91Ó°ÊÓ

Bank of America's Consumer Spending 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 (US Airways Attaché, December 2003 ). 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, the sample mean difference was \(d=\$ 850,\) and the sample standard deviation was \(s_{d}=\$ 1123\) 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. b. Use a .05 level of significance. Can you conclude that the population means differ? What is the \(p\) -value? c. Which category, groceries or dining out, has a higher population mean annual credit card charge? What is the point estimate of the difference between the population means? What is the \(95 \%\) confidence interval estimate of the difference between the population means?

Step by step solution

01

Formulate Hypotheses

We start by identifying the null and alternative hypotheses.- Null Hypothesis (\(H_0\)): There is no difference between the population mean credit card charges for groceries and for dining out, \( \mu_1 = \mu_2 \).- Alternative Hypothesis (\(H_a\)): There is a difference in the population mean credit card charges for groceries and dining out, \( \mu_1 eq \mu_2 \). This is a two-tailed test because we want to see if there is any kind of difference, not specifically if one is greater than the other.

02

Find the Test Statistic

We use the sample mean difference \( d = 850 \) and sample standard deviation \( s_d = 1123 \) to calculate the test statistic using the formula:\[ t = \frac{d}{s_d / \sqrt{n}} \]where \( n = 42 \) is the sample size. First, calculate the standard error \( SE = \frac{1123}{\sqrt{42}} \approx 173.45 \). Then,\[ t = \frac{850}{173.45} \approx 4.90 \].

03

Determine Critical Value and P-Value

Next, check the critical value for a two-tailed t-test with \( n-1 = 41 \) degrees of freedom at \( \alpha = 0.05 \). From a t-distribution table, the critical t-value is approximately \( t_{0.025, 41} \approx 2.019 \).Since the calculated \( t \) value of 4.90 is greater than 2.019, we reject the null hypothesis.For P-value, since \( t \) is well beyond our critical value, the \( p \)-value will be less than 0.05, confirming our decision to reject.

04

Identify Which Category Has Higher Mean

The positive sample mean difference \( d = 850 \) indicates that groceries (population 1) have higher expenditures compared to dining out (population 2).

05

Calculate Point Estimate and Confidence Interval

The point estimate for the difference between the population means is \( d = 850 \).To calculate the 95% confidence interval, we use:\[ CI = d \pm t_{critical} \times SE \]Thus, the confidence interval is:\[ 850 \pm 2.019 \times 173.45 \approx (500.61, 1199.39) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Mean

The sample mean is a fundamental concept in statistics, often used to provide a central value of a set of data. In simple terms, it's the average of the data points you have collected from a sample. When conducting research or experiments, we usually can't collect data from the entire population due to constraints like time and cost. Instead, we take a sample, which is a smaller, manageable group that represents the population.

• To find the sample mean, add up all the sample values.
• Then, divide by the number of values (i.e., the sample size).

For instance, in the exercise, the sample mean difference in annual credit card charges between groceries and dining out was $850. This means, on average, groceries cost more on credit cards than dining out across the surveyed accounts.

Standard Deviation

Standard deviation is a statistical measure that helps us understand how spread out the data in a sample is around the mean. It's like a friend who tells you how consistent or varied your data points are from the average.

• If the standard deviation is small, it means the data points are close to the mean.
• If large, the data points are more spread out.

In our exercise, the standard deviation of $1123 suggests there is a fair amount of variability in the credit card charges among the sampled accounts. Understanding this spread is crucial when interpreting results because it affects how confident we can be about our mean estimate.

Confidence Interval

A confidence interval provides a range of values within which we expect the true population parameter to lie, with a certain level of confidence, usually expressed as a percentage like 95%. It's like saying, "I’m 95% sure that the true population mean falls within this range."

• It is calculated using the sample mean, standard deviation, and critical values from a t-distribution table when sample sizes are small.
• The margin of error adds or subtracts from the sample mean to create the confidence interval.

From the exercise, the 95% confidence interval for the difference in costs between groceries and dining out was calculated to be approximately from $500.61 to $1199.39. This interval suggests that the true difference in spending likely lies within this range, acting as a safety net for our estimates.

T-Test

A t-test is a statistical test used to determine if there is a significant difference between the means of two groups. It's particularly useful when dealing with small sample sizes and when the population standard deviation is unknown.

• It's like having a judge who evaluates whether the difference between group means is due to chance or is statistically significant.
• The results are determined by comparing the calculated t-value to the critical t-value from the t-distribution table.

In the exercise, we wanted to test if the difference in credit card spending on groceries and dining out is significant. With a t-value of approximately 4.90, far exceeding the critical value of 2.019, we rejected the null hypothesis. This indicates that the difference in mean spending is statistically significant, not just due to random chance.