# StanCrunch Assignment 3 – Answered

1. ### Formulating the Question

How are students distributed across the classes of freshman, sophomore, junior, and senior)? Does the proportion of females change across the classes?

1. ### Methodology Used to Answer the Posted Question

#### Variable of Interest

The variable of interest in this case will gender and class. The paper will examine the possibility of selecting a female from the population who belong to the freshman classes.

#### Confidence Interval

The confidence interval to be used is 95%. To obtain confidential interval, we will need to calculate the sample mean, and the sample’s standard deviation. The confidence interval is computed by obtaining the marginal error. In this case marginal error is computed as Za/2 * σ/√(n).  Where Za/2 is the confidence coefficient, with a representing the confidence level, σ representing the standard deviation, where n represents the sample size. The coefficient value is 95% which is equivalent to 0.95 when divided by 2 it results to 0.475. This based on z tables is equivalent to 1.96. Therefore the confidence interval will be obtained by applying this formula Za/2 * σ/√(n) with the final value being given as  ± X

#### Null and Alternative Hypothesis

My Null hypothesis is: A random selection from the population will result to a female from freshman classes

Alternative hypothesis: A random selection from the population will result to a male from other classes other than freshman.

#### One-Sample versus Two-Sample Test

One-sample test involving one group sample taking different tests on similar condition or taking similar test on different conditions. On the other hand, two-sample test involves two different samples taking similar on different conditions. This research will be comprised of one sample test with 30 sample items which comprises of students from four classes that include freshman, senior, junior and sophomore, and from two genders that include male and female.

#### Type of Test (Upper, Lower or Two Tailed Test)

An upper tailed test is used when the mean is greater than the tested value x, while the lower tailed test is used when the mean is less than the tested value x. However a two tailed test is used when the analysis which is to test the value of x when the mean value is less than and more than the test value. These tests are normally used to evaluate the null hypothesis. The analysis will involve a two tailed sample with chances of having a positive and a negative occurrence on the upper and lower tail. The graphical technique that will be used to address to describe this data is graph charts.

#### Level of Analysis to Use

The level of significance defines the condition under which the null –hypothesis is accepted or rejected. In this case the level of significance will be 0.05, thus, the p – value is below the α level or the significance level. In this case, α level is the chance of rejecting the null hypothesis provided that it is true.

#### Condition for Conditional Interval and Null Hypothesis to be satisfied

For the conditional interval to be satisfied, the selected value must all in the accepted range. In this case, the selected value must fall in 95% interval and not within the minority 5%. This will make a selection acceptable. For a null hypothesis to be acceptable it must show a higher chance of being selected as compared to the alternative hypothesis. In this regard, the p-value must be far higher than 50% which demonstrate that the null hypothesis option has a higher chance of being selected.

1. #### Implementing the methodology

The Upper and Lower Bound

The confidence interval will be statistically evaluated as

 Column1 Mean 27263.77 Standard Error 2194.782 Median 30927.5 Mode #N/A Standard Deviation 12021.32 Sample Variance 1.45E+08 Kurtosis -0.64822 Skewness -0.60581 Range 42831 Minimum 3094 Maximum 45925 Sum 817913 Count 30 Confidence Level (95.0%) 4488.834

In this regard, the confidence interval will be X ± 4488.8. Therefore, the upper bound will be equivalent to Mean + 4488.0 while the lower bound will be equivalent to X – 4488.8.