91Ó°ÊÓ

The uniform length of nails is very important to a carpenter-the length of the nails being used are matched to the materials being fastened together, thereby making a small standard deviation an important property of the nails. A sample of 35 randomly selected 2-inch nails is taken from a large quantity of Nails, Inc.'s, recent production run. The resulting length measurements have a mean length of 2.025 inches and a standard deviation of 0.048 inch. a. Determine whether an assumption of normality is reasonable. Explain. b. Is the sample evidence sufficient to reject the idea that the nails have a mean length of 2 inches? Use \(\alpha=0.05\). c. Is there sufficient evidence, at the 0.05 level, to show that the length of nails from this production run has a standard deviation greater than the advertised 0.040 inch? d. Write a short report outlining the findings and recommendations as to whether or not the carpenter should use these nails for an application that requires 2 -inch nails.

Step by step solution

01

Assumption of Normality

In this step, it's important to note that the Central Limit Theorem states that if the sample size is large enough (generally n >= 30 is accepted), then the distribution of the sample can be assumed to be normal. Since the sample size is 35, which is greater than 30, an assumption of normality is reasonable.

02

Test the hypothesis: Mean Length

To test whether or not the mean length is 2 inches, a Z-test can be used because both the sample size is large and the standard deviation is known. First, set up the null hypothesis: \( H_0: \mu = 2 \) and the alternative hypothesis \( H_1: \mu ≠ 2 \). The Z statistic can then be calculated with the formula: \( Z = \frac{{\bar{X} - \mu}}{{\sigma/\sqrt{n}}} \), where \( \bar{X} = 2.025 \) is the sample mean, \( \mu = 2 \) is the population mean, \( \sigma = 0.048 \) is the standard deviation, and \( n = 35 \) is the sample size.

03

Reject or Fail to Reject the Null Hypothesis

Calculate the p-value from the Z statistic. If the p-value is less than the significance level \( \alpha = 0.05 \), reject the null hypothesis. This means there is enough evidence to claim that the mean length of the nails is different from 2 inches.

04

Chi-square test: Standard Deviation

To determine if the standard deviation is greater than 0.040, a chi-square test can be performed. Form the null hypothesis \( H_0: \sigma = 0.040 \) and the alternative hypothesis \( H_1: \sigma > 0.040 \). The chi-square statistic ( \( \chi^2 \) ) can be found using the formula: \( \chi^2 = \frac{(n - 1)s^2}{\sigma^2} \), where \( n = 35 \) is sample size, \( s = 0.048 \) is the sample standard deviation, and \( \sigma = 0.040 \) is the hypothesized standard deviation. A p-value can then be found. If the p-value < 0.05, reject the null hypothesis, providing evidence that the standard deviation is greater than 0.040.

05

Conclusion Report

Write a report summarizing the above findings. The report should discuss if there is evidence to suggest that the nail lengths are normally distributed, if the mean length is significantly different from 2 inches, and whether the standard deviation is significantly greater than 0.040 inches. Based on these findings, provide a recommendation for the carpenter.

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Key Concepts

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

Normal Distribution

In statistics, the normal distribution is a crucial concept. It represents a perfectly symmetric distribution where most observations cluster around the mean, creating the well-known bell curve. Understanding this distribution is critical, especially when applying hypothesis tests.
A practical feature of the normal distribution is that it's defined by its mean and standard deviation. The Central Limit Theorem (CLT) backs up the assumption of normality. It states that the sampling distribution of the sample mean will approximate a normal distribution as the sample size grows, typically n ≥ 30 allows for this assumption.
In the context of the nail length exercise, given a sample size of 35, we can reasonably assume the distribution of nail lengths is normal. This assumption helps streamline further statistical evaluations.

Standard Deviation

Standard deviation measures the amount of variation or dispersion in a set of values. It's a vital statistical tool to understand how spread out the data points are around the mean.
The formula for standard deviation is derived from the variance, where variance is the average of the squared differences from the Mean. When we take the square root of the variance, we get the standard deviation. A smaller standard deviation indicates that the data points are close to the mean, which was a desirable property for the nails in the exercise.
In the nail length scenario, the given standard deviation is 0.048 inches. This statistic becomes part of the hypothesis testing process to determine if it aligns with the advertised 0.040 inches.

Z-Test

The Z-test is a statistical test used to determine if there is a significant difference between the sample mean and the known population mean. It's applicable when the sample size is large and the population variance is known.
The Z statistic is calculated using the formula: \[ Z = \frac{{\bar{X} - \mu}}{{\sigma/\sqrt{n}}} \]
where

• \( \bar{X} \) is the sample mean,
• \( \mu \) is the population mean,
• \( \sigma \) is the standard deviation, and
• \( n \) is the sample size.

If the calculated p-value is less than the chosen significance level (commonly 0.05), we reject the null hypothesis. In the case of the nails, the Z-test helps us understand whether the average length of the nails significantly deviates from 2 inches.

Chi-Square Test

The chi-square test is used for testing the variability within a set of data. Specifically, it checks if the observed variance is significantly different from a known value, making it suitable for the exercise concerning standard deviation differences.
It uses the formula for the chi-square statistic:\[ \chi^2 = \frac{(n - 1)s^2}{\sigma^2} \]
where

• \( n \) is the sample size,
• \( s \) is the sample standard deviation, and
• \( \sigma \) represents the value of the standard deviation under the null hypothesis.

With a significance level of 0.05, if the p-value derived from the chi-square statistic is below this threshold, it implies sufficient evidence to reject the null hypothesis. For the nail length exercise, this test aids in establishing whether the standard deviation of nail lengths exceeds the advertised 0.040 inches.