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Atkins Diet Difference Ten people went on an 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 Normal.

Short Answer

Expert verified
We fail to reject the null hypothesis since the calculated t-value (1.67) is less than the critical value (1.833), hence there's not enough evidence to conclude that the mean weight loss is greater than 0.

Step by step solution

01

Hypothesis Formulation

Formulate the null hypothesis (H0) and alternative hypothesis (H1). H0: The mean weight loss is equal to 0 (no change). H1: The mean weight loss is greater than 0 (indicating an average weight loss).
02

Data Analysis

Calculate the sample mean (XÌ„) and the standard deviation (S). Given the weight loss data, perform these calculations: XÌ„ = (3 + 8 + 10 + 0 + 4 + 6 + 6 + 4 + 2 - 2)/10 = 41/10 = 4.1 pounds. S = sqrt(((3-4.1)^2 + (8-4.1)^2 + (10-4.1)^2 + (0-4.1)^2 + (4-4.1)^2 + (6-4.1)^2 + (6-4.1)^2 + (4-4.1)^2 + (2-4.1)^2 + (-2-4.1)^2)/9 = sqrt(60.9) = 7.805. Here, the sample size n=10.
03

T-Statistic Calculation

Calculate the t-score using the formula t = (X̄ - μ)/(s/√_n) = (4.1 - 0)/(7.805/sqrt(10)) = 4.1/(7.805/sqrt(10)) = 1.67.
04

Critical Value and Conclusion

Find the critical value for a one-tailed test with a 0.05 level of significance and 9 degrees of freedom (n-1). The critical value obtained from t-distribution tables is around 1.833. Since the calculated t-value (1.67) is less than the critical value (1.833), we fail to reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a fundamental concept in statistical hypothesis testing. It's a statement we make about the population parameter which indicates no effect or no difference, and in this case, it means that the mean weight loss for the dieters on the Atkins diet is zero (no weight loss). In the exercise, we denote it as H0: The mean weight loss is equal to 0. It acts as a baseline assumption that the alternative hypothesis challenges.

When conducting a hypothesis test, we gather sample data and calculate appropriate test statistics to decide whether to reject the null hypothesis. The decision is based on comparing the calculated test statistic to a critical value and seeing which hypothesis the data support more strongly.
Alternative Hypothesis
The alternative hypothesis (H1 or Ha) proposes what we would consider to be true if we find evidence against the null hypothesis. In this scenario, the alternative hypothesis is that the mean weight loss is greater than 0, which would suggest that the Atkins diet did lead to weight loss. Formally, it's written as H1: The mean weight loss is greater than 0.

The alternative hypothesis is not assumed to be true, but rather it is something to be proved through statistical evidence. In the given exercise, we would look for evidence in the weight loss data to support the claim that the Atkins diet has a positive effect on weight loss.
T-Test
The t-test is a statistical test used to compare the mean of a sample to a known value (often the population mean), when the population standard deviation is unknown and the sample size is small. It's useful when we have to work with approximations of the standard deviation from the sample itself. In this context, the t-test analyzes whether the mean weight loss of the dieting group significantly differs from 0.

To conduct a t-test, we calculate a t-statistic, which measures the size of the difference relative to the variation in the sample data. The calculated t-value is then compared to critical values from the t-distribution, which gives us a p-value to assess the significance of our results.
Significance Level
The significance level, often denoted by the Greek letter alpha (α), is a threshold we set to decide whether our test statistic is extreme enough to reject the null hypothesis. It represents the probability of incorrectly rejecting the null hypothesis, known as a Type I error. In this exercise, the significance level is set at 0.05 or 5%, which is a common choice in many statistical tests.

A significance level of 0.05 implies that we would accept a 5% chance of concluding that there is an effect (in this case, a mean weight loss) when there is none. If the p-value calculated from the test statistic is less than α, we reject the null hypothesis, indicating that our results are statistically significant.
Sample Mean
The sample mean, denoted as XÌ„ (pronounced 'x-bar'), is the average value of the sample data. It serves as an estimate for the population mean when the full population data isn't available. In hypothesis testing, the sample mean is used to calculate the test statistic. For the Atkins diet exercise, the sample mean was calculated to be 4.1 pounds. This value was then used in the t-test formula to determine whether the weight loss was significant compared to the null hypothesis assumption of no weight loss.
Standard Deviation
Standard deviation, symbolized as S or SD, is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are more spread out. In the exercise, we calculate the sample standard deviation to estimate the variability of weight loss in our sample, which is necessary for determining the t-statistic and ultimately testing our hypothesis about the Atkins diet's effectiveness.

The formula for calculating the standard deviation involves taking the square root of the variance. It is a critical component of the t-test as it helps to understand the context of the sample mean — how much individual data points deviate from that mean, which affects the reliability of the mean as an estimator of the population parameter.

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Most popular questions from this chapter

Deflated Footballs? Colts In the 2015 AFC Championship game, there was a charge the New England Patriots deflated their footballs for an advantage. The Patriots' opponents during the championship game were the Indianapolis Colts. Measurements of the Colts footballs were taken. The balls should be inflated to between \(12.5\) and \(13.5\) pounds per square inch (psi). The measurements were \(12.70,12.75,12.50,12.55,12.35,12.30,12.95\), and \(12.15 \mathrm{psi}\) (Source: http://online.wsj.com/public/resources/documents/ Deflategate.pdf) a. Test the hypothesis that the population mean is less than \(12.5\) psi using ? significance level of \(0.05\). State clearly whether the Colts' balls are deflated or not. Assume the conditions for a hypothesis test are satisfied. b. Use the data to construct a \(95 \%\) confidence interval for the mean psi for the Colts' footballs. How does this confidence interval support your conclusion in part a?

Ages A study of all the students at a small college showed a mean age of \(20.7\) and a standard deviation of \(2.5\) years. Are these numbers statistics or parameters? Explain. b. Label both numbers with their appropriate symbol (such as \(\bar{x}, \mu, s\), or \(\sigma)\).

Travel Time to School A random sample of \(50 \mathrm{1} 2\) th-grade students was asked how long it took to get to school. The sample mean was \(16.2\) minutes, and the sample standard deviation was \(12.3\) minutes. (Source: AMSTAT Census at School) a. Find a \(95 \%\) confidence interval for the population mean time it takes 12 th-grade students to get to school. b. Would a \(90 \%\) confidence interval based on this sample data be wider or narrower than the \(95 \%\) confidence interval? Explain. Check your answer by constructing a \(90 \%\) confidence interval and comparing this width of the interval with the width of the \(95 \%\) confidence interval you found in part a.

Student Ages The mean age of all 2550 students at a small college is \(22.8\) years with a standard deviation is \(3.2\) years, and the distribution is right- skewed. A random sample of 4 students' ages is obtained, and the mean is \(23.2\) with a standard deviation of \(2.4\) years. a. \(\mu=? \quad \sigma=? \quad \bar{x}=? \quad s=?\) b. Is \(\mu\) a parameter or a statistic? \(c\). Are the conditions for using the CLT fulfilled? What would be the shape of the approximate sampling distribution of many means, cach from a sample of 4 students? Would the shape be right-skewed, Normal, or left-skewed?

Construct two sets of body temperatures (in degrees Fahrenheit, such as \(96.2{ }^{\circ} \mathrm{F}\) ), one for men and one for women, such that the sample means are different but the hypothesis test shows the population means are not different. Each set should have three numbers in it.

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