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Eight obese persons were placed on a diet for 1 month, and their weights, at the beginning and at the end of the month, were recorded: $$ \begin{array}{ccc} & {\text { Weights }} \\ & & \\ \text { Subjects } & \text { Initial } & \text { Final } \\ \hline 1 & 310 & 263 \\ 2 & 295 & 251 \\ 3 & 287 & 249 \\ 4 & 305 & 259 \\ 5 & 270 & 233 \\ 6 & 323 & 267 \\ 7 & 277 & 242 \\ 8 & 299 & 265 \end{array} $$ Estimate the mean weight loss for obese persons when placed on the diet for a 1 -month period. Use a \(95 \%\) confidence interval and interpret your results. What assumptions must you make so that your inference is valid?

Short Answer

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Question: Estimate the mean weight loss for obese persons when placed on the diet for 1 month with a 95% confidence interval, given the following data on initial weights and final weights after one month: (310, 263), (295, 251), (287, 249), (305, 259), (270, 233), (323, 267), (277, 242), and (299, 265). Answer: We can estimate with 95% confidence that the mean weight loss for obese persons when placed on the diet for a 1-month period is between 35.07 to 47.18 pounds. To make our inference valid, we must assume the sample is representative, the weight losses are normally distributed, and the weights are independent observations.

Step by step solution

01

Calculate weight loss for each individual

Subtract the final weight from the initial weight for each person to get the weight loss: $$ \begin{array}{ccc} \text { Subjects } & \text { Weight Loss } \\ \hline 1 & 310 - 263 = 47 \\ 2 & 295 - 251 = 44 \\ 3 & 287 - 249 = 38 \\ 4 & 305 - 259 = 46 \\ 5 & 270 - 233 = 37 \\ 6 & 323 - 267 = 56 \\ 7 & 277 - 242 = 35 \\ 8 & 299 - 265 = 34 \\ \end{array} $$
02

Calculate the sample mean and sample standard deviation.

Add the weight losses together and divide by the number of subjects in the sample (8) to find the sample mean (\(\overline{x}\)) and then use the formula for the unbiased sample standard deviation (s): $$ \overline{x} = \frac{47 + 44 + 38 + 46 + 37 + 56 + 35 + 34}{8} = 41.125 \\ s = \sqrt{\frac{\Sigma (x_i -\overline{x})^2}{(n-1)}} = \sqrt{\frac{(47-41.125)^2 + (44-41.125)^2 + (38-41.125)^2 + (46-41.125)^2 + (37-41.125)^2 + (56-41.125)^2 + (35-41.125)^2 + (34-41.125)^2}{(8 - 1)}} \\ s \approx 7.11 $$
03

Calculate the 95% confidence interval.

To find the 95% confidence interval, use the t-distribution since the population standard deviation is unknown. In this case, the sample size (n) is 8. So, the degrees of freedom is 7. Using the t-distribution with 7 degrees of freedom, the critical t-value (or t-star) for a two-tailed test at \(\alpha = 0.05\) is 2.365. The margin of error (ME) calculation is: $$ ME = t* \frac{s}{\sqrt{n}} = 2.365 \frac{7.11}{\sqrt{8}} \approx 6.05 $$ The confidence interval (CI) is given by the formula: $$ \text{CI} = \overline{x} \pm ME = 41.125 \pm 6.05 $$ So, the 95% confidence interval is approximately (35.07, 47.18).
04

Interpretation and Assumptions

We can interpret our results as follows: We are 95% confident that the mean weight loss for obese persons when placed on the diet for a 1-month period is between 35.07 to 47.18 pounds. To make our inference valid, we must assume the sample is representative and that the weight losses are normally distributed. Additionally, we assume that the weights are independent observations.

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Key Concepts

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

Weight Loss Study
In this weight loss study, eight participants followed a specific diet for one month, and both their initial and final weights were recorded. The change in their weights, calculated as the difference between the initial and final values, helps us understand the diet's effectiveness. Weight loss is not only about healthier lifestyles but also involves statistical analysis to provide proof of its effectiveness. Here, the focus is on estimating the average weight loss for all obese individuals subjected to the diet. This estimate will help to identify whether the program is working as intended and how much weight participants can expect to lose on average. Clarifying this requires not just the raw data but a deeper analysis including statistical concepts.
Sample Standard Deviation
The `sample standard deviation` (denoted as \(s\)) measures the amount of variability or dispersion in the sample of weight losses recorded. It provides insights into how much the weight losses of the individuals in the study differ from the mean weight loss. A smaller standard deviation would suggest more consistent results across the group, while a larger one would indicate more variability.For this study, we calculate \(s\) using the formula:\[s = \sqrt{\frac{\Sigma (x_i - \overline{x})^2}{(n-1)}}\]Where \(\Sigma (x_i - \overline{x})^2\) represents the sum of the square of differences between each weight loss and the mean weight loss, and \(n-1\) is used for the sample size adjustment. This calculation is crucial as it directly influences the width of our confidence interval later on.
T-Distribution
The `t-distribution` is a vital statistical tool when dealing with small sample sizes, as in this study with eight participants. We use the t-distribution because the population standard deviation is unknown, and instead, we use the sample standard deviation. The t-distribution adjusts for this uncertainty, providing a more accurate reflection of expected variability when the sample size is small. It's more spread out than the normal distribution, accommodating the extra uncertainty. Practically, we refer to it to determine the critical t-value (or t-star), which is essential for constructing the confidence interval. In this study, with 7 degrees of freedom (n-1), the critical t-value for a 95% confidence interval is approximately 2.365. This value is combined with the sample standard deviation to compute the margin of error, which is then used to calculate the confidence interval for the average weight loss.
Statistical Inference
`Statistical inference` involves making conclusions about a population based on a sample. In this exercise, we infer the average weight loss for obese individuals on a specific diet program.The basis of this inference is the confidence interval, which estimates the range within which the true mean weight loss lies with a given level of confidence, in this case, 95%. The confidence interval is given by:\[\text{CI} = \overline{x} \pm ME\]where ME (margin of error) is calculated using the critical t-value and sample standard deviation.However, this inference comes with assumptions:
  • The sample should be representative of the entire population of obese individuals.
  • The distribution of weight loss should be approximately normal.
  • Each observation should be independent.
By adhering to these assumptions, the results will be more likely to reflect the true characteristics of the larger group.

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Most popular questions from this chapter

A paired-difference experiment was conducted using \(n=10\) pairs of observations. a. Test the null hypothesis \(H_{0}:\left(\mu_{1}-\mu_{2}\right)=0\) against \(H_{\mathrm{a}}:\left(\mu_{1}-\mu_{2}\right) \neq 0\) for \(\alpha=.05, \bar{d}=.3,\) and \(s_{d}^{2}=\) .16. Give the approximate \(p\) -value for the test. b. Find a \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right)\). c. How many pairs of observations do you need if you want to estimate \(\left(\mu_{1}-\mu_{2}\right)\) correct to within .1 with probability equal to \(.95 ?\)

Find the critical value(s) of \(t\) that specify the rejection region in these situations: a. A two-tailed test with \(\alpha=.01\) and \(12 d f\) b. A right-tailed test with \(\alpha=.05\) and \(16 d f\) c. A two-tailed test with \(\alpha=.05\) and \(25 d f\) d. A left-tailed test with \(\alpha=.01\) and \(7 d f\)

The EPA limit on the allowable discharge of suspended solids into rivers and streams is 60 milligrams per liter ( \(\mathrm{mg} /\) l) per day. A study of water samples selected from the discharge at a phosphate mine shows that over a long period, the mean daily discharge of suspended solids is \(48 \mathrm{mg} /\), but day-to-day discharge readings are variable. State inspectors measured the discharge rates of suspended solids for \(n=20\) days and found \(s^{2}=39(\mathrm{mg} / \mathrm{I})^{2}\). Find a \(90 \%\) confidence interval for \(\sigma^{2}\). Interpret your results.

A chemical manufacturer claims that the purity of his product never varies by more than \(2 \% .\) Five batches were tested and given purity readings of \(98.2,97.1,98.9,97.7,\) and \(97.9 \%\) a. Do the data provide sufficient evidence to contradict the manufacturer's claim? (HINT: To be generous, let a range of \(2 \%\) equal \(4 \sigma .)\) b. Find a \(90 \%\) confidence interval for \(\sigma^{2}\).

An experiment was conducted to compare the densities (in ounces per cubic inch) of cakes prepared from two different cake mixes. Six cake pans were filled with batter \(\mathrm{A}\), and six were filled with batter B. Expecting a variation in oven temperature, the experimenter placed a pan filled with batter \(\mathrm{A}\) and another with batter \(\mathrm{B}\) side by side at six different locations in the oven. The six paired observations of densities are as follows: $$ \begin{array}{lrrrrrr} \text { Location } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Batter A } & .135 & .102 & .098 & .141 & .131 & .144 \\ \hline \text { Batter B } & .129 & .120 & .112 & .152 & .135 & .163 \end{array} $$ a. Do the data present sufficient evidence to indicate a difference between the average densities of cakes prepared using the two types of batter? b. Construct a \(95 \%\) confidence interval for the difference between the average densities for the two mixes.

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