/*! 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 31 A recent study revealed the typi... [FREE SOLUTION] | 91Ó°ÊÓ

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A recent study revealed the typical American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens revealed they consumed the following amounts of coffee, reported in cups, yesterday. \(\begin{array}{lllllllllll}3.1 & 3.3 & 3.5 & 2.6 & 2.6 & 4.3 & 4.4 & 3.8 & 3.1 & 4.1 & 3.1 & 3.2\end{array}\) At the .05 significance level, do these sample data suggest there is a difference between the national average and the sample mean from senior citizens?

Short Answer

Expert verified
There is no statistically significant difference between senior citizens' coffee consumption and the national average.

Step by step solution

01

State the Hypotheses

First, define the null and alternative hypotheses. The null hypothesis, \( H_0 \), states that there is no difference between the coffee consumption of senior citizens and the national average, i.e., \( \mu = 3.1 \). The alternative hypothesis, \( H_a \), states that there is a difference, i.e., \( \mu eq 3.1 \).
02

Identify the Level of Significance

The level of significance is given as \( \alpha = 0.05 \). This is the probability threshold at which we will reject the null hypothesis if the evidence suggests a significant difference.
03

Calculate the Sample Mean and Standard Deviation

Calculate the sample mean (\( \bar{x} \)) and standard deviation (\( s \)) using the provided data. The sample data are \( 3.1, 3.3, 3.5, 2.6, 2.6, 4.3, 4.4, 3.8, 3.1, 4.1, 3.1, 3.2 \). The sample mean \( \bar{x} = \frac{1}{12} \sum x_i \approx 3.383 \), and the standard deviation can be calculated using the formula \( s = \sqrt{ \frac{1}{n-1} \sum (x_i - \bar{x})^2 } \approx 0.620 \).
04

Determine the Test Statistic

Use the t-test statistic formula: \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size. Thus, \( t = \frac{3.383 - 3.1}{0.620/\sqrt{12}} \approx 1.6007 \).
05

Find the Critical Value and Make Decision

Determine the critical t-value for \( n-1 = 11 \) degrees of freedom at the \( 0.05 \) significance level, which is approximately \( \pm 2.201 \) for a two-tailed test. Since the calculated t-value \( 1.6007 \) falls within the range \(-2.201 < t < 2.201\), we fail to reject the null hypothesis.
06

Conclusion

Since we failed to reject the null hypothesis, we conclude that there is not enough statistical evidence at the \( 0.05 \) significance level to suggest that the coffee consumption of senior citizens is different from the national average.

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Key Concepts

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

Sample Mean
In statistics, a sample mean is a summary statistic that provides the average value of a sample data set. It is crucial in hypothesis testing as it helps to compare the sample with the population. Here, we calculate the sample mean by summing up all observed values and then dividing by the number of observations.
In our example, the sample mean is calculated based on the 12 senior citizens' coffee consumption values:
  • 3.1, 3.3, 3.5, 2.6, 2.6, 4.3, 4.4, 3.8, 3.1, 4.1, 3.1, and 3.2
Adding these values yields 40.6. Divide this sum by 12, the number of observed samples, to get the sample mean: \( \bar{x} = \frac{40.6}{12} \approx 3.383 \).
By calculating the sample mean, we obtain a value that represents the average coffee consumption among the sampled senior citizens, which can then be compared with the national average of 3.1 cups per day.
T-test
The t-test is a statistical test used to analyze whether there is a significant difference between the means of two groups. It is particularly useful when the population standard deviation is unknown and the sample size is relatively small.
To perform a t-test, we follow these steps:
  • Compute the sample mean \( \bar{x} \) and the standard deviation \( s \)
  • Use the formula for the t-statistic: \\( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \)\, where:
    • \( \mu \) is the population mean (3.1 cups in our exercise)
    • \( s \) is the sample standard deviation
    • \( n \) is the sample size (12 senior citizens)
For our exercise, after plugging in the values, the t-statistic comes out to approximately 1.6007. The t-value helps us determine how far our sample mean deviates from the national average in terms of the standard error.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold that determines the cutoff for deciding whether to reject the null hypothesis. It is typically set at a conventional level such as 0.05, 0.01, or 0.10.
In the exercise, we've chosen a significance level of 0.05. This means that there is a 5% risk of concluding that there is a difference between the sample mean and the population mean when there is none.
The significance level helps to define the critical region boundaries for rejecting the null hypothesis. For our t-test:
  • If the calculated t-value falls outside this critical region, we reject the null hypothesis.
  • If it falls inside, as in our case (where the t-value is 1.6007 and within -2.201 to 2.201), we fail to reject the null hypothesis.
Thus, the significance level acts as a gatekeeper for making decisions about statistical evidence.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a claim that there is no effect or no difference that we are testing against. It serves as a starting point for statistical analysis.
For this exercise, the null hypothesis states that the mean coffee consumption of senior citizens (sample mean) is equal to the national average (population mean), mentioned in mathematical terms as \( \mu = 3.1 \).
The alternative hypothesis, \( H_a \), against which the null is tested, suggests there is a difference, \( \mu eq 3.1 \).
To make a decision on the null hypothesis, statistical tests, like the t-test, are performed to determine if the observed data provides enough evidence to conclude a significant difference.
  • If the test statistic falls into the critical region, we reject the null hypothesis.
  • If not, as in the case with a t-value of 1.6007, we conclude that our sample does not provide sufficient evidence to suggest a difference.
Thus, the null hypothesis is crucial in hypothesis testing as it frames the basic question our analysis seeks to address.

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Most popular questions from this chapter

The postanesthesia care area (recovery room) at St. Luke's Hospital in Maumee, Ohio, was recently enlarged. The hope was that the change would increase the mean number of patients served per day to more than \(25 .\) A random sample of 15 days revealed the following numbers of patients. \(\begin{array}{llllllllll}25 & 27 & 25 & 26 & 25 & 28 & 28 & 27 & 24 & 26 & 25 & 29 & 25 & 27 & 24\end{array}\) At the .01 significance level, can we conclude that the mean number of patients per day is more than 25 ? Estimate the \(p\) -value and interpret it.

Given the following hypotheses: $$\begin{array}{l}H_{0}: \mu=400 \\\H_{1}: \mu \neq 400\end{array}$$ A random sample of 12 observations is selected from a normal population. The sample mean was 407 and the sample standard deviation was 6 . Using the .01 significance level: a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

According to the Census Bureau, 3.13 people reside in the typical American household. A sample of 25 households in Arizona retirement communities showed the mean number of residents per household was 2.86 residents. The standard deviation of this sample was 1.20 residents. At the .05 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.13 persons?

During recent seasons, Major League Baseball has been criticized for the length of the games. A report indicated that the average game lasts 3 hours and 30 minutes. A sample of 17 games revealed the following times to completion. (Note that the minutes have been changed to fractions of hours, so that a game that lasted 2 hours and 24 minutes is reported at 2.40 hours.) \(\begin{array}{lllllllll}2.98 & 2.40 & 2.70 & 2.25 & 3.23 & 3.17 & 2.93 & 3.18 & 2.80 \\ 2.38 & 3.75 & 3.20 & 3.27 & 2.52 & 2.58 & 4.45 & 2.45 & \end{array}\) Can we conclude that the mean time for a game is less than 3.50 hours? Use the .05 significance level.

A recent article in Vitality magazine reported that the mean amount of leisure time per week for American men is 40.0 hours. You believe this figure is too large and decide to conduct your own test. In a random sample of 60 men, you find that the mean is 37.8 hours of leisure per week and that the standard deviation of the sample is 12.2 hours. Can you conclude that the information in the article is untrue? Use the .05 significance level. Determine the \(p\) -value and explain its meaning.

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