Mod 3. The t-test: Comparing Two Sets of Data-Case Study
I. INDEPENDENT t-Test
Research question “is there a statistical difference for mean heart rate between
participants either exposed to a hot- or cold environment?”
Assumptions Testing
1. Data level of measurement- what is the level of measurement for the data used in this
case study?
Interval and Ratio.
2. Would the amount of skewness and kurtosis in these variables affect the analysis? How
do you know? Give numerical values to support your conclusion.
On the Descriptives table, for the grouping variable “Mean of heart rate” hot, the
skewness statistic is .053 with Std. Error of .687. There is no skewness present when
calculating. The value reads: .077. The Kurtosis statistic for hot, referencing the
Descriptives table, is -.927 with Std. Error of 1.334. There is no sign of kurtosis as we
have a value of -.69 when calculated. Cold skewness statistic is .696 with a Std. Error of .
687. After calculations this shows a value of 1.01. Kurtosis statistic is -.698 with a Std.
Error of 1.334. After calculations we have a value of -.523. We use +-1.96 for skewness
and kurtosis. The amount of skewness and kurtosis present will not affect the analysis.
Table 1. PASTE TABLE with skewness and kurtosis BELOW
1
,Mod 3. The t-test: Comparing Two Sets of Data-Case Study
2
, Mod 3. The t-test: Comparing Two Sets of Data-Case Study
3. Was the assumption of normality met and how do you know?
The assumption of normality is not met for the grouping variable, cold. It is known
because skewness and kurtosis values fall within +/-1.96, however, the Test of Normality
table shows sig values and cold is does not met the assumption of normality. It is
below.05 with a value of .044. Grouping variable, hot, is .801 and the assumption of
normality is met.
Table 2. PASTE Tests of NORMALITY table BELOW
Figure 1. PASTE Histogram with normal curve for each variable BELOW
3
I. INDEPENDENT t-Test
Research question “is there a statistical difference for mean heart rate between
participants either exposed to a hot- or cold environment?”
Assumptions Testing
1. Data level of measurement- what is the level of measurement for the data used in this
case study?
Interval and Ratio.
2. Would the amount of skewness and kurtosis in these variables affect the analysis? How
do you know? Give numerical values to support your conclusion.
On the Descriptives table, for the grouping variable “Mean of heart rate” hot, the
skewness statistic is .053 with Std. Error of .687. There is no skewness present when
calculating. The value reads: .077. The Kurtosis statistic for hot, referencing the
Descriptives table, is -.927 with Std. Error of 1.334. There is no sign of kurtosis as we
have a value of -.69 when calculated. Cold skewness statistic is .696 with a Std. Error of .
687. After calculations this shows a value of 1.01. Kurtosis statistic is -.698 with a Std.
Error of 1.334. After calculations we have a value of -.523. We use +-1.96 for skewness
and kurtosis. The amount of skewness and kurtosis present will not affect the analysis.
Table 1. PASTE TABLE with skewness and kurtosis BELOW
1
,Mod 3. The t-test: Comparing Two Sets of Data-Case Study
2
, Mod 3. The t-test: Comparing Two Sets of Data-Case Study
3. Was the assumption of normality met and how do you know?
The assumption of normality is not met for the grouping variable, cold. It is known
because skewness and kurtosis values fall within +/-1.96, however, the Test of Normality
table shows sig values and cold is does not met the assumption of normality. It is
below.05 with a value of .044. Grouping variable, hot, is .801 and the assumption of
normality is met.
Table 2. PASTE Tests of NORMALITY table BELOW
Figure 1. PASTE Histogram with normal curve for each variable BELOW
3
