Inferential Statistics Analysis and Writeup

Inferential Statistics Analysis and Writeup

Assignment #3: Inferential Statistics Analysis and Writeup

Part A: Inferential Statistics Data Analysis Plan and Computation

Introduction:

Variables Selected:

Table 1: Variables Selected for 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:

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.

1. 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.

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.

2. 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:

STAT200: Assignment #3 – Inferential Statistics Analysis and Writeup Page 1 of 4

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