/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Lunch breaks are often considere... [FREE SOLUTION] | 91Ó°ÊÓ

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Lunch breaks are often considered too short, and employees frequently develop a habit of "stretching" them. The manager at Giant Mart randomly identified 22 employees and observed the lengths of their lunch breaks (in minutes) for one randomly selected day during the week: $$\begin{array}{lllllllllll}\hline 30 & 24 & 38 & 35 & 27 & 35 & 23 & 28 & 28 & 22 & 26 \\\34 & 29 & 25 & 28 & 34 & 24 & 26 & 28 & 32 & 29 & 40 \\\\\hline\end{array}$$ a. Show evidence that the normality assumptions are satisfied. b. Find the \(95 \%\) confidence interval for "mean length of lunch breaks" at Giant Mart.

Short Answer

Expert verified
The calculation and interpretation of a normality test such as Shapiro-Wilk test, determines if normality assumptions are satisfied. Then, by inserting the calculated values of sample mean, standard deviation into the confidence interval formula, one can find the range for mean length of lunch breaks with 95% confidence.

Step by step solution

01

Testing Normality Assumptions

Normality can be tested using Normal Quantile Plot or statistical tests like Shapiro-Wilk. By using the Shapiro-Wilk test, if p > 0.05, then the null hypothesis of the normality is accepted, which means the distribution can be assumed as normal.
02

Calculating Sample Mean and Standard deviation

Sample mean and standard deviation can be computed using following formulas: Mean, \( \mu = \frac{1}{n}\sum_{}^{}x_i \); Standard deviation, \( \sigma = \sqrt{\frac{1}{n}\sum_{}^{}(x_i - \mu)^2} \)
03

Confidence interval estimates

The confidence interval for the mean length of lunch breaks is calculated using the formula: \( \mu \pm Z(\frac{\alpha}{2}) \times \frac{\sigma} {\sqrt{n}} \), where \( Z(\frac{\alpha}{2}) \) is the z-score at \( 95% \) confidence, \( \sigma \) is the sample standard deviation and \( n \) is the sample size.

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Key Concepts

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

Normality Assumptions
In statistics, testing the normality assumptions is a crucial step when working with data distributions. This means checking whether your data follows a normal distribution pattern. A normal distribution is often depicted as a bell-shaped curve that is symmetric around the mean.
In the exercise, to show evidence that the normality assumptions are satisfied, one can utilize a few methods:
  • Visual inspections, such as a Normal Q-Q plot or histogram.
  • Statistical tests like the Shapiro-Wilk test, which is often preferred due to its usefulness with small sample sizes.
When conducting the Shapiro-Wilk test, if the p-value exceeds 0.05, we do not reject the null hypothesis. This suggests the data could follow a normal distribution. Ensuring data normality is essential as it affects many statistical tests and their outcomes.
Confidence Interval
The concept of a confidence interval involves estimating a range within which we expect the true population parameter, such as a mean, to lie. For instance, when determining the mean time employees spend on lunch breaks, a confidence interval provides a range that we can be certain contains the true mean with a particular level of confidence, typically 95%.
To calculate a confidence interval, we start with the sample mean and then adjust this estimate by adding and subtracting a margin of error. The formula used is:\[ \bar{x} \pm Z(\frac{\alpha}{2}) \times \frac{\sigma}{\sqrt{n}} \]where:
  • \( \bar{x} \) is the sample mean.
  • \( Z(\frac{\alpha}{2}) \) is the z-score associated with the desired confidence level.
  • \( \sigma \) is the standard deviation of the sample.
  • \( n \) is the sample size.
This range provides us with a high degree of certainty about where the true population mean lies based on our sample data.
Sample Mean
The sample mean is a simple yet fundamental concept in statistics. It is the average value of all the data points in a sample and is a crucial component in estimating the population mean. The sample mean is denoted by \( \bar{x} \) and can be calculated using the formula: \[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]
where:
  • \( n \) is the number of data points in the sample.
  • \( x_i \) represents each individual data point.
Calculating the sample mean is often one of the first steps in analyzing a data set. It's used in the creation of the confidence interval and in various other statistical methods. It provides a summary measure of the central tendency of the data, making it indispensable for statistical analysis.
Shapiro-Wilk Test
The Shapiro-Wilk test is a statistical test that helps determine if a sample comes from a normally distributed population. It is particularly useful when you have a small sample size, as was the case in our example with 22 employees.
Here's how it works:
1. **Null Hypothesis**: The data is normally distributed.
2. **Alternative Hypothesis**: The data is not normally distributed.

To perform the test, you calculate a test statistic based on your data, which is then compared to a critical value. If the resulting p-value from the test is greater than 0.05, you accept the null hypothesis that the data is drawn from a normal distribution.
This test is powerful and sensitive, making it one of the most trusted statistical tests for normality assumptions among statisticians.

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Most popular questions from this chapter

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The dry weight of a cork is another quality that does not affect the ability of the cork to seal a bottle, but it is a variable that is monitored regularly. The weights of the no. 9 natural corks \((24 \mathrm{mm}\) in diameter by \(45 \mathrm{mm}\) in length) have a normal distribution. Ten randomly selected corks were weighed to the nearest hundredth of a gram. Dry Weight (in grams) $$\begin{array}{rrrrrrrrr}3.26 & 3.58 & 3.07 & 3.09 & 3.16 & 3.02 & 3.64 & 3.61 & 3.02 & 2.79\end{array}$$ a. Does the preceding sample present sufficient reason to show that the standard deviation of the dry weights is different from 0.3275 gram at the 0.02 level of significance? A different random sample of 20 is taken from the same batch. Dry Weight (in grams) $$\begin{array}{llllllllll}\hline 3.53 & 3.77 & 3.49 & 3.24 & 3.00 & 3.41 & 3.33 & 3.51 & 3.02 & 3.46 \\\2.80 & 3.58 & 3.05 & 3.51 & 3.61 & 2.90 & 3.69 & 3.62 & 3.26 & 3.58 \\\\\hline\end{array}$$ b. Does the preceding sample present sufficient reason to show that the standard deviation of the dry weights is different from 0.3275 gram at the 0.02 level of significance? c. What effect did the two different sample standard deviations have on the calculated test statistic in parts a and b? What effect did they have on the \(p\) -value or critical value? Explain. d. What effect did the two different sample sizes have on the calculated test statistic in parts a and b? What effect did they have on the \(p\) -value or critical value? Explain.

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