A researcher records the crying time (in minutes) of infants given oral administ

Question

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) minutes; in the tap water group, it was 7.3±1.6 (M±SD) minutes. 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.

Explanation / Answer

Solution:- C) Crying time was significantly reduced among infants given the sucrose solution prior to a heel prick.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 - 2 = 0

Alternative hypothesis: 1 - 2 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]

SE = 0.80

DF = 26

t = [ (x1 - x2) - d ] / SE

t = - 5.0

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 26 degrees of freedom is more extreme than -5.0; that is, less than -5.0 or greater than 5.0.

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

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