Chapter 9

We can ____ the standard error by substituting the sample variance or standard deviation in pace of the unknown population value

estimate

A df increases, you should find that the critical t values become ____, ultimately reaching +1.96.

smaller and smaller

Reporting r^2

t(8)= -2.67, p <.05, r^2 0.47

Treatment causes scores to increase or decrease (i.e., vary) If we can measure how much of the variability is explained by the treatment, we will obtain a measure of the size of the ____

treatment effect

To maximize the similarity from one version to another, we will use ____ in the formula for all of the different t statistics

variance

On average, what value is expected for the t statistic when the null hypothesis is true?

0

Is a confidence interval an adequate substitute for Cohen's d or r^2?

No Confidence intervals are affected by sample size But they can still be used in a research report to provide a description of the size of the treatment effect

What's in the numerator and denominator of t?

Numerator: difference between sample data (M) and population hypothesis (m) Denominator: estimated standard error/how much difference is reasonable to expect between sample mean and population mean

A researcher fails to reject the null hypothesis with a regular two tailed test using alpha = .05. If the researcher uses a directional (one-tailed) test with the same data and the same alpha, then what decision would be made?

Possibly reject H0 if the treatment effect is in the predicted direction

A researcher predicts that a treatment will increase scores. To test the treatment effect, a sample of n = 16 is selected from a population with mu = 80 and a treatment is administered to the individuals in the sample. After treatment, the sample mean is M = 78 with s^2 = 16. If the researcher uses a one-tailed test with alpha = .05, then what decision should be made?

Reject H0, with alpha = .05, but not with alpha = .01 2/1 = 2 .05 is 1.753 (reject H0) .01 is 2.947 (fail to reject H0)

What influences effect size?

Sample size has no effect Sample variance DOES have an effect High variance reduces likelihood of rejecting null hypothesis AND reduces measures of effect size

E.g., M = 46, m = 50, s^2 = 20.25 What's estimated Cohen's d?

Sample standard deviation Sqrt(20.25) = 4.5 Cohen's d (46-50)/4.5 = -0.8889 Indicates a large treatment effect

A sample of n = 9 scores is selected from a population with a mu = 70, and a treatment is administered to the sample. After treatment, the sample has M = 74 and SS = 288. If a hypothesis test with a t statistic is used to evaluate the treatment effect, then what value will be obtained for the t statistic?

Sample variance = ss/n-1 = 288/8 = 36 Estimated SE= sqrt(sample variance/n) = sqrt(36/9) = 2 t= M-mu/SM = 4/2 = 2 t=2

E.g., n = 9, SS = 288. Compute sample variance and estimated standard error for the sample mean.

Sample variance SS/N-1 = 288/8=36 Estimated standard error Sqrt(36/9)=2

A sample is selected from a population and a treatment is administered to the sample. For a hypothesis test with a t statistic, if there is a 5-point difference between the sample mean and the original population mean, which set of sample characteristics is most likely to lead to a decision that there is a significant treatment effect?

Small variance for a large sample

Confidence intervals are based on the observation that a sample mean tends to provide a reasonably ____ estimate of the population mean

accurate E.g., if we obtain a sample mean of M = 86, we can be reasonably confident that the population mean is around 86

A hypothesis test produces a t statistic of t = 2.30. If the researcher is using a two-tailed test with an alpha = .05, how large does the sample have to be in order to reject the null hypothesis?

at least n = 10

To have a smaller, more precise interval, you must give up ____

confidence Smaller t values

An alternative technique for describing a treatment effect is to compute an estimate of the population mean after treatment Estimating an unknown population mean involves constructing a

confidence interval

A sample is selected from a population with a mean of mu = 75, and a treatment is administered to the individuals in the sample. The researcher intends to use a t statistic to evaluate the effect of the treatment. If the sample mean is M = 79, then which of the following outcomes produces the largest value for Cohen's d? a. n = 4 and s^2 = 30 b. n = 16 and s^2 = 30 c. n = 25 and s^2 = 30 d. All three samples would produce the same value for Cohen's d

d. All three samples would produce the same value for Cohen's d

For t statistics, this relationship is typically expressed in terms of degrees of freedom (df; n-1) As df value ____, the better a t statistic approximates a z-score

increases

When the obtained difference between the data and the hypothesis (numerator) is much greater than expected (denominator), we obtain a ____ value for t (positive or negative) In this case, we conclude that the data are ____ with the hypothesis, and our decision is to ____

large not consistent reject H0

A sample is selected from a population with a mu = 30 and a treatment is administered to the sample. If the treatment is expected to increase scores and a t statistic is used for a one-tailed hypothesis test, then which of the following is the correct null hypothesis?

mu < 30

When the sample size (and degrees of freedom) is sufficiently large, the difference between a t distribution and the normal distribution becomes ____

negligible

The greater the sample size (n), the larger the degrees of freedom (n-1), and the better the t distribution approximates the ____

normal distribution

All you need to compute a t statistic is a ____ and ____ from the unknown population Thus, a t test can be used in situations for which the null hypothesis is obtained from a theory, a logical prediction, or wishful thinking

null hypothesis a sample

The ____ and the ____ both have a large effect on the t statistic because they both effect the standard error, which is the denominator of the formula

number of scores in the sample magnitude of sample variance

In the previous chapter, we used a sample mean to test hypotheses about an unknown population mean Summarized: 1) Sample mean (M) is expected to approximate its population mean (mu). This permits us to use the sample mean to test a hypothesis about the _____ 2) The standard error (sigmaM) provides a measure of how much difference is reasonable to expect between a ____ and ____ 3) To test the hypothesis, we compare the obtained sample (M) with the hypothesized population mean (mu) by computing a ____ 4) If the z-score falls in the critical region, the obtained difference between the data and the hypothesis is significantly greater than would be expected by ____

population mean sample mean (M) and the population mean (mu) z-score test statistic chance

A z-score requires that we know the value of the ____, which is needed to compute standard error However, the standard deviation for the population is not known In fact, the whole reason for conducting a hypothesis test is to gain knowledge about an unknown population

population standard deviation (or variance)

t statistic is used to test a hypothesis about an unknown population mean, when the value of is ____ unknown

population standard deviation/variance

Percentage of variance accounted for by the treatment is notated by

r^2

*If your sample t statistic is greater than the value listed, you can be certain that the data are in the critical region and confidently ____

reject the null hypothesis

*Large variance means you are less likely to obtain a ____ treatment effect In general, large variance is bad for inferential statistics Large variance means that the scores are widely scattered, which makes it much more difficult to see any consistent patterns or trends in the data *The larger the sample is, the smaller the ____

significant error

he shape of the t distribution changes with the degrees of freedom As df gets large, the t distribution gets closer to a normal z-score distribution However, the t distribution is flatter/has more variability than a normal z distribution, especially when df values are ____

small

Thus, a confidence interval consists of an interval of values around a sample mean, and we can be reasonably confident that the unknown population mean is located ____

somewhere in the interval

A sample of n = 4 scores has an SS = 48. What is the estimated standard error for the same mean?

sqrt(sample variance/n) sample variance = SS/n-1 = 48/3 = 16 Sqrt(16/4) = 2

Use a unit normal table to locate proportions associated with z-scores Use a ____ table to find proportions for t statistics

t distribution

Now we can substitute the estimated standard error in the denominator of the z-score formula The result is a new test statistic called ____

t statistic

A researcher uses a sample of n = 25 individuals to evaluate the effect of a treatment. The hypothesis test uses a alpha = .05 and produces a significant result with t = 2.15 How would this result be reported in the literature?

t(24) = 2.15, p < .05

A hypothesis test simply determines whether the treatment effect is greater than chance, where "chance" is measured by ____

the standard error

For hypothesis testing using the t statistic, the population mean with no treatment is the value specified by the null hypothesis However, the population mean with treatment and standard deviation are both unknown There, we use the mean and standard deviation for the ____ as estimates of unknown parameters

treated sample

An alternative method for measuring effect size: Determine how much of the variability in the score is explained by the ____ effect

treatment

If you select all of the possible samples of a particular size (n) and compute the t statistic for each sample mean, then the entire set of t values will form a t distribution T distributions are bell-shaped, symmetrical and have a mean of zero. However, t distributions are more ____ than the normal distribution

variable

____ Indicates that the estimated value is computed from sample data rather than from the actual population

SM (rather than sigma M)

To gain more confidence in your estimate Increase the ____ of the interval

width Larger t values

A researcher obtains a sample of n = 9 volunteers who agree to spend at least 15 minutes using an eReader during the hour before sleeping and then take a standardized cognitive alertness test the next morning For the general population, scores on the test average = 50 from a normal distribution The sample of research participants had an average score of M = 46 with SS = 162 1) 2) 3) 4)

1) State hypotheses and select an alpha level Null hypothesis: late-night reading from a light-emitting screen has no effect on alertness the following morning H0 = 50 (same as the general population) Alternative hypothesis: reading from a screen does affect alertness the next morning H1 does not equal 50 Significance level a = .05 for two tails 2) Locate critical region Df = n - 1 = 9 - 1 = 8 Greater than +2.306 or less than -2.306 3) Calculate the test statistic Calculate sample variance SS/n-1 = 162/8 = 20.25 Use sample variance and sample size to compute estimated standard error Sqrt(sample variance/ n) = sqrt(20.25/9) = 1.5 Compute the t statistic M - m / SM = (46-50)/1.50 = -2.6667 4) Make a decision This value falls into the critical region on the left-hand side of the distribution Reject H0

The numbers in the table are the values of t that separate the tail from the main body of the distribution For a df of 3, 5% of the distribution is located in the tail beyond both +2.353 and -2.353 Thus, a total proportion of ___ is contained in the two tails beyond + 2.353

10%

A sample of n=4 scores is selected from a population with an unknown mean. The sample has a mean of M = 40 and a variance of s^2 = 16. What is the 90% CI for mu?

100-90 = 10% of distribution left over Divide between two tails 10/2 = 5 =.05 Df = 3 Look in t table df = 3, across to .05 find value of 2.353 M +2.353(SM) SM = sqrt(16/4)=2 40 +2.353(2)

A sample of n = 25 is selected from a population with mu = 40, and a treatment is administered to each individual in the sample. After treatment, the sample mean is M = 44 with a sample variance of s^2 = 100. Based on this information, what is the is the size of the treatment effect as measured by Cohen's d?

44-40/10 = .4 d= 0.40

Thus a problem that could arise: A small treatment effect could be "statistically significant" if the sample size was large enough Correction?

All results from a hypothesis test should be accompanied by a report of the size of effect: Cohens d

The bigger the sample, the smaller the interval

As sample size increases, standard error decreases, and the interval gets smaller

To determine how well a t statistic approximates a z-score, we must determine how well the sample variance approximates the population variance

As sample size increases, the better the sample variance represents the population variance AND the better the t statistic approximates the z-score

If late night reading has no effect on alertness in the morning, m = 50 Almost all of the scores are to the left of m = 50; 4-point effect with M = 46 This shift to the left is treatment effect Late night reading has pushed scores away from 50 and has increased the size of the deviations What happens if the treatment effect is removed? Simply add 4 points to each score Adjusted scores will be centered at 50, indicating no treatment effect Deviations are noticeably smaller How much of the variability was reduced when the treatment effect was removed?

Compute the SS for each set of scores Total variability (including treatment effect) is SS = 306 However, when the treatment effect is removed, the variability is reduced to SS = 162 The difference 306 - 162 = 144 points is the amount of variability that is accounted for by the treatment effect In percentage: 144/ 306 = 0.4706 (47%)

Directional test/One-tailed test E.g., n =9, M = 46, SS = 162 State Hypotheses H0 = mScreen reading = > 50 H1 = mScreen reading = < 50 Select significance level a = .05 Locate critical region Calculate the test statistic Make a decision

Df = 8, go over to a = .05, t value = 1.860 Therefore, if we obtain a sample mean of less than 50 and the t statistic is beyond d the -1.860 critical boundary on the left-hand side, we will reject null and conclude that reading from a light-emitting screen before bedtime significantly lowers alertness the next morning Calculate the test statistic Calculate sample variance SS/n-1 = 162/8 = 20.25 Use sample variance and sample size to compute estimated standard error Sqrt(sample variance/n ) = sqrt(20.25/9) = 1.5 Compute the t statistic M - m / SM = (46-50)/1.50 = -2.6667 This value falls into the critical region on the left-hand side of the distribution Reject H0 -2.6667 falls beyond -1.860 t(8) = -2.67, p < .05, one tailed

Why is the t distribution flatter and more variable?

For z-scores, the bottom of the formula (population variance) does not vary, provided that all of the samples are the same size and selected from the same population i.e., all of the z-scores have the same standard error in the denominator because the population variance and sample size are the same for every sample For t statistics, the bottom of the formula varies from one sample to another *The sample variance changes from one sample to the next, so the estimated standard error also varies

Difference between t statistic and z-score?

T statistic uses sample variance: s^2 Z-score uses population variance: sigma^2

In what circumstances is the t statistic used instead of a z-score for a hypothesis test?

The t statistic is used when the population variance (or standard deviation) is unknown

Two basic assumptions are necessary for hypothesis tests with the t statistic

The values in the sample must consist of independent observations The population sampled must be normal

Why use variance to estimate standard error?

Variance is an unbiased statistic On average, the sample variance provides an accurate and unbiased estimate of the population variance

Solution?

When the population standard deviation (or variance) is not known, we use the corresponding sample value in its place

Construct a confidence interval for: n = 9, df = 8, M = 46, SM = 1.50 For this example we will use 80%, which means that we will construct the confidence interval so that we are 80% confident the unknown population mean is actually contained in the interval

With a df of 8, the middle 80% of the distribution is bound by t values + 1.397 and - 1.397 46 +(1.397*1.50) = 48.0955 46+(-1.397*1.50)=43.9045 80% sure the population mean is between 43.9045 and 48.0955

The problem with z-Scores?

Z-score formula requires more information that is usually available

t statistic: substitute for ____

a z-score