Hits: 0
Inferential Statistics Analysis Identifying Information
Student (Full Name):
Class: Intro to STAT 200
Instructor:
Date: 27 July 2021
Part A: Inferential Statistics Data Analysis Plan and Computation
Introduction:
Variables Selected:
Table 1: Variables Selected for Analysis Inferential Statistics Analysis
| Variable Name in the Data Set | Variable Type | Description | Qualitative or Quantitative |
| Variable 1: Marital Status | Socioeconomic | Marital Status of Head of Household | Qualitative |
| Variable 2: Food | Expenditure | Total Amount of Annual Expenditure on Food | Quantitative |
| Variable 3: Entertainment | Expenditure | Total Amount of Annual Expenditure on Entertainment | Quantitative |
Data Set Description and Method Used for Analysis:
Inferential Statistics Analysis The data set contains information from 30 households, where a survey responder provided the requested information; it is all self-reported information. The data set file has 8 variables in which I have chosen 3, and I will use Microsoft Excel, web applets, and a TI Calculator to analyze my data.
- Confidence Interval Analysis:
Table 2: Confidence Interval Information and Results
| Name of Variable: Food |
| State the Random Variable and Parameter in Words:
x = annual expenditure on food µ = mean annual expenditure on food |
| Confidence interval method including confidence level and rationale for using it:
Confidence Interval for One Population Mean (t-Interval) at 95% since the larger the sample size is, the closer to the true population. Inferential Statistics Analysis |
| State and check the assumptions for confidence interval:
A random sample of 30 household’s annual expenditure on food was taken. This was stated in the data set. The population of the annual expenditure on food is normally distributed. The sample size is 30 or more. |
| Method Used to Analyze Data: Web applet and TI calculator |
| Find the sample statistic and the confidence interval:
x = 255,146.01 x ̅ = 8504.867 s = 1645.068 n = 30 df = n-1 = 30-1 = 29 C = 95% = 2.045 E = tc s/√n = 2.045*1645.068/√30 = 614.21 x ̅-E<µ< x ̅+E = 8504.867-614.21< µ<8504.867+614.21 = 7890.6 < µ<9119.1 |
| Statistical Interpretation: There is a 95% chance that 7890.6 < µ<9119.1 contains the mean annual expenditure on food. |
- Hypothesis Testing:
Table 3: Two Sample Hypothesis Test Analysis
| Research Question: Is there enough evidence to show that the annual expenditures on entertainment for married household is greater than not married household? Test at the 5% level.
|
| Two Sample Hypothesis Test that Will Be Used and Rationale for Using It:
Hypothesis Test for Independent t-Test (2-Sample t-Test) in order to compare the two independent groups on whether there is evidence of the two means being significantly different. |
| State the Random Variable and Parameters in Words:
x1= annual expenditures on entertainment for married household x2= annual expenditures on entertainment for not married household µ1= mean annual expenditures on entertainment for married household µ2= mean annual expenditures on entertainment for not married household |
| State Null and Alternative Hypotheses and Level of Significance:
Ho: µ1= µ2 or Ho: µ1- µ2= µd=0 Ha: µ1> µ2 or Ha: µ1- µ2= µd>0 α = 0.05 |
| Method Used to Analyze Data: Excel |
| Find the sample statistic, test statistic, and p-value:
x ̅_1=$125.53 s1=$49.61 x ̅_2= $95.80 s2= $9.43 s_1^2= 49.61^2= 2461.15 s_2^2= 9.43^2= 88.92 t = (x1-x2)-(µ1-µ2)/√s_1^2/n1+s_2^2/n2 = (125.53-95.80)-0/√2461.15/15+88.92/15 = 2.28 df= (164.08+5.93)^2/164.08^2/15-1+5.93/15-1 = 15.01 p value: Excel: =T.DIST= (2.28,15.01,true)= 0.98 |
| Conclusion Regarding Whether or Not to Reject the Null Hypothesis:
Since the p-value is greater than the level of significance, we fail to reject Ho. There is not enough evidence to show that the annual expenditures on entertainment for married household is greater than not married household.
|
Part B: Results Write Up
Confidence Interval Analysis:
Two Sample Hypothesis Test Analysis:
Discussion: