Z-分数

Measuring z-Scores

A z-score, or a standard score, tells us how many standard deviations from the mean an observation falls, and in what direction. We can also think of this as the population standard deviation.

$$z= \frac{observation- mean}{S_x}$$

A positive z-score indicates that the observation falls $z$ standard deviations to the right of the mean, and a negative z-score indicates that the observation $z$ standard deviations to the left of the mean. Z-scores are standardizations of raw scores.

How This Concept Applies

A z-Score allows us to describe an observation in terms of how many standard deviations it is from the average. This is known as standardizing. In other words, calculating the z-Score allows us an easy way to compare apples to oranges.

Let's take a look at an example

Sally earned a score of $86$ on her calculus test. The calculus class mean was $80$, and the standard deviation was $6.07.$ On her chemistry test, Sally earned a score of $82.$ The chemistry class mean was a $76,$ and the standard deviation was $4.$

On which test did Sally perform better relative to the rest of her class?

Finding and Interpreting the z-Scores

First, let's find the z-Score for Sally's calculus test:

$$z= \frac{observation- mean}{S_x} = \frac{86-80}{6.07}=.99$$

Now, let's find the z-Score for Sally's chemistry test:

$$z= \frac{observation- mean}{S_x} = \frac{82-76}{4}=1.5$$

Since both z-Scores are positive, it means that they are $z$ standard deviations higher than the mean, or to the right of the mean. In this case, Sally's calculus score was $.99$ standard deviations higher than the mean calculus score, and Sally's chemistry score was $1.5$ standard deviations higher than the mean chemistry score.

Since $1.5>.99$, Sally performed better on her chemistry test than she did on her calculus test.

How To Calculate A Confidence Interval (CI)

Depending on our sample size, our confidence interval formula will depend on values from either our Z-table or T-table.

Additionally, our CI formulas will change slightly based on whether or not our parameter of interest is either a mean or a proportion, or if we are comparing one-sample or two-samples.

In general, if you do not know the population standard deviation, you will use a t-score. It gets a little more complicated if you do know the population standard deviation, then it gets a little more complicated.

$$\text{Know population standard deviation} \rightarrow \text{sample size > 30} \rightarrow \text{z-score}$$ $$\text{Know population standard deviation} \rightarrow \text{sample size < 30} \rightarrow \text{t-score}$$

Online, you will find a lot of awesome z-score tables and t-score tables.

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Z-table Confidence Interval Formulas:

• one-sample mean z-table CI formula: $\bar{x} \pm Z^* \cdot (\frac{\sigma}{\sqrt{n}})$

• one-sample proportion z-table CI formula: $\hat{p} \pm Z^* \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

• two-sample mean z-table CI formula: $(\bar{x_1} - \bar{x_2}) \pm Z^* \cdot \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$

• two-sample proportion z-table CI formula: $\hat{p_1} - \hat{p_2} \pm Z^* \cdot \sqrt{\frac{\hat{p_1}(1-\hat{p_1})}{n_1} + \frac{\hat{p_2}(1-\hat{p_2})}{n_2}}$

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t-table Confidence Interval Formulas:

Search t-table with confidence level and degrees of freedom ($df$)

$df = n - 1$ for a one-sample problem

or

$df = n_1 + n_2 -2$ for a two-sample problem

• one-sample mean CI t-table formula: $\bar{x} \pm t^* \cdot (\frac{\sigma}{\sqrt{n}})$

• two-sample mean CI t-table formula: $(\bar{x_1} - \bar{x_2}) \pm t^* \cdot \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}}$

such that $s_p^2 = \frac{(n_1 - 1) s_1^2 + (n_2 - 1) s_2^2}{n_1 + n_2 -2}$

Note: There are several different methods that can be used to calculate the confidence interval for small one and two-sample proportions that are not listed here.

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Such that:

$n$ = Number of members in sample

$n_1$ = Number of members in sample #1

$n_2$ = Number of members in sample #2

$\bar{x}$ = Sample Mean

$\bar{x_1}$ = Sample #1 Mean

$\bar{x_2}$ = Sample #2 Mean

$\Delta$ = hypothesized difference between the two population means

$\hat{p} = \frac{x}{n}$

$\sigma$ = Standard Deviation

$\sigma_1$ = Standard Deviation for Sample #1

$\sigma_2$ = Standard Deviation for Sample #2

$s_p$ = pooled estimate of the variance between two sample means

$α = 1 - \frac{confidence level}{100}$

confidence level = 90%, 95%, 99%, etc...

Some common $Z^*$ values are 1.645, 1.96, and 2.576 for confidence levels 90%, 95%, and 99% respectively. More can be found on a Z-table.

$t^*$ values can be found on a t-table based on the desired confidence level and degrees of freedom.

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Hit "Continue" for a CI calculation walkthrough for a one-sample mean and one-sample proportion problem

Confidence interval for one-sample mean

Question: The average score obtained by 10 students in a test is 75. Standard deviation of the sample is 8. Find the confidence interval at 95% confidence level.

Step 1) State our given variables

Given:

$n$ = 10 --> Therefore $df = 10-1 = 9$

$x$ = sample mean = 75

$\sigma$ = standard deviation = 8

confidence level = 95%

Step 2) Plug values into appropriate CI formula

Because our sample of n = 10 is less than 30, we use a t-table.

$x \pm t^* \cdot (\frac{\sigma}{\sqrt{n}})$

$75 \pm t^* \cdot (\frac{8}{\sqrt{10}})$

$75 \pm 2.262 \cdot (\frac{8}{\sqrt{10}})$

Hence our CI is: [69.2775,80.7225]

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Confidence interval for one-sample proportion

Question: We survey 76 veterinarians from several local veterinary clinics. 19 reported that they are left-handed. Thus 19/76 = 0.25 is our estimate of the proportion of left-handed veterinarians. We want to determine how accurate this estimate is by calculating a confidence interval to give us a confident understanding of the percentage of veterinarians that are left-handed. What is the confidence interval for $\alpha=0.05$ based on this data?

Step 1) State our given variables

$x = numberSucesses$

$n = sampleSize = 76$

$\hat{p} = \frac{x}{n} = 19/76 = 0.25$

$\alpha = 0.05$ --> $\alpha = 1 - \frac{confidence level}{100}$

Therefore, $confidence Level = 95$%

Therefore $Z^* = 1.96$

Step 2) Plug in values into appropriate CI formula

$\hat{p} \pm Z^* \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$0.25 \pm 1.96 \cdot \sqrt{\frac{0.25(1-0.25)}{76}}$

[0.1526,0.3474] is our confidence interval

So provided this data came from a random sample, we are fairly confident that the true percentage of left-handed veterinarians is between .15 and .35

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Hit "Continue" for a CI calculation walkthrough for a two-sample mean and a two-sample proportion problem

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Confidence interval for two-sample mean

Question: What is the 95% confidence interval for these given values?

Step 1) State our given variables

$\bar{x_1} = 21$

$\bar{x_2} = 12.1$

$n_1 = 13$

$n_2 = 17$

$s_1^2 = 4.9$

$s_2^2 = 5.6$

$s_p^2 = \frac{(n_1 - 1) s_1^2 + (n_2 - 1) s_2^2}{n_1 + n_2 -2} = \frac{(13 - 1) 4.9 + (17 - 1) 5.6}{13 + 17 -2} = 5.3$

$df = n_1 + n_2-2 = 13 + 17 -2 = 28$

From t-table --> $t^* = 2.045$

Step 2: Plug in values into appropriate CI formula

$(\bar{x_1} - \bar{x_2}) \pm t^* \cdot \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}}$

$( 21 - 12.1) \pm 2.045 \cdot \sqrt{\frac{5.3}{13} + \frac{5.3}{17}}$

[7.165, 10.634] is our confidence interval

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Confidence interval for two-sample proportion

Question: In two different space camp groups (#1 and #2), space camp participants were asked whether or not they LOVED mathematics. There were 96 participants in Group #1 and 123 participants in Group #2. 18 participants in Group #1 and 60 in Group #2 reported that they LOVED math. What is the 99% confidence interval for the difference between the two proportions?

Step 1) State our given variables

$n_1 = 96$

$n_2 = 123$

$x_1 = 18$

$x_2 = 60$

confidence level $= 99$

Therefore Z* = 2.576

$\hat{p_1} = \frac{x_1}{n_1} = \frac{60}{123} = 0.4878$

$\hat{p_2} = \frac{x_2}{n_2} = \frac{18}{96} = 0.1875$

Step 2: Plug in values into appropriate CI formula

$\hat{p_1} - \hat{p_2} \pm Z^* \cdot \sqrt{\frac{\hat{p_1}(1-\hat{p_1})}{n_1} + \frac{\hat{p_2}(1-\hat{p_2})}{n_2}}$

$(0.4878- 0.1875) \pm 2.576 \cdot \sqrt{\frac{0.1875(1-0.1875)}{96} + \frac{0.4878(1-0.4878)}{123}}$

[0.1453,0.4553] is our confidence interval

This means that we are 99% confident that the true value of the difference between the two population proportions lies between .1435 and .4553

(Video) How To Find Z-Score Probabilities With Z-Table

by eHowEducation

In this video by eHowEducation, you will learn how to use the z-table to calculate z-score probabilities.

What Z-Scores Are Used For

A Z-Table is used to find a z-score, which is the number of standard deviations from the mean a data point is. Because of this, Z-scores can tell us things like how far away a particular test score is from the class average (mean) and whether or not it is lower or higher than it.

Visually, the z-score is also the area under a standard normal curve between the mean (zero) and any z-score.

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Note: The reason we standardize the curve and even need the z-table is that sometimes our original data distribution is too complicated and has large ranges and we would rather "standardize" it so that the mean is centered at 0 and not like... 72.543.

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Z-Scores tell us whether a particular score is equal to the mean, below the mean, or above the mean of all the scores in a data set.

If a Z-Score….

• Equals 0 it is equal to the mean (because the normal curve is standardized)

• Is + it is above the mean

• Is - it is below the mean

• Is equal to +1 it is one standard deviation above the mean

• Is equal to +2 it is two standard deviations above the mean

• Is equal to -1 it is one standard deviation below the mean

• Is equal to -2 it is two standard deviations below the mean