Question

Part A: Inferential Statistics Data Analysis Plan and Computation Variables Selected: Table 1: Variables Selected for Analysis Variable Name in the Data Set Variable Type Variable1 Variable2 Variable 3 Data Analysis: 1. Confidence Interval Analysis: For one expenditure 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 (Note: Format follows Kozak outline): Table 2: Confidence Interval Information and Name of Variable: State the Random Variable and Parameter in Words: Confidence interval method including confidence level and rationale for using it State and check the assumptions for confidence interval Method Used to Analyze Data Find the sample statistic and the confidence interval: Statistical Interpretation: 2. Hypothesis Testing: Using the second expenditure variable (with socioeconomic variable as the grouping variable for making two groups), select and run the appropriate method for making decisions about two parameters relative to observed statistics (i.e, two sample hypothesis testing method) and complete the following table (Note: Format follows Kozak outline): Table 3: Two Sample Hypothesis Test Analysis Research Question: Two Sample Hypothesis Test that Will Be Used and Rationale for Using It: State the Random Variable and Parameters in Words: State Null and Alternative Hypotheses and Level of Significance Method Used to Analyze Data Find the sample statistic, test statistic, and p-value Conclusion Regarding Whether or Not to Reject the Null Hypothesis: Part B: Results Write Up Confidence Interval Analysis Two Sample Hypothesis Test Analysis:

Answer 1

Answer:

By using given data,

We have to consider three variables which are marital status, food and fruit.

Marital status is categorical variable having married and un married are two categories.

Food and fruit are quantitative variables.

1. Confidence Interval Analysis: For one expenditure 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

Here we use Confidence interval methof for variable food.

95% confidence interval for population mean (mu) is,

Xbar - E < mu < Xbar + E

where Xbar is sample mean and

E is margin of error.

E = Zc * (sd / sqrt(n))

For 95% confidence Zc = 1.96

Now we have to find sample mean and standard deviation of the data.

Xbar = 8464.83

sd = 1794.65

n = 30

E = 1.96 * (8464.83 / sqrt(30)) = 3029.10

lower limit = Xbar - E = 8464.83 - 3029.10 = 5435.73

upper limit = XBar + E = 8464.83 + 3029.10 = 11493.94

95% confidence interval for population mean is (5435.73, 11493.94)

SO population mean is lies between these two limits.

2. Hypothesis Testing: Using the second expenditure variable (with socioeconomic variable as the grouping variable for making two groups), select and run the appropriate method for making decisions about two parameters relative to observed statistics (i.e., two sample hypothesis testing method) and complete the following table

Now we use second variable as fruit and grouping variable as marital status.

We can test here,

H0 : mu1 = mu2 Vs H1 : mu1 not= mu2

where mu1 and mu2 are two population means for malee and females.

Assume alpha = level of significance = 0.05

We can use here two sample t-test assuming equal variances.

We can do this test in MINITAB.

ENTER data into MINITAB sheet --> Stat --> Basic statistics --> 2-Sample t --> Both samples are in one column --> Samples : fruit --> Sample ids : marital status --> Options --> Confidence levle : 95.0 --> Hypothesized difference : 0.0 --> Alternative hypothesis : not equal --> Assume equal variances --> ok --> ok

————— 06-08-2018 11:03:00 ————————————————————

Welcome to Minitab, press F1 for help.

Two-Sample T-Test and CI: Fruit, marital status

Two-sample T for Fruit

marital
status N Mean StDev SE Mean
0 15 1115 107 28
1 15 1226 1782 460

Difference = ? (0) - ? (1)
Estimate for difference: -111
95% CI for difference: (-1055, 833)
T-Test of difference = 0 (vs ?): T-Value = -0.24 P-Value = 0.811 DF = 28
Both use Pooled StDev = 1262.2626

Test statistic = -0.24

P-value = 0.811

P-value > alpha

Accept H0 at 5% level of significance.

COnclusion : Two population means do not differ.