Diet and iron status

Which statistical test should be performed to analyze the data in scenario 1?

Ans: one-way ANOVA

In question 1 you were asked to identify which statistical test should be performed to analyze the data in scenario 1. Why did you choose this statistical test?

According to the significance value of Kolmogorov Simonov test normality test, it is greater than 0.05 so null hypothesis is accepted which depicted that the population the sample represents is normally-distributed. To investigate variances observed in all conditions for the 3 different dietary patterns, one-way ANOVA, a parametric test, is chosen. The test is needed for determining if there are any statistically significant variances between the means of three or more independent variables. For those who need assistance in navigating this complex analysis, seeking statistics dissertation help can provide valuable insights and guidance.

The p value obtained from statistical analysis

Ans: .000

When interpreting the statistical analysis the p value obtained may be referred to as

Ans: p<0.05

It was appropriate to perform a post-hoc test.

Ans: yes

A clear summary version of the data using an appropriate graph or table

Ans: H0: there is no statistically significant difference between the three types of dietary patterns.

H1: there is statistically significant difference between the three types of dietary patterns.

The ANOVA table shows that the p-value of the analysis is reported as 0.000 and it is much lesser than 0.05. It means that obtainable null hypothesis is rejected which means that there is statistically significant variance in the three types of 3 different dietary patterns.

Here, Group A denotes 1, Group B denotes 2, Group C denotes 3

Energy intake and birth weight

Which statistical test should be performed to analyze the data in scenario 2?

Ans: Mann Whitney test

Statistical test should be performed to analyze the data in scenario 2.

Ans: According to the significance value of Kolmogorov Simonov test and Shapiro Wilk

The p value obtained from statistical analysis

Ans: .014

When interpreting the statistical analysis the p value obtained may be referred to as

A clear summary version of the data using an appropriate graph or table

Null hypothesis (H0): the mean of change of mean of consuming energy of pregnant women carrying male fetuses is not equal to the mean of consuming energy of pregnant women carrying female fetuses.

Alternative hypothesis (H1): Means are not equal

According to the result of Mann Whitney test, the associated p-value of t-test of equality of mean is .778 (U statistics = 294) which is greater than p value 0.05. So null hypothesis is accepted which depicted that means the mean of consuming energy of pregnant women carrying male fetuses is equal to the mean of consuming energy of pregnant women carrying female fetuses.

So. The theory proved wrong.

Here 1 = women with male babies 2 = women with female babies

A novel anti-tumor agent

Which statistical test should be performed to analyze the data in scenario 3?

Ans: Chi squared

Statistical test should be performed to analyze the data in scenario 3.

Ans: For determining the effectiveness of anticancer agent, Chi square test has been chosen. The chi-square test for independence is used for discovering any type of association or relationship between two categorical variables.

The p value obtained from statistical analysis

Ans: .012

When interpreting the statistical analysis the p value obtained may be referred to as

Ans: p<0.05

It was appropriate to perform a post-hoc test.

Ans: No

A clear summary version of the data using an appropriate graph or table

Null hypothesis (H0): Two independent variable are independent of each other.

Alternative hypothesis (H1): Two independent variable are independent of each other.

As per the table of chi square test, the value of the test statistic is 6.282 with 1 degree of freedom and the corresponding p-value of the test statistic is p =.012 which is p-value is lesser than chosen significance level (α = 0.05). So null hypothesis is rejected and it can be established that there is enough evidence to prove an association between anticancer agent and its effectiveness in curing cancer.

The importance of study

Which statistical test should be performed to analyze the data in scenario 4?

Ans: Pearson

The p value obtained from statistical analysis

Ans: Hours spent revising = .000 and Hours spent sleeping = .796

When interpreting the statistical analysis the p value obtained may be referred to as

Ans: p<0.01

It was appropriate to perform a post-hoc test.

Ans: No

A clear summary version of the data using an appropriate graph or table

Null hypothesis = the relation between two variable is not statistically significant.

According to result of Pearson correlation test, corresponding p-value is .000 for hours spent revising which is p-value is lesser than chosen significance level (α = 0.01). So, null hypothesis is rejected which depicted that the association between marks and hours spent revising is statistically significant. Corresponding p-value is .796 for hours spent sleeping which is p-value is greater than chosen significance level (α = 0.01). So, null hypothesis is rejected which depicted that the association between marks and hours spent revising is not statistically significant. According to the result, a strong and positive linear relationship between marks and hours spent revising as the correlation coefficient value is reported as .844.

So, it can be said that hours spent revising is being predictive for the exam mark.

Antioxidants in red wine

Which statistical test should be performed to analyze the data in scenario 5?

Ans: Paired T test

In question 1 you were asked to identify which statistical test should be performed to analyze the data in scenario 5. Why did you choose this statistical test?

According to the significance value of Kolmogorov Simonov test and Shapiro Wilk test normality tests, it is greater than 0.05 so null hypothesis is accepted which depicted that the population the sample represents is normally-distributed. To investigate variances observed in all conditions for the antioxidant capacity of LDL of same woman, paired t test, a parametric test, is chosen. The test is needed for determining if there are any statistically significant variances between the means of two paired observations of the dependent samples.

The p value obtained from statistical analysis

Ans: .000

When interpreting the statistical analysis the p value obtained may be referred to as

Ans: p<0.05

It was appropriate to perform a post-hoc test.

Ans: No

A clear summary version of the data using an appropriate graph or table

Ho: The mean difference between two cases is equal to zero.

Alternative hypothesis (H1): The mean difference between two cases is not equal to zero.

According to the result of independent sample test, the associated p-value of t-test of equality of mean (t (9) = 8.599) is .000 that is lesser than significance level 0.05. So null hypothesis is rejected which depicted that the true mean difference between the rate of the antioxidant capacity of LDL is not equal to zero.

In precisely it can be said that there was a statistically significant increase of antioxidant capacity of LDL at the time of consuming Vodka from consuming red wine and the increase in vodka is reported as 3.96 ± 0.72 to 4.82 ± 0.86 (p < 0.0005).