Exercise 29 Statistices

We conduct an independent sample t-test using Excel, and obtain the following output (see t-test-height)

2) Compute the standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable?

8. The researchers state that the sample for their study was 28 women with a diagnosis of OA, and that 18 were randomly assigned to the intervention group and 10 were randomly assigned to the control group. Discuss the study strengths and/or weaknesses in this statement.

What does the t value for the Physical Component Score tell you about men and women post MI? If this result was consistent with previous research, how might you use this knowledge in your practice?

We can use z-table instead of t-table because the standard deviation comes from the population.

where n is the sample size, t is the t-student value for the respective level of

The t-test is a parametric analysis technique used to determine significant differences between the scores obtained from two groups. The t-test uses the standard deviation to estimate the standard error of the sampling distribution and examines the differences between the means of the two groups. Since the t-test is considered fairly easy to calculate, researchers often use it in determining differences between two groups. When interpreting the results of t-tests, the larger the calculated t ratio, in absolute value, the greater the difference between the two groups. The significance of a t ratio can be determined by comparison with the critical values in a

Gender – whether there was a difference in performance between genders; used for comparison between male and female participants

The null hypothesis was that the female and male shoe sizes have an equal mean while the alternative hypothesis was that female and male shoe sizes do not have an equal mean. With the degrees of freedom being 33, the t-statistic is -8.27. The probability that -8.27 is ≤-1.69 is 7.5×10-10 for the one-tailed test. Also, the probability that -8.27 is ≤ ±2.03. is 1.5×10-9 for the two-tailed test. Due to both probabilities being under the alpha value of 0.05, the null hypothesis is rejected, and the alternative hypothesis is accepted at the 95% confidence level.

It tells that the t-statistic with 97 degrees of freedom was 2.14, and the corresponding p-value was less than .05, specifically around 0.035. Therefore, it is appropriate to conclude the research study was statistically significant.

However, treatment four, 0.1296 (±0.608), represents that the mean was extraneous from what it should be (Table 1). The t-tests show how different the mean is in each treatment.

groups. The paired t-test is used when two measures of the same interest are taken on

For this independent t test, the mean GPAs of 64 females and 41 males were compared. The variables used are (1) gender, and (2) GPA. The predictor, or independent, variable is gender. And the outcome variable is GPA. Gender can only have two values, male or female; this

Since 3.27 the t statistic is in the rejection area to the right of =1.701, the level of

In this case, level of significance, α was not provided. Therefore, the analysis will be evaluated based on two α values which are: