How To Find P Value In Spss

Using SPSS for t Tests

This tutorial will show yous how to use SPSS version 12.0 to perform one-sample t-tests, contained samples t-tests, and paired samples t-tests.

This tutorial assumes that you take:

Downloaded the standard class information set (click on the link and relieve the data file)
Started SPSS (click on Start | Programs | SPSS for Windows | SPSS 12.0 for Windows)

Ane Sample t-Tests

One sample t-tests can be used to determine if the mean of a sample is different from a particular value. In this case, we will determine if the mean number of older siblings that the PSY 216 students have is greater than 1.

We volition follow our customary steps:

Write the null and culling hypotheses beginning:
H₀: µ_{216 Students} ≤ 1
H₁: µ_{216 Students} > 1
Where µ is the hateful number of older siblings that the PSY 216 students have.
Determine if this is a i-tailed or a two-tailed test. Considering the hypothesis involves the phrase "greater than", this must be a one tailed exam.
Specify the α level: α = .05
Decide the appropriate statistical examination. The variable of involvement, older, is on a ratio scale, so a z-score test or a t-test might be appropriate. Considering the population standard difference is not known, the z-test would be inappropriate. Nosotros will use the t-test instead.
Summate the t value, or allow SPSS do information technology for you!
The command for a i sample t tests is found at Clarify | Compare Ways | Ane-Sample T Examination (this is shorthand for clicking on the Analyze bill of fare item at the top of the window, so clicking on Compare Means from the drop down carte du jour, and One-Sample T Test from the popular up carte.):

The 1-Sample t Test dialog box will appear:

Select the dependent variable(southward) that you want to test past clicking on information technology in the left mitt pane of the Ane-Sample t Test dialog box. Then click on the arrow button to move the variable into the Test Variable(s) pane. In this case, motility the Older variable (number of older siblings) into the Test Variables box:

Click in the Test Value box and enter the value that yous volition compare to. In this example, nosotros are comparison if the number of older siblings is greater than 1, so we should enter 1 into the Examination Value box:

Click on the OK push button to perform the one-sample t test. The output viewer will appear. At that place are two parts to the output. The first function gives descriptive statistics for the variables that you moved into the Test Variable(southward) box on the One-Sample t Test dialog box. In this example, nosotros get descriptive statistics for the Older variable:

This output tells the states that we accept 46 observations (Due north), the mean number of older siblings is 1.26 and the standard divergence of the number of older siblings is 1.255. The standard mistake of the hateful (the standard deviation of the sampling distribution of means) is 0.185 (ane.255 / square root of 46 = 0.185).

The second part of the output gives the value of the statistical examination:

The second column of the output gives us the t-examination value: (one.26 - 1) / (1.255 / square root of 46) = 1.410 [if you do the calculation, the values volition not match exactly considering of round-off error). The third column tells us that this t examination has 45 degrees of freedom (46 - 1 = 45). The fourth column tells us the two-tailed significance (the 2-tailed p value.) But we didn't want a 2-tailed exam; our hypothesis is 1 tailed and there is no option to specify a one-tailed test. Because this is a one-tailed test, look in a tabular array of critical t values to decide the critical t. The critical t with 45 degrees of freedom, α = .05 and one-tailed is 1.679.
Determine if we can decline the aught hypothesis or not. The determination rule is: if the ane-tailed disquisitional t value is less than the observed t AND the means are in the right guild, and then we can decline H₀. In this instance, the critical t is 1.679 (from the table of critical t values) and the observed t is ane.410, so we neglect to reject H₀. That is, at that place is insufficient evidence to conclude that the mean number of older siblings for the PSY 216 classes is larger than 1.

If we were writing this for publication in an APA journal, nosotros would write information technology every bit:
A t test failed to reveal a statistically reliable divergence betwixt the hateful number of older siblings that the PSY 216 class has (1000 = 1.26, s = one.26) and ane, t(45) = 1.410, p < .05, α = .05.

Contained Samples t-Tests
Unmarried Value Groups

When two samples are involved, the samples tin can come from unlike individuals who are not matched (the samples are independent of each other.) Or the sample tin come from the aforementioned individuals (the samples are paired with each other) and the samples are non independent of each other. A third alternative is that the samples tin come from unlike individuals who have been matched on a variable of involvement; this type of sample will not exist contained. The form of the t-test is slightly unlike for the independent samples and dependent samples types of two sample tests, and SPSS has split up procedures for performing the two types of tests.

The Independent Samples t-exam can be used to see if 2 means are different from each other when the two samples that the means are based on were taken from different individuals who have not been matched. In this example, we volition determine if the students in sections one and two of PSY 216 take a different number of older siblings.

We will follow our customary steps:

Write the cipher and alternative hypotheses beginning:
H₀: µ_{Section 1} = µ_{Section 2}
H₁: µ_{Section 1} ≠ µ_{Section 2}
Where µ is the mean number of older siblings that the PSY 216 students have.
Make up one's mind if this is a 1-tailed or a two-tailed exam. Because the hypothesis involves the phrase "different" and no ordering of the means is specified, this must be a two tailed exam.
Specify the α level: α = .05
Make up one's mind the appropriate statistical test. The variable of interest, older, is on a ratio scale, so a z-score test or a t-test might be appropriate. Because the population standard deviation is not known, the z-test would be inappropriate. Furthermore, in that location are different students in sections 1 and 2 of PSY 216, and they take non been matched. Because of these factors, nosotros will utilize the independent samples t-test.
Calculate the t value, or let SPSS exercise it for you!
The command for the independent samples t tests is establish at Analyze | Compare Means | Independent-Samples T Test (this is shorthand for clicking on the Clarify bill of fare detail at the pinnacle of the window, and and so clicking on Compare Ways from the driblet down menu, and Independent-Samples T Test from the pop up carte.):

The Independent-Samples t Exam dialog box will appear:

Select the dependent variable(s) that you desire to test by clicking on it in the left hand pane of the Independent-Samples t Test dialog box. Then click on the upper arrow push to move the variable into the Test Variable(s) pane. In this example, movement the Older variable (number of older siblings) into the Test Variables box:

Click on the independent variable (the variable that defines the ii groups) in the left hand pane of the Independent-Samples t Exam dialog box. And so click on the lower arrow button to move the variable in the Grouping Variable box. In this example, move the Section variable into the Grouping Variable box:

Y'all demand to tell SPSS how to define the two groups. Click on the Ascertain Groups button. The Define Groups dialog box appears:

In the Group one text box, type in the value that determines the beginning group. In this case, the value of the 10 AM section is 10. So y'all would type ten in the Group one text box. In the Group ii text box, type the value that determines the second grouping. In this example, the value of the 11 AM section is xi. Then you would type eleven in the Group 2 text box:

Click on the Go on button to close the Define Groups dialog box. Click on the OK push in the Contained-Samples t Test dialog box to perform the t-test. The output viewer will appear with the results of the t test. The results have 2 main parts: descriptive statistics and inferential statistics. Starting time, the descriptive statistics:

This gives the descriptive statistics for each of the two groups (every bit defined by the group variable.) In this example, there are 14 people in the 10 AM department (N), and they have, on boilerplate, 0.86 older siblings, with a standard deviation of 1.027 older siblings. At that place are 32 people in the 11 AM department (N), and they have, on boilerplate, 1.44 older siblings, with a standard difference of ane.318 older siblings. The final column gives the standard mistake of the hateful for each of the 2 groups.

The 2nd role of the output gives the inferential statistics:

The columns labeled "Levene'south Exam for Equality of Variances" tell us whether an assumption of the t-examination has been met. The t-test assumes that the variability of each group is approximately equal. If that assumption isn't met, then a special form of the t-examination should be used. Look at the column labeled "Sig." under the heading "Levene'south Examination for Equality of Variances". In this example, the significance (p value) of Levene's test is .203. If this value is less than or equal to your α level for the examination (usually .05), then you can reject the null hypothesis that the variability of the ii groups is equal, implying that the variances are unequal. If the p value is less than or equal to the α level, and then you should utilise the bottom row of the output (the row labeled "Equal variances not causeless.") If the p value is greater than your α level, then y'all should use the centre row of the output (the row labeled "Equal variances assumed.") In this example, .203 is larger than α, so we will assume that the variances are equal and we will use the centre row of the output.

The column labeled "t" gives the observed or calculate t value. In this instance, assuming equal variances, the t value is 1.461. (We can ignore the sign of t for a two tailed t-test.) The column labeled "df" gives the degrees of freedom associated with the t examination. In this case, there are 44 degrees of freedom.

The column labeled "Sig. (2-tailed)" gives the two-tailed p value associated with the test. In this case, the p value is .151. If this had been a 1-tailed test, we would demand to look up the disquisitional t in a tabular array.
Determine if nosotros can pass up H₀: As before, the decision rule is given by: If p ≤ α , and so reject H₀. In this case, .151 is not less than or equal to .05, and then nosotros fail to pass up H₀. That implies that nosotros failed to observe a difference in the number of older siblings between the ii sections of this class.

If we were writing this for publication in an APA periodical, we would write information technology as:
A t exam failed to reveal a statistically reliable difference between the mean number of older siblings that the 10 AM department has (M = 0.86, s = 1.027) and that the 11 AM section has (Thou = 1.44, s = 1.318), t(44) = 1.461, p = .151, α = .05.

Independent Samples t-Tests
Cut Point Groups

Sometimes you want to perform a t-exam but the groups are defined by a variable that is not dichotomous (i.eastward., it has more than 2 values.) For example, you may want to run across if the number of older siblings is different for students who have higher GPAs than for students who accept lower GPAs. Since there is no single value of GPA that specifies "higher" or "lower", we cannot proceed exactly equally nosotros did before. Earlier proceeding, decide which value yous will use to divide the GPAs into the higher and lower groups. The median would be a proficient value, since half of the scores are above the median and one-half are below. (If you lot do not remember how to calculate the median meet the frequency command in the descriptive statistics tutorial.)

Write the null and alternative hypotheses kickoff:
H₀: µ_{lower GPA} = µ_{higher GPA}
H₁: µ_{lower GPA} ≠ µ_{College GPA}
Where µ is the mean number of older siblings that the PSY 216 students take.
Determine if this is a ane-tailed or a two-tailed examination. Because the hypothesis involves the phrase "different" and no ordering of the ways is specified, this must be a ii tailed test.
Specify the α level: α = .05
Determine the appropriate statistical test. The variable of interest, older, is on a ratio scale, so a z-score test or a t-test might be appropriate. Because the population standard deviation is not known, the z-test would exist inappropriate. Furthermore, different students have higher and lower GPAs, and then we have a between-subjects design. Because of these factors, we volition use the independent samples t-exam.
Calculate the t value, or allow SPSS do it for you lot.
The command for the independent samples t tests is found at Analyze | Compare Means | Independent-Samples T Test (this is shorthand for clicking on the Analyze menu item at the top of the window, and and so clicking on Compare Ways from the drop downwards menu, and Independent-Samples T Test from the popular up menu.):

The Independent-Samples t Examination dialog box volition appear:

Select the dependent variable(southward) that you desire to test past clicking on it in the left hand pane of the Independent-Samples t Test dialog box. Then click on the upper arrow push to move the variable into the Exam Variable(s) pane. In this instance, move the Older variable (number of older siblings) into the Test Variables box:

Click on the independent variable (the variable that defines the ii groups) in the left manus pane of the Contained-Samples t Test dialog box. So click on the lower arrow button to move the variable in the Group Variable box. (If there already is a variable in the Grouping Variable box, click on information technology if information technology is not already highlighted, and so click on the lower arrow which should be pointing to the left.) In this example, move the GPA variable into the Group Variable box:

You need to tell SPSS how to define the two groups. Click on the Define Groups push. The Define Groups dialog box appears:

Click in the circumvolve to the left of "Cut Point:". Then type the value that splits the variable into two groups. Group one is divers every bit all scores that are greater than or equal to the cut point. Group two is defined as all scores that are less than the cut point. In this example, use 3.007 (the median of the GPA variable) as the cut point value:

Click on the Continue button to close the Define Groups dialog box. Click on the OK button in the Independent-Samples t Exam dialog box to perform the t-test. The output viewer will appear with the results of the t exam. The results have two main parts: descriptive statistics and inferential statistics. First, the descriptive statistics:

This gives the descriptive statistics for each of the two groups (as defined by the grouping variable.) In this example, at that place are 23 people with a GPA greater than or equal to 3.01 (North), and they have, on average, i.04 older siblings, with a standard difference of 1.186 older siblings. In that location are 23 people with a GPA less than 3.01 (N), and they have, on average, 1.48 older siblings, with a standard departure of 1.310 older siblings. The terminal column gives the standard error of the mean for each of the two groups.

The 2nd function of the output gives the inferential statistics:

Every bit before, the columns labeled "Levene'due south Test for Equality of Variances" tell united states of america whether an assumption of the t-examination has been met. Look at the column labeled "Sig." under the heading "Levene's Test for Equality of Variances". In this example, the significance (p value) of Levene'south test is .383. If this value is less than or equal to your α level for this exam, then y'all tin can pass up the naught hypothesis that the variabilities of the 2 groups are equal, implying that the variances are unequal. In this example, .383 is larger than our α level of .05, so we will assume that the variances are equal and we will use the middle row of the output.

The column labeled "t" gives the observed or calculated t value. In this example, assuming equal variances, the t value is 1.180. (We can ignore the sign of t when using a two-tailed t-test.) The column labeled "df" gives the degrees of freedom associated with the t test. In this example, at that place are 44 degrees of liberty.

The cavalcade labeled "Sig. (2-tailed)" gives the two-tailed p value associated with the examination. In this case, the p value is .244. If this had been a one-tailed exam, we would need to look upward the critical t in a table.
Decide if we tin turn down H₀: As earlier, the decision rule is given by: If p ≤ α , then decline H₀. In this example, .244 is greater than .05, so we fail to decline H₀. That implies that there is not sufficient show to conclude that people with higher or lower GPAs take different number of older siblings.

If nosotros were writing this for publication in an APA journal, we would write information technology as:
An equal variances t test failed to reveal a statistically reliable deviation between the mean number of older siblings for people with college (M = one.04, south = 1.186) and lower GPAs (M = ane.48, s = 1.310), t(44) = 1.xviii, p = .244, α = .05.

Paired Samples t-Tests

When two samples are involved and the values for each sample are collected from the same individuals (that is, each private gives us 2 values, one for each of the 2 groups), or the samples come from matched pairs of individuals so a paired-samples t-examination may be an appropriate statistic to use.

The paired samples t-examination tin be used to determine if two means are different from each other when the 2 samples that the means are based on were taken from the matched individuals or the aforementioned individuals. In this example, we will make up one's mind if the students have different numbers of younger and older siblings.

Write the cipher and alternative hypotheses:
H₀: µ_older = µ_younger
H₁: µ_older ≠ µ_younger
Where µ is the mean number of siblings that the PSY 216 students have.
Make up one's mind if this is a one-tailed or a two-tailed examination. Considering the hypothesis involves the phrase "different" and no ordering of the means is specified, this must exist a two tailed exam.
Specify the α level: α = .05
Determine the advisable statistical test. The variables of interest, older and younger, are on a ratio calibration, so a z-score test or a t-examination might exist appropriate. Considering the population standard deviation is not known, the z-test would be inappropriate. Furthermore, the aforementioned students are reporting the number of older and younger siblings, nosotros have a within-subjects design. Because of these factors, nosotros will use the paired samples t-examination.
Allow SPSS calculate the value of t for y'all.
The command for the paired samples t tests is found at Analyze | Compare Means | Paired-Samples T Test (this is autograph for clicking on the Analyze menu item at the top of the window, and then clicking on Compare Means from the drib downwardly bill of fare, and Paired-Samples T Test from the popular upwardly menu.):

The Paired-Samples t Test dialog box will announced:

You must select a pair of variables that represent the two conditions. Click on one of the variables in the left paw pane of the Paired-Samples t Test dialog box. Then click on the other variable in the left mitt pane. Click on the pointer button to motion the variables into the Paired Variables pane. In this example, select Older and Younger variables (number of older and younger siblings) and and so click on the arrow button to move the pair into the Paired Variables box:

Click on the OK button in the Paired-Samples t Examination dialog box to perform the t-exam. The output viewer will appear with the results of the t test. The results take iii primary parts: descriptive statistics, the correlation between the pair of variables, and inferential statistics. Start, the descriptive statistics:

This gives the descriptive statistics for each of the two groups (as defined past the pair of variables.) In this instance, there are 45 people who responded to the Older siblings question (N), and they have, on average, 1.24 older siblings, with a standard deviation of 1.26 older siblings. These same 45 people also responded to the Younger siblings question (N), and they have, on average, 1.13 younger siblings, with a standard deviation of 1.20 younger siblings. The last cavalcade gives the standard error of the mean for each of the two variables.

The 2nd part of the output gives the correlation betwixt the pair of variables:

This again shows that there are 45 pairs of observations (N). The correlation betwixt the two variables is given in the third cavalcade. In this instance r = -.292. The concluding cavalcade requite the p value for the correlation coefficient. Equally ever, if the p value is less than or equal to the alpha level, so you can reject the null hypothesis that the population correlation coefficient (ρ) is equal to 0. In this case, p = .052, so we neglect to reject the zip hypothesis. That is, there is insufficient evidence to conclude that the population correlation (ρ) is different from 0.

The 3rd part of the output gives the inferential statistics:

The column labeled "Mean" is the difference of the ii means (1.24 - one.13 = 0.11 in this case (the difference is due to round off error).) The next column is the standard deviation of the difference between the two variables (1.98 in this example.)

The column labeled "t" gives the observed or calculated t value. In this example, the t value is 0.377 (you can ignore the sign.) The column labeled "df" gives the degrees of freedom associated with the t test. In this instance, there are 44 degrees of liberty. The column labeled "Sig. (2-tailed)" gives the two-tailed p value associated with the examination. In this case, the p value is .708. If this had been a ane-tailed examination, we would demand to expect up the critical value of t in a tabular array.
Make up one's mind if we can reject H₀: As earlier, the conclusion dominion is given past: If p ≤ α, and then reject H₀. In this case, .708 is non less than or equal to .05, and so nosotros neglect to pass up H₀. That implies that at that place is insufficient bear witness to conclude that the number of older and younger siblings is different.

If nosotros were writing this for publication in an APA periodical, we would write it as:
A paired samples t test failed to reveal a statistically reliable deviation between the mean number of older (M = 1.24, south = 1.26) and younger (Yard = ane.13, s = 1.20) siblings that the students have, t(44) = 0.377, p = .708, α = .05.

Source: https://academic.udayton.edu/gregelvers/psy216/spss/ttests.htm

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