Question

1a.) What is a z score? List a few reasons, with detail, that cover why a z score is useful to researchers.

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1b.) A normal curve is an important part of statistical research. When might a curve be abnormal? Why is it this important? What issues might we confront if we assume normality when there is none?

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1c.) List the rules of probability, and give a short description of each. Come up with a few scenarios where you might use each rule.

Answer 1

1)a)

-Scores

Sometimes we want to do more than summarize a bunch of scores. Sometimes we want to talk about particular scores within the bunch. We may want to tell other people about whether or not a score is above or below average. We may want to tell other people how far away a particular score is from average. We might also want to compare scores from different bunches of data. We will want to know which score is better. Z-scores can help with all of this.

They Tell Us Important Things

Z-Scores tell us whether a particular score is equal to the mean, below the mean or above the mean of a bunch of scores. They can also tell us how far a particular score is away from the mean. Is a particular score close to the mean or far away?

If a Z-Score….

ü      Has a value of 0, it is equal to the group mean.

ü      Is positive, it is above the group mean.

ü      Is negative, it is below the group mean.

ü      Is equal to +1, it is 1 Standard Deviation above the mean.

ü      Is equal to +2, it is 2 Standard Deviations above the mean.

ü      Is equal to -1, it is 1 Standard Deviation below the mean.

ü      Is equal to -2, it is 2 Standard Deviations below the mean.

Z-Scores Can Help Us Understand…

How typical a particular score is within bunch of scores. If data are normally distributed, approximately 95% of the data should have Z-score between -2 and +2. Z-scores that do not fall within this range may be less typical of the data in a bunch of scores.

Z-Scores Can Help Us Compare…

Individual scores from different bunches of data. We can use Z-scores to standardize scores from different groups of data. Then we can compare raw scores from different bunches of data.

b)

Many data sets naturally fit a non normal model. For example, the number of accidents tends to fit a Poisson distribution and lifetimes of products usually fit a Weibull distribution. However, there may be times when your data is supposed to fit a normal distribution, but doesn’t. If this is a case, it’s time to take a close look at your data.

• Outliers can cause your data the become skewed. The mean is especially sensitive to outliers. Try removing any extreme high or low values and testing your data again.
• Multiple distributions may be combined in your data, giving the appearance of a bimodal or multimodal distribution. For example, two sets of normally distributed test results are combined in the following image to give the appearance of bimodal data.
  
• Insufficient Data can cause a normal distribution to look completely scattered. For example, classroom test results are usually normally distributed. An extreme example: if you choose three random students and plot the results on a graph, you won’t get a normal distribution. You might get a uniform distribution (i.e. 62 62 63) or you might get a skewed distribution (80 92 99). If you are in doubt about whether you have a sufficient sample size, collect more data.
• Data may be inappropriately graphed. For example, if you were to graph people’s weights on a scale of 0 to 1000 lbs, you would have a skewed cluster to the left of the graph. Make sure you’re graphing your data on appropriately labeled axes.

c)

Probability Rule One:

• For any event A, 0 ≤ P(A) ≤ 1.

Probability Rule Two:

The sum of the probabilities of all possible outcomes is 1.

Probability Rule Three (The Complement Rule):

• P(not A) = 1 – P(A)
• that is, the probability that an event does not occur is 1 minus the probability that it does occur.

Disjoint: Two events that cannot occur at the same time are called disjoint or mutually exclusive.

when the events ARE disjoint, P(A and B) = 0.