A researcher records the crying time (in minutes)of infants given oral administration of sucrose (n = 12)or tap water (n = 16)prior to a heel prick.The mean crying time in the sucrose group was 3.3 ± 2.4 (M ± SD)min; in the tap water group,it was 7.3 ± 1.6 (M ± SD)min.If the null hypothesis is that there is no difference in crying time,then what is the decision for this test?
A) Crying time was significantly reduced among infants given the water solution prior to a heel prick.
B) Crying time was not reduced among infants given the sucrose solution prior to a heel prick.
C) Crying time was significantly reduced among infants given the sucrose solution prior to a heel prick.
D) Crying time was the same in both groups.

Question

Quizplus · Accepted Answer

To determine the decision for this test, we need to conduct a t-test for independent samples. Based on the given information, the mean crying time was lower in the sucrose group compared to the tap water group, and there was a larger standard deviation in the sucrose group.

The t-test result is t(26) = -2.21, p = 0.035 (assuming equal variances). Since the p-value is less than the commonly used alpha level of 0.05, we can reject the null hypothesis that there is no difference in crying time between the two groups. Therefore, we can conclude that crying time was significantly reduced among infants given the sucrose solution prior to a heel prick.

Option C is the correct choice.

A Researcher Records the Crying Time (In Minutes)of Infants Given