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Atkins Diet Difference Ten randomly selected people went on a Atkins diet for a month. The weight losses experienced (in pounds) were $$ 3,8,10,0,4,6,6,4,2, \text { and }-2 $$ The negative weight loss is a weight gain. Test the hypothesis that the mean weight loss was more than 0 , using a significance level of \(0.05\). Assume the population distribution is

Short Answer

Expert verified
If the calculated t-value is greater than the critical t-value, we conclude that the mean weight loss was more than 0, using a significance level of 0.05. Otherwise, we conclude that the mean weight loss was not significantly more than 0.

Step by step solution

01

Compute the sample mean and sample standard deviation

First, sum up all the weight losses and divide by the total number of data points (10 in this case) to get the sample mean (\(\overline{x}\)). Use the formula \(s = \sqrt{\frac{\sum (x_i - \overline{x})^2}{n-1}}\) where \(x_i\) are the data points, to compute the sample standard deviation.
02

Calculate the t statistic

Use the formula \(t = \frac{\overline{x} - \mu}{s/\sqrt{n}}\) where \(\mu = 0\) is the suspected population mean, \(s\) is the sample standard deviation, and \(n\) is the number of data points, to compute the t statistic.
03

Determine the critical t-value

Refer to a t-distribution table to find the critical t-value. The degrees of freedom are \(n-1\), and alpha/2 (the desired level of significance) is \(0.05\)/2.
04

Decision rule

If the absolute value of the calculated t statistic is greater than the critical t-value, then reject the null hypothesis in favor of the alternative hypothesis. Otherwise, do not reject the null hypothesis.

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Key Concepts

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

Atkins Diet
The Atkins Diet is a popular eating plan that restricts carbohydrate intake to help individuals lose weight. It is based on the idea that by reducing carbs, the body is forced to burn stored fat for energy, leading to weight loss. This diet typically includes a higher intake of proteins and fats, which are believed to be more satiating. It is divided into phases, starting with a very low carb intake and gradually increasing the quantity allowed as a person approaches their target weight.

While it has been effective for many, it's important to consult with a healthcare provider before beginning any diet regimen, especially those with strict limitations like Atkins. The effectiveness of the Atkins Diet, like any weight loss strategy, can vary based on individual body responses and adherence to the dietary guidelines.
Weight Loss Statistics
Understanding the statistics behind weight loss involves analyzing data to draw meaningful conclusions. In this context, statistical measures such as mean and standard deviation are vital. The mean tells us the average weight loss among the participants, while the standard deviation explains how much variation exists from the average.

To compute these, one adds up all individual weight loss values and divides by the count of participants to get the mean. The standard deviation is calculated using the formula:
\[ s = \sqrt{\frac{\sum (x_i - \overline{x})^2}{n-1}} \]

where \(x_i\) are the observed values, \(\overline{x}\) is the mean, and \(n\) is the number of data points.
  • This statistical measure helps to understand the spread of weight loss data and if the diet effectiveness is consistent across different individuals.
T-Test
A T-Test is a statistical method used to determine if there is a significant difference between the means of two groups. In our case, it helps to test the hypothesis regarding the mean weight loss from the Atkins Diet.

To conduct a T-test, one calculates the t statistic using the formula:
\[ t = \frac{\overline{x} - \mu}{s/\sqrt{n}} \]

where \(\overline{x}\) represents the sample mean, \(\mu\) is the hypothesized population mean (in this exercise, 0), \(s\) is the sample standard deviation, and \(n\) is the number of observations. This helps quantify whether any observed effect can be attributed to chance or if it's statistically significant.
  • If the calculated t-value exceeds a critical value found in t-distribution tables, the null hypothesis can be rejected.
Significance Level
The significance level, often denoted as alpha (\(\alpha\)), is a threshold used in hypothesis testing to determine the probability of rejecting a true null hypothesis. In simpler terms, it is the risk one is willing to take for making a Type I error, which is rejecting the null hypothesis when it is actually true.

In hypothesis testing exercises like this one, a common significance level used is \(0.05\). So, if the probability of the observed data, given that the null hypothesis is true, is less than the significance level, the null hypothesis is rejected.

Setting the significance level dictates the stringency of the test and how confidently one can support the alternative hypothesis. It is crucial because it guides the decision-making process in hypothesis testing by establishing the critical regions where the null hypothesis would be rejected.
  • This value, together with the degrees of freedom, helps to find the critical t-value from a t-distribution table.

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Most popular questions from this chapter

Hamburgers (Example 9) A hamburger chain sells large hamburgers. When we take a sample of 30 hamburgers and weigh them, we find that the mean is \(0.51\) pounds and the standard deviation is \(0.2\) pound. a. State how you would fill in the numbers below to do the calculation. with Minitab. b. Report the confidence interval in a carefully worded sentence. Normal.

GPAs (Example 11) In finding a confidence interval for a random sample of 30 students GPAs, one interval was \((2.60,3.20)\) and the other was \((2.65,3.15)\). a. One of them is a \(95 \%\) interval and one is a \(90 \%\) interval. Which is which, and how do you know? b. If we used a larger sample size \((n=120\) instead of \(n=30\) ). would the \(95 \%\) interval be wider or narrower than the one reported here?

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 \(\sigma\) ).

Babies Weights (Example 2) Some sources report that the weights of full-term newborn babies have a mean of 7 pounds and a standard deviation of \(0.6\) pound and are Normally distributed. a. What is the probability that one newborn baby will have a weight within \(0.6\) pound of the mean-that is, between \(6.4\) and \(7.6\) pounds, or within one standard deviation of the mean? b. What is the probability the average of four babies' weights will be within \(0.6\) pound of the mean; will be between \(6.4\) and \(7.6\) pounds? c. Explain the difference between a and \(\mathrm{b}\).

Carrots The weights of four randomly chosen bags of horse carrots, each bag labeled 20 pounds, were \(20.5,19.8,20.8\), and \(20.0\) pounds. Assume that the distribution of weights is Normal. Find a \(95 \%\) confidence interval for the mean weight of all bags of horse carrots. 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. \(95 \%\) of all sample means based on samples of the same size will be between and ii. I am \(95 \%\) confident that the population mean is between and iii. We are \(95 \%\) confident that the boundaries are and b. Can you reject a population mean of 20 pounds? Explain.

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