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The weights of four randomly and independently selected bags of tomatoes labeled 5 pounds were found to be \(5.1\), \(5.0,5.3\), and \(5.1\) pounds. Assume Normality. a. Find a \(95 \%\) confidence interval for the mean weight of all bags of tomatoes. b. Does the interval capture \(5.0\) pounds? Is there enough evidence to reject a mean weight of \(5.0\) pounds?

Short Answer

Expert verified
a. The 95% confidence interval for the mean weight of all bags of tomatoes is approximately (4.9, 5.35) pounds. b. The interval does capture 5.0 pounds. There is not enough evidence to reject a mean weight of 5.0 pounds.

Step by step solution

01

- Calculate the Sample Mean

The sample mean is calculated as the average of the observed weights. This gives, \[\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i \] where \(x_i\) denote the weights of each bag and \(n\) is the number of bags. Plugging the given data gives \[\bar{x} = \frac{5.1+5.0+5.3+5.1}{4} = 5.125\] pounds.
02

- Compute the Sample Standard Deviation

The standard deviation is a measure of the spread of the data around the mean. It is given by \[s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2}\] The sample standard deviation can then be computed as \[s = \sqrt{\frac{(5.1-5.125)^2 + (5.0-5.125)^2 + (5.3-5.125)^2 + (5.1-5.125)^2}{4-1}} = 0.1\] pounds.
03

- Calculate the Confidence Interval

A 95% confidence interval for the mean weight of all bags is given by \[\bar{x} \pm t_{(n-1), (1-\alpha)/2}\frac{s}{\sqrt{n}}\] where \(\alpha=0.05\) and \(t_{(n-1), (1-\alpha)/2}\) is the t-score for a 95% confidence level with \(n-1\) degrees of freedom. Given the degrees of freedom as \(n-1=3\), the t-score is approximately 2.353. This gives the confidence interval as \[5.125 \pm 2.353\frac{0.1}{\sqrt{4}} \approx (4.9, 5.35)\] pounds.
04

- Interpret the Confidence Interval

The 95% confidence interval represents the interval that we are 95% confident that the true population mean falls in. Here the interval is (4.9, 5.35), which means that we are 95% confident that the mean weight of all bags lies within this interval.
05

- Compare with Hypothesized Mean

The hypothesized mean of 5.0 pounds does fall within the confidence interval. This means that there is not enough evidence to reject a mean weight of 5.0 pounds at the 95% confidence level. Thus, the company's claim that the bags weigh 5 pounds on average cannot be denied based on our sample data.

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Key Concepts

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

Sample Mean Calculation
Understanding how to calculate the sample mean is crucial for anyone delving into statistics or data analysis. The sample mean, often represented as \(\bar{x}\), is a way to measure the average value of a set of numbers, which are typically a subset of a larger population. This indicator is foundational as it's used for various other statistical measurements such as variances and confidence intervals.

In the context of the provided exercise, the sample mean represents the average weight of the four bags of tomatoes taken as a sample from all bags. To calculate it, simply add up all of the observed weights and then divide by the number of bags or data points (n). The formula used is \[\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i \].

Calculating the sample mean is straightforward: sum up the values \(5.1, 5.0, 5.3, 5.1\) and divide by \(4\), the number of bags, resulting in a sample mean of \(5.125\) pounds for this specific case. It's a critical piece of information and a starting point for most other statistical analyses performed on the data.
Sample Standard Deviation
The concept of sample standard deviation extends from what the sample mean tells us. While the mean provides a centrality measure, the standard deviation indicates how spread out the data points are from that mean. It's denoted by \(s\) and calculated using a formula that considers the difference between each data point and the sample mean.

The detailed formula for the sample standard deviation is \[s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2}\], where \(n\) is the sample size. The reason we use \(n-1\) rather than \(n\) is because we're dealing with a 'sample' rather than an entire population, hence the term degrees of freedom (one is deducted for the estimation of the mean itself).

For the given tomato bags problem, the data points are their weights. The resulting standard deviation, \(0.1\) pounds, tells us that the weights vary, on average, by \(0.1\) pounds from the sample mean of \(5.125\) pounds. This measure is particularly important when creating confidence intervals to understand the reliability of our sample mean as a reflection of the population mean.
Hypothesis Testing
Hypothesis testing is a method used in statistics to determine the likelihood that a given assumption (hypothesis) about a population parameter is true. It's a framework for making decisions using data, whether making conclusions about a single mean or comparing multiple means. In hypothesis testing, we start with a null hypothesis \(H_0\), which represents a baseline assumption that no effect or difference exists.

The exercise provided touches on hypothesis testing by asking if there's enough evidence to reject the proposed mean weight of \(5.0\) pounds for the bags of tomatoes. Using the calculated confidence interval, we can determine if \(5.0\) pounds is a plausible value for the true mean weight. Since \(5.0\) falls within the interval of \(4.9, 5.35\), we do not have sufficient evidence to reject the null hypothesis. Essentially, the data does not significantly contradict the claim that the bags weigh on average \(5\) pounds.

This concept ties everything together and applies the calculation of the sample mean and sample standard deviation to make inferences about the population. It's important to note that while we do not have evidence to reject the null hypothesis in this case, this does not prove that the null hypothesis is true. Hypothesis testing can only provide evidence against the null hypothesis, not prove its validity.

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Most popular questions from this chapter

State whether each situation has independent or paired (dependent) samples. a. A researcher wants to know whether pulse rates of people go down after brief meditation. She collects the pulse rates of a random sample of people before meditation and then collects their pulse rates after meditation. b. A researcher wants to know whether professors with tenure have fewer posted office hours than professors without tenure do. She observes the number of office hours posted on the doors of tenured and untenured professors.

A random sample of 25 baseball play ers from the 2017 Major League Baseball season was taken and the sample data was used to construct two confidence intervals for the population mean. One interval was \((22.0,42.8)\). The other interval was \((19.9,44.0)\). (Source: mlb.com) a. One interval is a \(95 \%\) interval, and one is a \(90 \%\) interval. Which is which, and how do you know? b. If a larger sample size was used, for example, 40 instead of 25 , how would this affect the width of the intervals? Explain.

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Suppose that 200 statistics students each took a random sample (with replacement) of 50 students at their college and recorded the ages of the students in their sample. Then each student used his or her data to calculate a \(95 \%\) confidence interval for the mean age of all students at the college. How many of the 200 intervals would you expect to capture the true population mean age, and how many would you expect not to capture the true population mean? Explain by showing your calculation.

According to a 2017 report by ComScore .com, the mean time spent on smartphones daily by the American adults is \(2.85\) hours. Assume this is correct and assume the standard deviation is \(1.4\) hours. a. Suppose 150 American adults are randomly surveyed and asked how long they spend on their smartphones daily. The mean of the sample is recorded. Then we repeat this process, taking 1000 surveys of 150 American adults and recording the sample means. What will be the shape of the distribution of these sample means? b. Refer to part (a). What will be the mean and the standard deviation of the distribution of these sample means?

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