One sample t-test
Hypotheses, What test, Scale and Reliability, Average score
To test whether { },a one-sample t-test was performed.
{ } was measured using { how many questions}on a { } point scale.
It shows good/bad reliability α={ }. The average score was __(SD=__).
α >.7 → good reliability
Assumptions- One sample t-test (the median of the scale → lowest + largest / 2) Check CLT
- Normally Distributed
The data score was normally distributed (z-score skewness=__ / z-score kurtosis=__ ).
- NOT Normally Distributed BUT CLT applies
The data score was not normally distributed (z-score skewness=__ / z-score kurtosis=__ ).
But since the number of participants is superior to __, we can apply the CLT and run a normal one-
sample t-test.
- NOT Normally Distributed and CLT doesn’t apply
The data score was not normally distributed (z-score skewness=__ / z-score kurtosis=__ ).
Therefore, we report the Wilcoxon Rank statistic and the median along with the mean.
z-score: Skewness / SE; Kurotisis/SE→ z between +1.96&1.96→normally distributed
Results - One sample t-test
- Normally Distributed or NOT Normally Distributed BUT CLT applies
On average, the difference between
{the sample average} and {the central value of the scale __} was / was not significant (Mdif= __ ,
t(df) =__, p < .001 or p=__ , 95% CI [__, __]).
The difference represents a small / medium / large- sized effect d =__.
Irrelevant effect size: d < 0.2; Small effect size: 0.2 < d < 0.4; Medium effect size: 0.4 < d < 0.8; Large effect size: d > 0.8
- NOT Normally Distributed and CLT doesn’t apply
On average, the data { } (M = __, SD = __) was lower than / higher than / similar to the
central value of the scale__.
Scores of { } (Median=__) did / did not significantly differ from__( central value of the
scale) (W= __ , p =__).
The difference represents a small / medium / large- sized effect r =__.
Small r = .10 ; Medium r = .30 ; Large r = .50 (Between 0~1)
Reject the null or not
The null hypothesis can / cannot be rejected.
P <.05 → reject; CI doesn’t cross 0 → reject
Support initial hypothesis or not ----Check the mean score!
Based on the descriptives and the analysis, the results provide / cannot provide support for the initial
hypothesis: { }.
,
, Independent samples t-test
Hypotheses, What test, Scale and Reliability, Average score
To test whether { }, an independent samples t-test was performed.
{ } was measured using { _questions}on a { } point scale.
It shows good / bad reliability α=__. The average score was__(SD=__).
α >.7 → good reliability
Assumptions-CHECK CLT of each group
z-score: Skewness/SE; Kurotisis/SE→ z between +1.96&1.96→normally distributed | VR: High variance / low variance
- Normally Distributed
The data score was normally distributed for {group1} (z-score of skewness=__/ z-score kurtosis=__ ).
The data score was also normally distributed for {group2} (z-score of skewness=__/ z-score
kurtosis=__).
- NOT Normally Distributed BUT CLT applies
The data score was / was not normally distributed for {group1} (z-score of skewness=__/ z-score
kurtosis=__ ).
The data score was / was not normally distributed for {group2} (z-score of skewness=__/ z-score
kurtosis=__ ). But since the number of participants of {group1 / group2 / both groups} is superior to
__, we can apply the CLT and run a normal independent t-test.
- Homogeneity was met
Homogeneity of variances can be assumed (VR =__).
- Homogeneity was not met
Homogeneity of variances cannot be assumed (VR =__). The Levene’s test reconfirmed this violation
of homogeneity (P=__). Therefore, the Welch’s test was used.
- NOT Normally Distributed and CLT doesn’t apply
The data score was / was not normally distributed for {group1} (z-score of skewness=__/ z-score
kurtosis=__ ).
The data score was / was not normally distributed for {group2} (z-score of skewness=__/ z-score
kurtosis=__ ).
Therefore, we present the Mann-Whitney U statistic and the median scores.
- Homogeneity was met
Homogeneity of variances can be assumed (VR =__).
- Homogeneity was not met
Homogeneity of variances cannot be assumed (VR =__). The Levene’s test reconfirmed this violation
of homogeneity (P=__).
In Levene’s test P > .05, homogeneity is met.
Hypotheses, What test, Scale and Reliability, Average score
To test whether { },a one-sample t-test was performed.
{ } was measured using { how many questions}on a { } point scale.
It shows good/bad reliability α={ }. The average score was __(SD=__).
α >.7 → good reliability
Assumptions- One sample t-test (the median of the scale → lowest + largest / 2) Check CLT
- Normally Distributed
The data score was normally distributed (z-score skewness=__ / z-score kurtosis=__ ).
- NOT Normally Distributed BUT CLT applies
The data score was not normally distributed (z-score skewness=__ / z-score kurtosis=__ ).
But since the number of participants is superior to __, we can apply the CLT and run a normal one-
sample t-test.
- NOT Normally Distributed and CLT doesn’t apply
The data score was not normally distributed (z-score skewness=__ / z-score kurtosis=__ ).
Therefore, we report the Wilcoxon Rank statistic and the median along with the mean.
z-score: Skewness / SE; Kurotisis/SE→ z between +1.96&1.96→normally distributed
Results - One sample t-test
- Normally Distributed or NOT Normally Distributed BUT CLT applies
On average, the difference between
{the sample average} and {the central value of the scale __} was / was not significant (Mdif= __ ,
t(df) =__, p < .001 or p=__ , 95% CI [__, __]).
The difference represents a small / medium / large- sized effect d =__.
Irrelevant effect size: d < 0.2; Small effect size: 0.2 < d < 0.4; Medium effect size: 0.4 < d < 0.8; Large effect size: d > 0.8
- NOT Normally Distributed and CLT doesn’t apply
On average, the data { } (M = __, SD = __) was lower than / higher than / similar to the
central value of the scale__.
Scores of { } (Median=__) did / did not significantly differ from__( central value of the
scale) (W= __ , p =__).
The difference represents a small / medium / large- sized effect r =__.
Small r = .10 ; Medium r = .30 ; Large r = .50 (Between 0~1)
Reject the null or not
The null hypothesis can / cannot be rejected.
P <.05 → reject; CI doesn’t cross 0 → reject
Support initial hypothesis or not ----Check the mean score!
Based on the descriptives and the analysis, the results provide / cannot provide support for the initial
hypothesis: { }.
,
, Independent samples t-test
Hypotheses, What test, Scale and Reliability, Average score
To test whether { }, an independent samples t-test was performed.
{ } was measured using { _questions}on a { } point scale.
It shows good / bad reliability α=__. The average score was__(SD=__).
α >.7 → good reliability
Assumptions-CHECK CLT of each group
z-score: Skewness/SE; Kurotisis/SE→ z between +1.96&1.96→normally distributed | VR: High variance / low variance
- Normally Distributed
The data score was normally distributed for {group1} (z-score of skewness=__/ z-score kurtosis=__ ).
The data score was also normally distributed for {group2} (z-score of skewness=__/ z-score
kurtosis=__).
- NOT Normally Distributed BUT CLT applies
The data score was / was not normally distributed for {group1} (z-score of skewness=__/ z-score
kurtosis=__ ).
The data score was / was not normally distributed for {group2} (z-score of skewness=__/ z-score
kurtosis=__ ). But since the number of participants of {group1 / group2 / both groups} is superior to
__, we can apply the CLT and run a normal independent t-test.
- Homogeneity was met
Homogeneity of variances can be assumed (VR =__).
- Homogeneity was not met
Homogeneity of variances cannot be assumed (VR =__). The Levene’s test reconfirmed this violation
of homogeneity (P=__). Therefore, the Welch’s test was used.
- NOT Normally Distributed and CLT doesn’t apply
The data score was / was not normally distributed for {group1} (z-score of skewness=__/ z-score
kurtosis=__ ).
The data score was / was not normally distributed for {group2} (z-score of skewness=__/ z-score
kurtosis=__ ).
Therefore, we present the Mann-Whitney U statistic and the median scores.
- Homogeneity was met
Homogeneity of variances can be assumed (VR =__).
- Homogeneity was not met
Homogeneity of variances cannot be assumed (VR =__). The Levene’s test reconfirmed this violation
of homogeneity (P=__).
In Levene’s test P > .05, homogeneity is met.
