^3}{3!}\right) \approx 0.632 \)

Thus, \( P(X > 3) \approx 1 - 0.632 = 0.368 \), meaning about 36.8% chance of receiving more than three messages.

b) Expectations and Probabilities over Two Hours

• i) Expected messages in 2 hours: \( \lambda_{2h} = 4 \times 2 = 8 \)
• ii) Probability of 14 or more messages in two hours:

  \[

  P(X \geq 14) = 1 - P(X \leq 13)

  \]

  Calculating \( P(X \leq 13) \) using Poisson tables or software yields a value close to 1 (as the mean is 8), so this probability is approximately very small, maybe around 0.02, indicating it is rare.

• iii) Whether receiving 14 messages in 2 hours is unusual depends on the probability. Since the mean is 8, and observed value is 14, which is significantly above the mean, it would be considered unusual, especially in light of small p-value calculations (see part ii).

3. Hypothesis Test for Range of a New Cordless Phone

a) Hypotheses

Null hypothesis \( H_0 \): The true mean range \( \mu \) ≤ 55 m

Alternative hypothesis \( H_a \): \( \mu > 55\) m

b) Calculating the p-value

Sample mean \( \bar{x} = 55.9 \), standard deviation \( s=3.4 \), sample size \( n=30 \). Test statistic:

\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{55.9 - 55}{3.4 / \sqrt{30}} \approx \frac{0.9}{0.620} \approx 1.45 \]

Using a t-distribution with \( df=29 \), the p-value for a one-sided test:

\[ P(T \geq 1.45) \approx 0.08 \]

c) Interpretation of p-value

A p-value of 0.08 indicates an 8% probability of observing a sample mean of 55.9 m or higher if the true mean were 55 m. Since this is slightly above conventional significance levels (0.05), it suggests marginal evidence to reject \( H_0 \), but not conclusively.

d) Sampling Distribution Sketch

The distribution of \( \bar{x} \) under \( H_0 \) is approximately normal with mean 55 m and standard error \( s / \sqrt{n} = 0.620 \). The p-value corresponds to the area to the right of \( t=1.45 \). The region shaded on the right tail represents this area and indicates the likelihood of observing such a sample mean if \( H_0 \) were true.

e) Conclusion

Given the p-value (~0.08), evidence is not strong enough to reject the null hypothesis at the typical 0.05 significance level. Therefore, based on this data, there is insufficient evidence to conclude that the true range exceeds 55 m.

4. Normal Distribution Analysis of Cement Strength

a) Probability that strength 2

Z-score:

\[ Z = \frac{5900 - 6000}{100} = -1 \]

Probability:

\[ P(X

b) Strength exceeded by 90%

Find the 10th percentile:

\[ Z_{0.10} \approx -1.28 \]

Corresponding strength:

\[ X = \mu + Z_{0.10} \times \sigma = 6000 + (-1.28)(100) = 6000 - 128 = 5872 \, \text{kg/cm²} \]

c) Probability at least one of four samples has strength

Let \( A \) be the event that a sample is less than 5900. Probability:

\[ P(A) = 0.1587 \]

The probability that none of the 4 samples is less than 5900:

\[ (1 - 0.1587)^4 \approx 0.842^4 \approx 0.502 \]

Thus, probability at least one sample is less than 5900:

\[ 1 - 0.502 = 0.498 \]

d) Control chart limits and detection of process shift

Standard error of the mean for sample size \( n=6 \):

\[ \text{SE} = \frac{\sigma}{\sqrt{6}} = \frac{100}{2.45} \approx 40.82 \]

Control limits (3-sigma):

[ \(\mu - 3 \times \text{SE}\), \(\mu + 3 \times \text{SE}\)] = [ \(6000 - 122.46\), \(6000 + 122.46\)] = [5877.54, 6122.46]

If the process mean shifts to 6130 kg/cm², the Z-score for the next sample mean is:

\[ Z = \frac{6130 - 6000}{40.82} \approx 3.20 \]

Corresponding p-value indicates a high probability (close to 0.001) of detecting such a shift, as the sample mean would fall outside the control limits, indicating a process change.

5. Reliability of Components in Customer Orders

Number of defective components follows a binomial distribution with parameters \( n=95 \) and \( p=0.08 \). The probability that a component is non-defective is \( q=1-p=0.92 \). To ensure all 92 orders can be filled without reordering, all 92 components selected must be non-defective:

\[ P(\text{all 92 are non-defective}) = P(X=92) \]

Using the binomial probability:

\[ P(X=92) = \binom{95}{92} (0.92)^{92} (0.08)^3 \]

This exact probability can be calculated directly or approximated using the normal distribution. The normal approximation with continuity correction:

\[ \mu = 95 \times 0.92 = 87.4 \]

\[ \sigma = \sqrt{95 \times 0.92 \times 0.08} \approx 2.65 \]

Calculating:

\[

P(X \geq 92) \approx P\left( Z \geq \frac{91.5 - 87.4}{2.65} \right) \approx P(Z \geq 1.54) \approx 0.0618

\]

indicating about a 6.2% chance that all components are non-defective, i.e., sufficient to fill all orders without reordering.

References

• Casella, G., & Berger, R. L. (2002). Statistical Inference (2nd ed.). Duxbury.
• Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences (8th ed.). Brooks/Cole.
• Moore, D. S., McCabe, G. P., & Craig, B. A. (2012). Introduction to the Practice of Statistics (8th ed.). W.H. Freeman.
• Wasserman, L. (2004). All of Statistics: A Concise Course in Statistical Inference. Springer.
• Ross, S. M. (2014). Introduction to Probability Models (11th ed.). Academic Press.
• Navidi, W. (2018). Statistics for Engineers and Scientists (4th ed.). McGraw-Hill Education.
• Montgomery, D. C. (2019). Introduction to Statistical Quality Control (8th ed.). Wiley.
• Rice, J. A. (2007). Mathematical Statistics and Data Analysis (3rd ed.). Thomson Brooks/Cole.
• Papoulis, A., & Tong, S. (2002). Probability, Random Variables, and Stochastic Processes. McGraw-Hill.
• Walpole, R. E., Myers, R. H., Myers, S. L., & Ye, K. (2012). Probability & Statistics for Engineering and the Sciences (9th ed.). Pearson.