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Oranges A statistics instructor randomly selected four bags of oranges, each bag labeled 10 pounds, and weighed the bags. They weighed \(10.2,10.5,10.3\), and \(10.3\) pounds. Assume that the distribution of weights is Normal. Find a \(95 \%\) confidence interval for the mean weight of all bags of oranges. Use technology for your calculations. a. Decide whether each of the following three statements is a correctly worded interpretation of the confidence interval, and fill in the blanks for the correct option(s). i. I am \(95 \%\) confident that the population mean is between ii. There is a \(95 \%\) chance that all intervals will be between iii. I am \(95 \%\) confident that the sample mean is between b. Does the interval capture 10 pounds? Is there enough evidence to reject the null hypothesis that the population mean weight is 10 pounds? Explain your answer.

Short Answer

Expert verified
The 95% confidence interval for the mean weight of all bags of oranges is (10.166, 10.484). The correct interpretation of the confidence interval is 'I am 95% confident that the population mean is between'. The interval does not contain the '10 pounds', hence, there may be enough evidence to reject the null hypothesis that the population mean weight is 10 pounds.

Step by step solution

01

Calculate the Sample Mean

Firstly, calculate the mean of the sample weights. So, add all the weights up and divide by the number of weights. The values are 10.2, 10.5, 10.3, 10.3, and there are 4 weights in total. The mean (represented as \( \overline{X} \)) is therefore (10.2 + 10.5 + 10.3 + 10.3) / 4 = 10.325 pounds.
02

Calculate the Sample Standard Deviation

The Sample Standard Deviation (S) calculates the dispersion of the sample data. Using the sample's weights with the formula for standard deviation in statistics, you will have S = \(\sqrt{\frac{(10.2-10.325)^2 + (10.5-10.325)^2 + (10.3-10.325)^2 + (10.3-10.325)^2}{3}} \approx 0.1 pounds.
03

Calculate the Confidence Interval

The margin of error can be calculated using the formula E = \(t*\frac{S}{\sqrt{n}}\). For a 95% confidence interval, and with a sample size of 4, the t-value is 3.182 (from t-table). Substituting the known values into the margin of error formula, E = 3.182 * (0.1 / \(\sqrt{4}\)) = 0.159. Then calculate the confidence interval by subtracting and adding the margin of error to the mean: \( \overline{X} -E = 10.325 - 0.159 = 10.166\) and \( \overline{X} +E = 10.325 + 0.159 = 10.484\). Thus, the 95% confidence interval for the mean weight of all bags of oranges is \(10.166, 10.484\).
04

Interpret the Confidence Interval and Hypothesis Testing

a. The correct interpretation of the confidence interval is the first statement: 'I am 95% confident that the population mean is between'. This indicates the level of confidence that the calculated interval of weights contains the true population mean of weights. \n b. The confidence interval (10.166, 10.484) does capture 10 pounds. However, as 10 pounds is not within the interval, we may have enough evidence to reject the null hypothesis that the population mean weight is 10 pounds. Because the interval does not contain 10, the weight indicated on the bag label, we may suggest that the weight is not accurately stated.

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Key Concepts

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

Sample Mean
The sample mean is a crucial concept in statistics, used to determine the average of a set of numerical data collected from a sample of a population. When the statistics instructor weighed the four bags of oranges, they obtained weights in pounds: 10.2, 10.5, 10.3, and 10.3. To find the sample mean, add all weights together and divide by the number of observations (in this case, 4). Thus, the sample mean is computed as:\[ \overline{X} = \frac{10.2 + 10.5 + 10.3 + 10.3}{4} = 10.325. \]The sample mean is an estimate of the population mean—the true average weight of all bags of oranges. It serves as a central point in further statistical calculations, like confidence intervals. Understanding and calculating the sample mean ensures you have a reliable starting point for more advanced analyses.
Sample Standard Deviation
Sample standard deviation provides insight into the variation or spread of sample data points. It tells us how much the individual measurements differ from the sample mean. Calculating standard deviation involves finding the square root of the average of squared deviations from the mean. For our orange bags, the calculations go as follows:\[ S = \sqrt{\frac{(10.2-10.325)^2 + (10.5-10.325)^2 + (10.3-10.325)^2 + (10.3-10.325)^2}{3}} \approx 0.1\]Given a sample size of 4, this means we divide by 3 (one less than the number of observations) to account for sample bias. Standard deviation is crucial because it informs us about the reliability of the mean. A smaller standard deviation indicates that the data points are close to the mean, while a larger one suggests more spread. It's a key component in creating confidence intervals, reflecting how much estimates might fluctuate if new samples are taken.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions regarding a claim, based on sample data. In the example with orange weights, the null hypothesis states that the mean weight of the bags is 10 pounds. The alternative hypothesis would claim that the mean is not 10 pounds.Using a 95% confidence interval (\(10.166, 10.484\)), we evaluate whether the null hypothesis can be rejected. Since the claimed mean weight of 10 pounds does not fall within this interval, we have sufficient evidence to reject the null hypothesis. This decision implies that the label might not accurately represent the true mean weight of the bags.Understanding how to conduct hypothesis testing helps in determining the correctness of claims and identifying when data does not support an assumed condition. It’s a vital part of verifying results in many real-world scenarios, including ensuring that product labels match actual content.

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Most popular questions from this chapter

Exam Scores The distribution of the scores on a certain exam is \(N(70,10)\), which means that the exam scores are Normally distributed with a mean of 70 and standard deviation of \(10 .\) a, Sketch the curve and label, on the \(x\) -axis, the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be between 50 and 90 . Shade the region under the Normal curve whose area corresponds to this probability.

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Weight A study of all the employees at an office showed a mean weight of \(60.4\) kilograms and a standard deviation of \(1.5\) kilograms. a. Are these numbers statistics or parameters? Explain. b. Label both numbers with their appropriate symbol (such as \(\bar{x}, \mu, s\), or \(\sigma\) ).

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