Examples of t-score in the following topics:
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- The mean anxiety score for women was 43.5, and the mean anxiety score for men was 47.9.
- This difference was significant; a t-test found a t-score of 2.34, and the p-value was 0.01.
- In terms of their scores on the Anxiety Scale, women $(M=3.68\text{, }SD=.70)$ were found to be significantly more anxious than men $(M=3.28\text{, }SD=.68)$, $t(61) = 2.34\text{, }p < .05.$
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- $z$-scores are also called standard scores, $z$-values, normal scores or standardized variables.
- If one only has a sample set, then the analogous computation with sample mean and sample standard deviation yields the Student's $t$-statistic.
- The conversion of a raw score, $x$, to a $z$-score can be performed using the following equation:
- $z$-scores for this standard normal distribution can be seen in between percentiles and $t$-scores.
- Define $z$-scores and demonstrate how they are converted from raw scores
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- Thus, a positive standard score represents a datum above the mean, while a negative standard score represents a datum below the mean.
- Standard scores are also called $z$-values, $z$-scores, normal scores, and standardized variables.
- The $z$-score is only defined if one knows the population parameters.
- If one only has a sample set, then the analogous computation with sample mean and sample standard deviation yields the Student's $t$-statistic.
- Includes: standard deviations, cumulative percentages, percentile equivalents, $Z$-scores, $T$-scores, and standard nine.
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- However, we will write T instead of Z, because we have a T T score (like Z score) small sample and are basing our inference on the t distribution:
- If the null hypothesis was true, the test statistic T would follow a t distribution with df = n − 1 = 29 degrees of freedom.
- We can use the t distribution method.
- The value T = 2.39 falls between the third and fourth columns.
- For instance, maybe SAT test takers tend to improve their score over time even if they don't take a special SAT class, or perhaps only the most motivated students take such SAT courses.
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- For each subject a difference score between their initial weight and final weight could be computed.
- A t test of whether the mean difference score differs significantly from 0 could then be computed.
- The mean difference score will equal the difference between the mean weight losses of the two groups (61.3 - 11.2 = 50.1).
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- As shown in the section on testing a single mean, the mean difference score is 4.96 which is significantly different from 0: t = 3.22, df = 23, p = 0.0038.
- This t test has various names including "correlated t test" and "related-pairs t test. "
- In general, the correlated t test is computed by first computing the differences between the two scores for each subject.
- This is because in correlated t tests, each difference score is a comparison of performance in one condition with the performance of that same subject in another condition.
- To see why the standard error of the difference between means is smaller in a correlated t test, consider the variance of difference scores.
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- Report the t and p values.
- The three scores per subject are their scores on three trials (a, b, and c) of a memory task.
- Their scores are shown below.
- Calculate the two-tailed t and p values using this t test.
- Calculate the one-tailed t and p values using this t test.
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- The average SAT score is 560, with a standard deviation of 75.
- If you know no information (you don't know the SAT score), it is best to make predictions using the average.
- If the students admitted all had SAT scores within the range of 480 to 780, the regression model may not be a very good estimate for a student who only scored a 350 on the SAT.
- For example, if no one before had received an exact SAT score of 650, we would predict his GPA by looking at the GPAs of those who scored 640 and 660 on the SAT.
- Let's say the highest SAT score of a student the college admitted was 780.
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- A college professor wants to compare her students' scores with the national average.
- Their scores have a standard deviation of $2.5$.
- She wants to know if her students scored significantly lower than the national average.
- i.e.: The null hypothesis is that her students scored on par with the national average.
- i.e.: The alternative hypothesis is that her students scored lower than the national average.
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- We have a sample of 10 scores.
- Therefore, we can perform a Student's $t$-test, with $n-1$, 9 degrees of freedom.
- $\displaystyle \overline { X } \sim \left( \mu ,\frac { s }{ \sqrt { n } } \right) =T\left( 65,\frac { 5.0111 }{ \sqrt { 10 } } \right)$
- There is no reason to think the score of one exam has any bearing on the score of another exam.
- $\displaystyle t=\frac { x-\mu }{ \frac { s }{ \sqrt { n } } } =\frac { 63-65 }{ 1.585 } =-1.2618$.