Statistical Hypothesis Test

So, we should reject the null hypothesis H0. At a 0.05 level of significance level, we conclude that there is a significant difference between the average height for females and the average height for the males.

Because the p-value of .035 is less than the significance level of .05, I will reject the null hypothesis at 5% level.

Testing allows the p-value that represents the probability showing that results are unlikely to occur by chance. A p-value of 5% or lower is statistically significant. The p value helps in minimizing Type I or Type II errors in the dataset that can often occur when the p value is more than the significance level. The p value can help in stopping the positive and negative correlation between the dataset to reject the null hypothesis and to determine if there is statistical significance in the hypothesis. Understanding the p value is very important in helping researchers to determine the significance of the effect of their experiment and variables for other researchers

Select one (1) project from your working or educational environment that you would use the hypothesis test technique. Next, propose the hypothesis structure (e.g., the null hypothesis, data collection process, confidence interval, test statistics, reject or not reject the decision, etc.) for the business process of the selected project. Provide a rationale for your response.

The null hypothesis is rejected since the p-value is below the significance level of 0.05.

When you perform a test of hypothesis, you must always use the 4-step approach: i. S1:the “Null” and “Alternate” hypotheses, ii. S2: calculate value of the test statistic, iii. S3: the level of significance and the critical value of the statistic, iv. S4: your decision rule and the conclusion reached in not rejecting or rejecting the null hypothesis. When asked to calculate p–value, S5, relate the p-value to the level of significance in reaching your conclusion.

If 9 t tests were conducted and the set alpha for this study is 0.05, then the alpha level that should be used to determine the differences between the two groups is 0.05/9=0.0056 and the resulting alpha will be used to determine significant differences.

Since the P-value (0.386) is greater than the significance level (0.05), we fail to reject the null hypothesis. The p-value implies the probability of rejecting a true null hypothesis.

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.

To test the null hypothesis, if the P-Value of the test is less than 0.05 I will reject the null hypothesis.

“Hypothesis testing is a decision-making process for evaluating claims about a population” (Bluman, 2013, p. 398). This process is used to determine if you will accept or reject the hypothesis. The claim is that the bottles contain less than 16 ounces. The null hypothesis is the soda bottles contain 16 ounces. The alternative hypothesis is the bottles contain less than 16 ounces. The significance level will be 0.05. The test method to be used is a t-score. The test statistic is calculated to be -11.24666539 and the P-value is 1.0.  The P-value is the probability of observing a sample statistic as extreme as the test statistic, assuming the null hypothesis is true. The T Crit value is 1.69912702. The calculations show there is enough evidence to support the claim that the soda bottles do

At the .01 significance level is there a difference in the mean amount purchased on an impulse at the two stores? Explain these results to a person who knows about the t test for a single sample but is unfamiliar with the t test for independent means.

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

COMMENTS argument is that because the average effect size for published research was equivalent to that of a medium effect, the reviewer 's decision to reject the bogus manuscript under the nonsignificant condition was "reasonable." Further examination of the Haase et al. (1982) article and our own analysis of published research, however, demonstrates that the power of the bogus study was great enough to detect effect sizes that are typical of research published in JCP, which was our intention when we designed the bogus study. First, although the median effect size (if) for all univariate statistical tests, significant and nonsignificant, reported by Haase et al. (1982) was .083, this index was steadily increasing at a rate of approximately .5% per year, so that the projected median if- in 1981 (the year our study was completed) would be .13. Importantly, an r)2 of .13 corresponds to an effect size (/) of .39, which Cohen (1977) designates as a large effect. A further examination of the Haase et al. (1982) data also lends support to our argument. Their analysis examined the strength of association for 11,044 univariate statistical tests derived from only 701 manuscripts; thus, each manuscript reported an average of more than 15 statistical tests. Since statistically significant and

With a P-value of 0.00, we have a strong level of significance. No additional information is needed to ensure that the data given is accurate.