Question

Part A: Inferential Statistics Data Analysis, Plan and Computation

Introduction:

Variables Selected: Any two variables out of Income, Age, Food, Meat, Bakery, Fruits.

Table 1: Quantitative Variables from the given Dataset spreadsheet Selected for Analysis

Variable Name	Description
Variable 1:
Variable 2:

Data Analysis:

1. Confidence Interval Analysis: For one quantitative variable, select and run the appropriate method for estimating a parameter, based on a statistic (i.e., confidence interval method) and complete the following Table 2.

Table 2: Confidence Interval Information and Results

Name of Variable:   Select one Quantitative Variable from Table 1

State confidence level and method, you will use to find Confidence Interval

State and check the assumptions for Confidence Interval:

Find the sample mean, sample standard deviation and t-value for sample statistic: Find Margin of Error and create the Confidence Interval:

Statistical Interpretation:

2. Hypothesis Testing: Using the second variable from Table 1, select data from the given Dataset for Married and Not Married people. Run two samples hypothesis testing method (Married and Not Married). Show results in the Table 3.

Table 3: Two Sample Hypothesis Test Analysis

Research Question: Does married person spend on … more than not married?

Describe in words parameters you will use in Hypothesis Testing test: X1 – X2 – S1 – S2 – n1 – n2 – m1 – m2 –

State Null and Alternative Hypotheses. Identify it as Two-Tailed or One-Tailed Test. Use Level of Significance a = 0.05

Calculate the sample statistic, test statistic, and p-value: Sample Statistics: x1 = … x2 = … S1 = … S2 = … n1 = … n2 = … Test Statistics t = x1-x2S12n1+S22n2 = … To find p-value you can use Excel function p-value = TDIST(t, df, 1) = …

Conclusion Regarding Whether or Not to Reject the Null Hypothesis:

.

Part B: Results Write Up

Confidence Interval Analysis: 95% Confidence Interval for Population Mean is (… , …)

meaning generally expenses on … is between … and …

Two Sample Hypothesis Test Analysis:
Describe in words results of Hypothesis Testing.

Answer 1

Sol:

Part - A:

The variables selected for an Inferential data analysis plan are Marital status, Food expenses and Housing expenses. The variable Marital status is qualitative, and the variables Food expenses and Housing expenses are quantitative. 1. Assume that the expenditure variable selected for parameter estimation using confidence interval is Food expenses. To obtain the confidence interval, compute the mean and standard deviation of the data of food expenses as shown in the steps below: x X 30 6900+6937+...+9035 30 258579 30 = 8619.3 30 (*; -17) S= 30-1 (6900 -8619.3)” +(6937 – 8619.3)” +...+(9035 – 8619.3)? 29 = 1824.88 As the sample size is 30 and the population standard deviation is unknown, therefore, a t- test statistic will be used to construct the confidence interval. The degrees of freedom for a t-test are obtained as follows: Degrees of freedom = n-1 = 30-1 = 29 As the level of significance is not provided, so consider it as 5%, that is, a = 0.05. The two-tailed critical value of the t-test statistic is obtained from the MS-Excel function as shown in the screenshot below: =T.INV.2T(0.05,29)

The above function yields a critical value of 2.0452. Now, the 95% confidence interval for the food expenses is calculated by substituting the sample values in the following formula: S CI = īsto 2 vn 1824.88 = 8619.3 + 2.0452 - V30 = 8619.3 +681.412 = (8619.3 -681.412,8619.3 +681.412) (7937.888,9300.712) Therefore, the required confidence interval is (7937.89, 9300.71). Hence, the required answers are: Name of variable: Food Expenses Random variable: The variable X assuming value of the Food expenses of people in the sample. Parameter: The average food expenses of people in population. Confidence level & reason: As the level of significance is not provided, so it is considered as 5%. In most of the situations, the inferential procedures are performed at 95% confidence level. Assumptions for confidence interval: The parent population is assumed to be normally distributed and a random sample should be selected from the population. Also, the sample values are independently collected. Method used: A t-test statistic will be used to construct the confidence interval. The reason of using the t-test statistic is the unknown population standard deviation. Confidence interval: The obtained confidence interval for the average food expenses is (7937.89, 9300.71). Interpretation: The 95% confidence interval implies that that if a large number of independent samples are taken then 95% of the sample means will lie in the obtained interval. 2.

Another expenditure variable is Housing expenses. Considering marital status as the grouping variable, a two-sample hypothesis procedure will be conducted to test the significance of the difference between Food expenses and Housing expenses. Research Question: Is there is significant difference between the housing expenses for married and unmarried people. Two-sample hypothesis test that will be used: As the population standard deviation is unknown, and the variables are quantitative, therefore, a two-sample t-test will be sued to answer the above framed research question. Null and alternative hypothesis and the level of significance:

The null hypothesis will assume that there is no significant difference between the average housing expenses of married and unmarried people and the alternative hypothesis will assume that there is a significant difference between the average housing expenses of married and unmarried people. If y, and I are the true mean housing expenses of married and unmarried people, then the hypothesis is formulated as follows: H:44 = 112 H:147 the As the level of significance is not provided, so test the above hypothesis at 5% level of significance, that is, a = 0.05. In most of the practical situations, the confidence level is taken to be 95%. Method used to analyze data: A two-sample test statistic will be sued to analyze the given data. The test statistic is calculated by the following formula: Ã - 2 t = 2 + S2 n2 n The above test statistic will follow student's t-distribution with degrees of freedom (n +n2-2). Computation of sample statistic, test statistic and p-value: Compute the sample mean and standard deviation by using the following formula: 15 . i=1 = 15 23063+23155+...+22867 15 = 23880 15 į (*; - ;) 2 15-1 (23063 – 23880)? +(23155 – 23880) +...+(22867 – 23880)? = 14

= 3337929 15 ΣΑ, i=1 15 18514+18619+...+18389 15 = 18522 15 (*; - 13) j=1 S2 15-1 (18514–18522) +(18619-18522)? + +...+(18389-18522) 14 = 22201 Now, compute the test statistic by using the following formula: 2 7 - 82 si пп, 23880-18522 3337929 18522 + 15 15 = 11.3268 Therefore, the calculated test statistic is 11.3268. The degrees of freedom for a t-test are obtained as follows: Degrees of freedom = n +n -2 = 15+15-2 = 30-2 = = 28 The p-value is obtained by using the MS-Excel function as shown in the screenshot below: =T.INV.2T(0.05,28)

The above function yields a p-value of 2.0484. Conclusion: As the p-value is greater than 0.05, the level of significance, therefore, we fail to reject the null hypothesis, and it is concluded that there is no significant difference between the average housing expenses of married and unmarried people.

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