/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 225 Restaurant Bill by Gender In the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Restaurant Bill by Gender In the study described in Exercise 6.224 the diners were also chosen so that half the people at each table were female and half were male. Thus we can also test for a difference in mean meal cost between females \(\left(n_{f}=24, \bar{x}_{f}=44.46, s_{f}=15.48\right)\) and males \(\left(n_{m}=24, \bar{x}_{m}=43.75, s_{m}=14.81\right) .\) Show all details for doing this test.

Short Answer

Expert verified
The t-value obtained is 0.101 with degrees of freedom 46. This information can be used to find the p-value which will be used to determine whether there is statistical evidence that the two means are different.

Step by step solution

01

Calculate the Sample Mean Difference

Calculate the difference between the sample means which is simply \( \bar{x}_{f} - \bar{x}_{m} = 44.46 - 43.75 = 0.71 \)
02

Calculate the Pooled Variance

Compute the pooled variance. You do this by summing the squares of the standard deviations of both samples divided by their respective sample sizes minus one then divided by the total sample size minus two. This gives us: \(S_p^2 = \frac{ ((n_{f}-1) * s_{f}^2 + (n_{m}-1) * s_{m}^2) }{ (n_{m}+n_{f}-2) } = \frac{ ((24-1)*15.48^2 + (24-1)*14.81^2)}{ (24+24-2) } = 232.08 \)
03

Calculate the Standard Error

Calculate the standard error using the pooled variance and the sample sizes. It represents the average amount that the sample mean 'misses' the population mean. It's given by: \(SE = \sqrt{ S_p^2*(1/n_{m} + 1/n_{f}) } = \sqrt{232.08 * (1/24 + 1/24) } = 7.06 \)
04

Compute the t-value

Compute the t-value. The t-value measures how many standard errors are between the sample mean and the null hypothesis. It's given by: \(t = \frac{ \bar{x}_{f} - \bar{x}_{m} } {SE} = \frac{0.71}{7.06} = 0.101 \)
05

Determine the Degrees of Freedom

The degrees of freedom for this two-sample t-test is \(df = n_{m} + n_{f} - 2 = 24 + 24 - 2 = 46 \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

t-test
A **t-test** is a statistical method that determines if there is a significant difference between the means of two groups. In our exercise, we are comparing the average meal costs of females and males to see if these differences are significant.
During hypothesis testing, a t-test helps determine whether these observed differences are due to random chance or reflect a real difference in the populations being compared.
In practice, you would:
  • Assume the null hypothesis that there is no difference between the groups.
  • Calculate the t-value, which is derived from the difference between the sample means, the pooled variance, and the sample sizes.
  • Compare the calculated t-value to a critical value from the t-distribution table to determine significance. A small t-value indicates a small difference between the groups, suggesting the null hypothesis may be true.
Understanding how to interpret the results of a t-test is crucial for determining the relationship between the sample data and the inferred population parameters.
pooled variance
**Pooled variance** is a method used to estimate the variance of two different groups in statistical analyses. This is particularly useful when comparing two groups that have different sample sizes or standard deviations.
The formula for pooled variance is \[S_p^2 = \frac{ ((n_{f}-1) \times s_{f}^2 + (n_{m}-1) \times s_{m}^2) }{ (n_{m}+n_{f}-2) }\]where:
  • \(n_f\) and \(n_m\) are the sample sizes for the female and male groups, respectively.
  • \(s_f\) and \(s_m\) are the standard deviations for each group.
By computing the pooled variance, we are essentially finding a weighted average of the two variances, which allows us to account for the variability in both groups when comparing their means.
In our exercise, this calculation produced a pooled variance of 232.08, which is then used in further calculations, such as determining the standard error and ultimately the t-value. The pooled variance helps maintain the precision of statistical comparisons by evenly weighing both groups' variances.
degrees of freedom
**Degrees of freedom** (df) is a concept used in various statistical tests, including the t-test, to approximate the variability of a sample. It helps determine the appropriate distribution needed for calculating statistical significance.
In the context of a two-sample t-test, the degrees of freedom can be calculated using the formula:\[df = n_{m} + n_{f} - 2\]where:
  • \(n_m\) and \(n_f\) are the respective sample sizes for males and females.
Knowing the degrees of freedom is crucial because it influences the critical value taken from a t-distribution table, against which the calculated t-value is compared.
In our scenario, the degrees of freedom were calculated to be 46. This number helps indicate how much information is available for estimating the population variance. A higher degrees of freedom usually leads to a more accurate inference about the population parameters. By understanding degrees of freedom, you can make more informed judgments about the statistical significance of your results.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Metal Tags on Penguins and Survival Data 1.3 on page 10 discusses a study designed to test whether applying metal tags is detrimental to penguins. One variable examined is the survival rate 10 years after tagging. The scientists observed that 10 of the 50 metal tagged penguins survived, compared to 18 of the 50 electronic tagged penguins. Construct a \(90 \%\) confidence interval for the difference in proportion surviving between the metal and electronic tagged penguins \(\left(p_{M}-p_{E}\right)\). Interpret the result.

Physician's Health Study In the Physician's Health Study, introduced in Data 1.6 on page 37 , 22,071 male physicians participated in a study to determine whether taking a daily low-dose aspirin reduced the risk of heart attacks. The men were randomly assigned to two groups and the study was double-blind. After five years, 104 of the 11,037 men taking a daily low-dose aspirin had had a heart attack while 189 of the 11,034 men taking a placebo had had a heart attack. \({ }^{39}\) Does taking a daily lowdose aspirin reduce the risk of heart attacks? Conduct the test, and, in addition, explain why we can infer a causal relationship from the results.

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To study the effect of women's tears on men, levels of testosterone are measured in 50 men after they sniff women's tears and after they sniff a salt solution. The order of the two treatments was randomized and the study was double-blind.

What Gives a Small P-value? In each case below, two sets of data are given for a two-tail difference in means test. In each case, which version gives a smaller \(\mathrm{p}\) -value relative to the other? (a) Both options have the same standard deviations and same sample sizes but: Option 1 has: \(\quad \bar{x}_{1}=25 \quad \bar{x}_{2}=23\) $$ \text { Option } 2 \text { has: } \quad \bar{x}_{1}=25 \quad \bar{x}_{2}=11 $$ (b) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same sample sizes but: Option 1 has: \(\quad s_{1}=15 \quad s_{2}=14\) $$ \text { Option } 2 \text { has: } \quad s_{1}=3 \quad s_{2}=4 $$ (c) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same standard deviations but: Option 1 has: \(\quad n_{1}=800 \quad n_{2}=1000\) $$ \text { Option } 2 \text { has: } \quad n_{1}=25 \quad n_{2}=30 $$

We examine the effect of different inputs on determining the sample size needed to obtain a specific margin of error when finding a confidence interval for a proportion. Find the sample size needed to give a margin of error to estimate a proportion within \(\pm 3 \%\) with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the confidence level desired.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.