Sample Solution

1. (a) State whether the following statements are True or False. Give reason in support of your answer:

Justification:

1. Small Subgroup Sizes: The $R$$R$RR$R$ – chart is most effective and commonly used when the subgroup size ($n$$n$nn$n$) is small, typically between 2 and 10. This is because the range is a simpler and more efficient calculation when dealing with smaller subgroup sizes.
2. Large Subgroup Sizes: When subgroup sizes are larger than 10, the $R$$R$RR$R$ – chart may not be the most appropriate choice for evaluating within-group variability due to the following reasons:
   • Sensitivity: The range is less sensitive to the spread of data as the subgroup size increases. It only considers the largest and smallest values and ignores the distribution of all the values in between.
   • Normality Assumption: The distribution of the range is highly dependent on the underlying distribution of the process data, and this dependency increases with subgroup size. For larger subgroups, if the data is not normally distributed, the control limits calculated for the $R$$R$RR$R$ – chart may not be accurate or reliable.
   • Alternative Charts: The $S$$S$SS$S$ – chart (standard deviation chart) is often recommended for larger subgroup sizes because it considers all data points in the subgroup, providing a more accurate and reliable measure of dispersion, especially when $n>10$$n>10$n > 10n > 10$n>10$.

Example:

• The range ($R$$R$RR$R$) would be calculated as the difference between the largest and smallest values: $R=13-5=8$$R=13-5=8$R=13-5=8R = 13 – 5 = 8$R=13-5=8$.
• However, this doesn’t give any insight into the variability of all the other values in the subgroup.
• If there were outliers or shifts within the subgroup, the $R$$R$RR$R$ – chart might not detect them effectively due to its insensitivity to the distribution of internal values.

Conclusion:

Justification:

1. Acceptance Number: The acceptance number in a single sampling plan is the maximum number of defective items allowed in the sample for the lot to still be accepted.
2. OC Curve Characteristics:
   • The x-axis of the OC curve represents the quality level of the lot (often the fraction or percentage nonconforming).
   • The y-axis represents the probability of accepting the lot at each quality level.

Explanation:

• Increasing the Acceptance Number: If we increase the acceptance number in a single sampling plan:
  • The probability of accepting a lot (even with a higher defect rate) increases because we are allowing more defective items in the sample before rejecting the lot.
  • Therefore, the OC curve becomes less steep and shifts upwards, indicating a higher probability of accepting lots even as their quality decreases.

Example:

• Plan A: Sample size = 50, Acceptance number = 1 (accept the lot if there is 1 or fewer defective item in the sample).
• Plan B: Sample size = 50, Acceptance number = 5 (accept the lot if there are 5 or fewer defective items in the sample).

• Plan A will have a steeper OC curve because the probability of accepting a lot decreases more rapidly as the quality level decreases (due to the lower acceptance number).
• Plan B will have a less steep OC curve because it is more lenient (due to the higher acceptance number), and the probability of accepting a lot decreases more slowly as the quality level decreases.

Conclusion:

Justification:

1. Additive Model: Assumes that the components of the time series (trend, seasonality, and residual) are added together.
   $$Y_t=T_t+S_t+R_t$$$$Y_t=T_t+S_t+R_t$$Y_(t)=T_(t)+S_(t)+R_(t)Y_t = T_t + S_t + R_t$$Y_t=T_t+S_t+R_t$$
   Where:
   • $Y_t$$Y_t$Y_(t)Y_t$Y_t$ is the observed data at time $t$$t$tt$t$,
   • $T_t$$T_t$T_(t)T_t$T_t$ is the trend component,
   • $S_t$$S_t$S_(t)S_t$S_t$ is the seasonal component, and
   • $R_t$$R_t$R_(t)R_t$R_t$ is the residual or error component.
     The additive model is typically used when the magnitude of the seasonal fluctuations does not depend on the level of the time series (i.e., the fluctuations are roughly constant over time).
2. Multiplicative Model: Assumes that the components of the time series are multiplied together.
   $$Y_t=T_t\times S_t\times R_t$$$$Y_t=T_t\times S_t\times R_t$$Y_(t)=T_(t)xxS_(t)xxR_(t)Y_t = T_t \times S_t \times R_t$$Y_t=T_t\times S_t\times R_t$$
   The multiplicative model is typically used when the magnitude of the seasonal fluctuations depends on the level of the time series (i.e., the fluctuations increase or decrease as the data values increase or decrease).

Explanation:

• Non-Constant Seasonal Effect: If the seasonal effect is not constant and varies with the level of the time series (e.g., the increase in sales during summer is proportionally larger when overall sales levels are higher), this implies a multiplicative relationship between the components.

Example:

• Scenario 1: Every summer, AC sales increase by approximately 1000 units, regardless of the overall sales level.
• Scenario 2: Every summer, AC sales double, regardless of the overall sales level.

Conclusion:

Justification:

1. Autocorrelation: Autocorrelation, also known as serial correlation, measures the linear relationship between lagged values of a time series. It does not account for the correlations at all intervening lags. In other words, it does not control for the potential influence of other time lags.
2. Partial Autocorrelation: Partial autocorrelation, on the other hand, measures the linear relationship between the observation at time $t$$t$tt$t$ and the observation at time $t-k$$t-k$t-kt-k$t-k$, controlling for the observations at all intervening time points. It essentially isolates the correlation at lag $k$$k$kk$k$ by removing the effect of correlations at shorter lags.

Explanation:

• If a researcher wants to understand the relationship between today’s unemployment rate and that of 5 years ago, without considering the intervening years, the partial autocorrelation function (PACF) would indeed be the more appropriate tool.
• The PACF will show the direct effect (correlation) of the unemployment rate 5 years ago on today’s rate, after removing the effects of the unemployment rate in the intervening 4 years. This allows the researcher to understand the isolated relationship between the unemployment rate at these two points in time, without the influence of the years in between.
• In contrast, the autocorrelation function (ACF) would show the total effect (correlation) of the unemployment rate 5 years ago on today’s rate, without controlling for the intervening years. This means that the ACF could be capturing indirect effects that are not of interest to the researcher in this scenario.

Example:

• Suppose we have a time series data of unemployment rates for several years.
• If we calculate the autocorrelation at lag 5, it will give us the correlation between the unemployment rate today and that of 5 years ago, but it will not control for the unemployment rates in the intervening years.
• If we calculate the partial autocorrelation at lag 5, it will give us the correlation between the unemployment rate today and that of 5 years ago, controlling for (or removing the effects of) the unemployment rates in the intervening years.

Conclusion:

Justification:

1. Parallel System Reliability: In a parallel configuration, the system will function if at least one of the components functions. The reliability of a parallel system, $R_s$$R_s$R_(s)R_s$R_s$, is calculated as:
   $$R_s=1-\prod _{i=1}^n(1-R_i)$$$$R_s=1-\prod _{i=1}^n (1-R_i)$$R_(s)=1-prod_(i=1)^(n)(1-R_(i))R_s = 1 – \prod_{i=1}^{n}(1 – R_i)$$R_s=1-\prod _{i=1}^n(1-R_i)$$
   where $R_i$$R_i$R_(i)R_i$R_i$ is the reliability of the $i^{th}$$i^{th}$i^(th)i^{th}$i^{th}$ component and $n$$n$nn$n$ is the number of components.
2. Improving Reliability: To improve the reliability of a parallel system most effectively, it is generally best to improve the component with the lowest reliability because it has the most room for improvement and will contribute the most to increasing the overall system reliability when improved.

Explanation:

• The current reliability of the system, using the formula for parallel reliability, is:
  $$R_s=1-(1-R_1)(1-R_2)(1-R_3)(1-R_4)$$$$R_s=1-(1-R_1)(1-R_2)(1-R_3)(1-R_4)$$R_(s)=1-(1-R_(1))(1-R_(2))(1-R_(3))(1-R_(4))R_s = 1 – (1 – R_1)(1 – R_2)(1 – R_3)(1 – R_4)$$R_s=1-(1-R_1)(1-R_2)(1-R_3)(1-R_4)$$
  Substituting the given reliabilities:
  $$R_s=1-(1-0.2)(1-0.4)(1-0.5)(1-0.8)=0.952$$$$R_s=1-(1-0.2)(1-0.4)(1-0.5)(1-0.8)=0.952$$R_(s)=1-(1-0.2)(1-0.4)(1-0.5)(1-0.8)=0.952R_s = 1 – (1 – 0.2)(1 – 0.4)(1 – 0.5)(1 – 0.8) =0.952$$R_s=1-(1-0.2)(1-0.4)(1-0.5)(1-0.8)=0.952$$
• If we improve the reliability of Component 1 (the component with reliability $0.2$$0.2$0.20.2$0.2$), it will have a larger impact on the overall system reliability than improving a component with higher reliability, because the unreliability $(1-R_1)$$(1-R_1)$(1-R_(1))(1 – R_1)$(1-R_1)$ of Component 1 is currently the largest among the components, and reducing it will have a larger impact on reducing the overall unreliability of the system.

Example:

Conclusion: