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A machine produces metal rods used in an automobile suspension system. A random sample of 15 rods is selected, and the diameter is measured. The resulting data (in millimeters) are as follows: \(\begin{array}{lllll}8.24 & 8.25 & 8.20 & 8.23 & 8.24 \\ 8.21 & 8.26 & 8.26 & 8.20 & 8.25 \\ 8.23 & 8.23 & 8.19 & 8.28 & 8.24\end{array}\) (a) Check the assumption of normality for rod diameter. (b) Calculate a \(95 \%\) two-sided confidence interval on mean rod diameter. (c) Calculate a \(95 \%\) upper confidence bound on the mean. Compare this bound with the upper bound of the two-sided confidence interval and discuss why they are different.

Short Answer

Expert verified
The data is normally distributed; the 95% confidence interval is [8.224, 8.252], and the upper bound is different due to the tail distribution.

Step by step solution

01

Check Normality Assumption

To assess the normality of the rod diameter dataset, we can use a visual inspection through a Q-Q plot and a formal statistical test like the Shapiro-Wilk test. For this exercise, generate a Q-Q plot. If the points roughly follow a straight line, the dataset is normally distributed. To conduct the Shapiro-Wilk test, compute the test statistic and p-value; if the p-value is greater than 0.05, we assume normality.
02

Calculate Sample Mean and Standard Deviation

First, calculate the sample mean \(\bar{x}\) as the sum of all sample values divided by the number of values: \( \bar{x} = \frac{8.24 + 8.25 + \, ... \, + 8.24}{15} \). This gives \(\bar{x} = 8.238\). Next, compute the sample standard deviation \(s\), using the formula \( s = \sqrt{\frac{1}{n-1} \sum (x_i - \bar{x})^2} \). After calculations, \(s \approx 0.026\).
03

Calculate Two-Sided 95% Confidence Interval

To find the 95% two-sided confidence interval, use the t-distribution since the sample size is small. The degrees of freedom \(df\) is \(n-1 = 14\). Use a t-table or calculator to find the t-value \(t_{0.025, 14}\). The confidence interval is computed as \( \bar{x} \pm t_{0.025, 14} \cdot \frac{s}{\sqrt{n}} \). Calculations yield the interval \( 8.238 \pm 0.014\), resulting in the interval \([8.224, 8.252]\).
04

Calculate 95% Upper Confidence Bound

For the 95% upper confidence bound, we use the t-distribution but only consider the upper part. The formula is \( \bar{x} + t_{0.05, 14} \cdot \frac{s}{\sqrt{n}} \). Compute to get \( 8.238 + 0.011\), resulting in the upper bound \(8.249\).
05

Compare Upper Bound of Two-Sided Interval and Upper Confidence Bound

The upper bound of the two-sided interval is 8.252, while the 95% upper confidence bound is 8.249. The upper confidence bound is lower because it only captures the upper tail with 5% significance, while the two-sided confidence interval splits the tails, capturing 2.5% in each.

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Key Concepts

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

Normality Assumption
In statistical analysis, the **normality assumption** is a key concept that underlies many statistical methods, including confidence intervals and hypothesis testing. This assumption states that the data follows a normal distribution, often visualized as the classic bell curve. Ensuring that your data meets this assumption is critical, as many statistical tests are only valid when this condition is satisfied.
Visual inspection techniques like Q-Q plots are often used, where data points are plotted against a theoretical normal distribution. If the points lie approximately along a straight line, it's an indication that the data is normally distributed.
However, visual inspection can sometimes be subjective. For a more objective test of normality, the Shapiro-Wilk test is often employed. This statistical test provides a test statistic and a p-value, where a p-value greater than 0.05 generally suggests that the data does not significantly deviate from a normal distribution.
Shapiro-Wilk Test
The **Shapiro-Wilk test** is a statistical test that helps in determining whether a dataset follows a normal distribution. It is particularly useful for smaller sample sizes, like the one in our exercise, which involves 15 metal rods.
Here's how it works:
  • First, the test computes a test statistic which ranges between 0 and 1.
  • Next, it provides a p-value, which is key in making a decision about normality.
A p-value greater than 0.05 often supports the assumption of normality, suggesting that any deviation from the normal distribution is statistically insignificant. This means that the data is potentially appropriate for further parametric testing, which depends on normality assumptions.
For example, if you conduct the Shapiro-Wilk test on the rod diameters in this exercise and find a p-value of 0.20, the assumption of normality stands. This result allows us to proceed with methods like the t-distribution for calculating confidence intervals.
t-Distribution
The **t-distribution** is crucial in statistics, especially when dealing with small sample sizes, typically less than 30. It is similar to the normal distribution but is slightly wider or has fatter tails, which accounts for increased variability when estimating a population parameter from a small sample.
Unlike the standard normal distribution, the t-distribution is defined by degrees of freedom (df), calculated as the sample size (n) minus one. For instance, with 15 rods, you have 14 degrees of freedom.
When calculating a confidence interval, the t-distribution helps determine the range in which we expect the population mean to fall. For a two-sided 95% confidence interval, you would find the critical t-value that corresponds to 2.5% in each tail (the remaining 5% is divided equally due to two tails).
  • If you’re calculating only an upper bound, like in the exercise, you focus solely on the upper tail with 5% significance.
Hence, the calculations differ slightly, leading to distinct results for two-sided intervals and single bounds.

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