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The standard error is a measure used to quantify the amount of variation or dispersion of a set of sample means. It essentially tells us how much the sample mean is expected to vary if you were to repeat the study multiple times. The standard error is calculated using the population standard deviation and the sample size.

For our problem at Camford University, we know:

• Population standard deviation (\( \sigma \)
• Sample size (\( n\)

The formula for the standard error of the sample mean (\( SEM \)) is:

• \[ SEM = \frac{\sigma}{\sqrt{n}} \]

In our example, the population standard deviation is 2.2, and the sample size is 121. Plugging these into the formula gives us:

• \[ SEM = \frac{2.2}{\sqrt{121}} = \frac{2.2}{11} = 0.2 \]

This means that the sample means will vary, on average, by 0.2 hours from the actual population mean.

Z-score is a measure of how many standard deviations an element is from the mean. It's used in statistics to determine how far a specific point is from the average. This is essential for calculating probabilities in a normal distribution.

The Z-score can be calculated using the formula:

• \[ Z = \frac{X - \mu}{SEM} \]

Where:

• \( X \) is the value we are examining
• \( \mu \) is the population mean
• \( SEM \) is the standard error of the mean

For example, when examining a sample mean of 5 or 6 hours, we subtract the population mean (5.5) and divide by the standard error (0.2). This calculation provides the Z-scores of -2.5 and 2.5 respectively. These Z-scores are then used to find probabilities using the standard normal distribution table.

Probability estimation involves calculating the chance of an event occurring within a specified range in a normal distribution. After we determine the Z-scores for our values of interest, the next step is to use them to find probabilities.

For example, the probability of finding a sample mean between 5 and 6 hours involves determining how likely it is for the sample mean to fall within this range. We translated these values into Z-scores and used the standard normal distribution table:

• \( Z = -2.5 \) translates to\( P(Z < -2.5) = 0.0062 \)
• \( Z = 2.5 \) translates to\( P(Z < 2.5) = 0.9938 \)
• The probability of the sample mean falling between 5 and 6 hours is then:\[ P(5 < X < 6) = 0.9938 - 0.0062 = 0.9876 \]

This means that there is a 98.76% chance that a sample mean between 5 and 6 hours will occur.

Sample mean analysis is crucial for understanding how sample data reflects the larger population. With a known population mean and standard deviation, one can predict how sample means tend to behave.

In the context of our exercise, we compare sample means to the population mean of 5.5 hours. Through various calculations, such as estimating probabilities or checking how unusual certain sample means are, we can evaluate if outcomes are typical or rare.

For instance, when assessing how unusual a sample mean greater than 6.5 hours is, we found it corresponds to a Z-score of 5. This Z-score indicates an extremely rare event, as the standard normal distribution shows a probability close to zero for such high Z-scores.

Overall, understanding these aspects allows Healthy Lifestyles Incorporated to make informed predictions and decisions based on their data.