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The weekly weight losses of all dieters on Diet I have a normal distribution with a mean of \(1.3\) pounds and a standard deviation of \(.4\) pound. The weekly weight losses of all dieters on Diet II have a normal distribution with a mean of \(1.5\) pounds and a standard deviation of \(.7\) pound. A random sample of 25 dieters on Diet I and another sample of 36 dieters on Diet II are observed. a. What is the probability that the difference between the two sample means, \(\bar{x}_{1}-\bar{x}_{2}\), will be within \(-.15\) to \(.15\), that is, \(-.15<\bar{x}_{1}-\bar{x}_{2}<.15 ?\) b. What is the probability that the average weight loss \(\bar{x}_{1}\) for dieters on Diet I will be greater than the average weight loss \(\bar{x}_{2}\) for dieters on Diet II? c. If the average weight loss of the 25 dieters using Diet I is computed to be \(2.0\) pounds, what is the probability that the difference between the two sample means, \(\bar{x}_{1}-\bar{x}_{2}\), will be within \(-.15\) to \(.15\), that is, \(-.15<\bar{x}_{1}-\bar{x}_{2}<.15 ?\) d. Suppose you conclude that the assumption \(-.15<\mu_{1}-\mu_{2}<.15\) is reasonable. What does this mean to a person who chooses one of these diets?

Step by step solution

01

Understand the given

We begin by understanding the information given. We are given two sets of data, representing the weekly weight losses of dieters on two different diets. Diet I has a mean of 1.3 pounds with a standard deviation of .4 while Diet II has a mean of 1.5 pounds with a standard deviation of .7. We have samples of sizes 25 and 36 respectively.

02

Find Difference and Standard Deviation of Difference

Remember that the difference in means for two normally distributed populations is also normally distributed. That difference can be calculated by \(\mu_{d}=\mu_{1}-\mu_{2}\). The standard deviation of the difference \(\sigma_{d}\) is given by \(\sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}}\). Using the provided data, the difference in means will be -0.2 and the standard deviation of the difference is \(\sqrt{\frac{.4^{2}}{25} + \frac{.7^{2}}{36}}=0.1516\)

03

Determine the Z-Scores

To find out the probabilities, we need to first determine the Z scores which is obtained by \(Z_{score}=\frac{X-\mu_{d}}{\sigma_{d}}\). The values -0.15 and 0.15 will convert to a Z scores of -0.33 to 0.33.

04

Find Probability of Difference in Means being Between -0.15 and 0.15

Use the standard normal table or calculator to find the probabilities associated with those Z scores. From a standard Z-table, we find that the probability of the difference in the sample means being between -0.15 and 0.15 is \(P(-0.33 \leq Z \leq 0.33 )=0.63.\)

05

Find Probability of Average Weight Loss of Diet I being Greater than Diet II

In this step we are looking for the probability that the average weight loss for dieters on Diet I will be greater than Diet II. This means \(\bar{x}_{1}-\bar{x}_{2} > 0\). Converting 0 to Z score we get \(Z= \frac{0-(-0.2)}{0.1516}=1.32\). From Z table lookup, we have \(P(Z>1.32)=0.0934\). Hence there's a 9.34% chance that Diet I will result in greater average weight loss than Diet II.

06

Calculate Probability Given Average Weight Loss for Diet I

We're given a new average weight loss for Diet I of 2.0 pounds. This will change the difference in means to 0.5 pounds. We can calculate the new Z scores for the range of -0.15 to 0.15 becoming 2.20 to 2.31. Looking up in the Z table we see that the probability is approximately negligible.

07

Interpretation of .15 difference assumption

If the assumption \(-.15<\mu_{1}-\mu_{2}<.15\) is reasonable, it means that there's no significant difference between the two diets. In other words, the person choosing between these diets would expect to lose about the same amount of weight on either one.

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Key Concepts

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

Understanding Normal Distribution

Normal distribution is a fundamental concept in statistics. It represents how data is spread out across different values. This bell-shaped curve shows that most of the data points cluster around a central region called the mean.

• In a normal distribution, the mean, median, and mode are all equal.
• The spread of the data is determined by the standard deviation.
• About 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.

In our exercise, the weight losses from Diet I and Diet II are normally distributed with different means and standard deviations. This shows that while most people on each diet have a weight loss close to the diet’s average, some will lose more or less according to the distribution spread.
This understanding helps when comparing the results of different diets using statistical methods like hypothesis testing.

Essentials of Probability Calculations

Calculating probability helps in making informed predictions about future events based on known data. In statistics, it is often used to determine the likelihood of a specific outcome occurring.

• Probability values range between 0 and 1, where 0 means impossible and 1 means certain.
• Probability distribution shows all the possible outcomes and their associated probabilities.

In the context of our exercise, we calculate the probability that the difference between the mean weight losses of two groups falls within a specific range (e.g., \(-0.15 < \bar{x}_{1} - \bar{x}_{2} < 0.15\)).
This involves using the standard deviation of the difference and converting the range into Z-scores, which we can then use to find probabilities from a standard normal distribution table.
Understanding probability calculations allows us to assess how likely certain outcomes are and make data-driven decisions in hypothesis testing.

Exploring Z-Scores

Z-scores are vital for understanding how far and in what direction a data point is from the mean. It is a way of standardizing scores on different scales.

• A Z-score tells us how many standard deviations a data point is from the mean.
• Z-scores are calculated using the formula: \( Z = \frac{X - \mu}{\sigma} \) , where \(X\) is a value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

In our problem, we use Z-scores to find the probability that the sample means will fall within a specific range. By converting our range of interest into Z-scores, we can look up these values in a Z-table to find probabilities.
Using Z-scores provides a clear picture of positioning within the distribution, making it easier to compare different data sets or assess the significance of results in hypothesis testing.