A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment. Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population 1) and the annual credit card charges for dining out (population 2). Using the difference data, with population 1 − population 2, the sample mean difference was d = $870, and the sample standard deviation was sd = $1,124. (a) Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out. H0: ?d ≤ 0 Ha: ?d > 0 H0: ?d < 0 Ha: ?d = 0 H0: ?d ≥ 0 Ha: ?d < 0 H0: ?d = 0 Ha: ?d ≠ 0 H0: ?d ≠ 0 Ha: ?d = 0 (b) Calculate the test statistic. (Round your answer to three decimal places.) What is the p-value? (Round your answer to four decimal places.) Can you conclude that the population means differ? Use a 0.05 level of significance. The p ≤ 0.05, therefore we cannot conclude that there is a difference between the annual population mean expenditures for groceries and for dining out. The p ≤ 0.05, therefore we can conclude that there is a difference between the annual population mean expenditures for groceries and for dining out. The p > 0.05, therefore we cannot conclude that there is a difference between the annual population mean expenditures for groceries and for dining out. The p > 0.05, therefore we can conclude that there is a difference between the annual population mean expenditures for groceries and for dining out. (c) What is the point estimate (in dollars) of the difference between the population means? $ What is the 95% confidence interval estimate (in dollars) of the difference between the population means? (Round your answers to the nearest dollar.) $ to $ Which category, groceries or dining out, has a higher population mean annual credit card charge? The 95% confidence interval ---Select--- contains or is completely below or is completely above zero. This suggests that the category with higher mean annual expenditure is ---Select--- groceries or dining out .

A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment. Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population 1) and the annual credit card charges for dining out (population 2). Using the difference data, with population 1 − population 2, the sample mean difference was d = $870, and the sample standard deviation was sd = $1,124. (a) Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out. H0: ?d ≤ 0 Ha: ?d > 0 H0: ?d < 0 Ha: ?d = 0 H0: ?d ≥ 0 Ha: ?d < 0 H0: ?d = 0 Ha: ?d ≠ 0 H0: ?d ≠ 0 Ha: ?d = 0 (b) Calculate the test statistic. (Round your answer to three decimal places.) What is the p-value? (Round your answer to four decimal places.) Can you conclude that the population means differ? Use a 0.05 level of significance. The p ≤ 0.05, therefore we cannot conclude that there is a difference between the annual population mean expenditures for groceries and for dining out. The p ≤ 0.05, therefore we can conclude that there is a difference between the annual population mean expenditures for groceries and for dining out. The p > 0.05, therefore we cannot conclude that there is a difference between the annual population mean expenditures for groceries and for dining out. The p > 0.05, therefore we can conclude that there is a difference between the annual population mean expenditures for groceries and for dining out. (c) What is the point estimate (in dollars) of the difference between the population means? $ What is the 95% confidence interval estimate (in dollars) of the difference between the population means? (Round your answers to the nearest dollar.) $ to $ Which category, groceries or dining out, has a higher population mean annual credit card charge? The 95% confidence interval ---Select--- contains or is completely below or is completely above zero. This suggests that the category with higher mean annual expenditure is ---Select--- groceries or dining out .

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Description: A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment. Using data from a sample of 42 credit card a

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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d = $870,

and the sample standard deviation was

s_d = $1,124.

(a)

Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

H₀: ?_d ≤ 0

H_a: ?_d > 0

H₀: ?_d < 0

H_a: ?_d = 0

H₀: ?_d ≥ 0

H_a: ?_d < 0

H₀: ?_d = 0

H_a: ?_d ≠ 0

H₀: ?_d ≠ 0

H_a: ?_d = 0

(b)

Calculate the test statistic. (Round your answer to three decimal places.)

What is the p-value? (Round your answer to four decimal places.)

Can you conclude that the population means differ? Use a 0.05 level of significance.

The p ≤ 0.05, therefore we cannot conclude that there is a difference between the annual population mean expenditures for groceries and for dining out.

The p ≤ 0.05, therefore we can conclude that there is a difference between the annual population mean expenditures for groceries and for dining out.

The p > 0.05, therefore we cannot conclude that there is a difference between the annual population mean expenditures for groceries and for dining out.

The p > 0.05, therefore we can conclude that there is a difference between the annual population mean expenditures for groceries and for dining out.

(c)

What is the point estimate (in dollars) of the difference between the population means?

What is the 95% confidence interval estimate (in dollars) of the difference between the population means? (Round your answers to the nearest dollar.)

$ to $

Which category, groceries or dining out, has a higher population mean annual credit card charge?

The 95% confidence interval ---Select--- contains or is completely below or is completely above zero. This suggests that the category with higher mean annual expenditure is ---Select--- groceries or dining out .

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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