Create standardized scores for all scale variables (price through alcohol). Which beverages have positive standardized scores on every variable? What does this mean?

Scale lacking.

The Z score is the raw score in terms standard deviations (SD). It is one type of standard score, cutting off the same percentages for any normal distribution.

Z = (score-mean)/SD

Find the mean first = sum of scores/number of scores

Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.

Standard deviation = square root of variance

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to any Z score.

To create standardized scores, you need to calculate the z-score for each variable. The z-score is a measure of how many standard deviations a particular observation is from the mean of the variable.

Here are the steps to calculate the z-score:

1. Calculate the mean (average) and standard deviation for each variable.
2. Subtract the mean from each observation, then divide by the standard deviation.

Once you have calculated the z-scores for all variables, you can determine which beverages have positive standardized scores on every variable.

Positive standardized scores indicate that a particular beverage has values above the mean for each variable. It means that the beverage has a higher price, sugar content, caffeine level, and alcohol content compared to the average of the dataset.

To identify the beverages with positive standardized scores on every variable, you would look for the data points that have z-scores greater than zero for price, sugar content, caffeine level, and alcohol content. These beverages would be considered as outliers with higher values for all the variables.

Overall, positive standardized scores on every variable suggest that these beverages are relatively expensive, have higher sugar content, more caffeine, and higher alcohol content compared to the average.