/*! 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 25 The article "Effect of Temperatu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 25

The article "Effect of Temperature on the pH of Skim Milk" (Journal of Dairy Research [1988]: \(277-280\) ) reported on a study involving \(x=\) temperature \(\left({ }^{\circ} \mathrm{C}\right)\) under specified experimental conditions and \(y=\) milk \(\mathrm{pH}\). The accompanying data (read from a graph) are a representative subset of that which appeared in the article: \(\sum x=678 \quad \sum y=104.54 \quad \sum x^{2}=36,056\) \(\sum y^{2}=683.4470 \quad \sum x y=4376.36\) Do these data strongly suggest that there is a negative linear relationship between temperature and \(\mathrm{pH}\) ? State and test the relevant hypotheses using a significance level of \(.01\).

Short Answer

Expert verified
The answer would need calculated correlation coefficient and hypothesis test result to give a precise answer. If the correlation coefficient is close to -1 and the hypothesis test rejects the null hypothesis, then it can be concluded that the data strongly suggests a negative linear relationship between temperature and pH level of skim milk. Please refer to the step-by-step solution for detailed execution.

Step by step solution

01

Calculate the number of data points

Let's denote the number of observations/data points as \(n\). This can be found by dividing the sum of x by the average x value. Unfortunately, the average value is not given, so we will assume \(n\) to be provided or can be calculated from provided data.
02

Calculate the correlation coefficient

To calculate the sample correlation coefficient \(r\) between x and y, the following formula is used: \[ r = \frac{n\(\sum xy\) - (\(\sum x\)* \(\sum y\))}{\sqrt{[n\(\sum x^2\) - (\(\sum x\))^2][n\(\sum y^2\) - (\(\sum y\))^2]}} \] Plug indata to calculate the correlation coefficient.
03

Apply the hypothesis test

Hypotheses for the test are: \[ H0: 蟻 = 0 \text{(There is no linear correlation.)} \] \[ H1: 蟻 鈮� 0 \text{(There is a linear correction.)} \] Where \(蟻\) is the population correlation coefficient. With the significance level (伪) of 0.01, one may reject the null hypothesis \(H_0\) if the absolute value of \(r\) is greater than the critical value from the t-distribution table with \(df = n - 2\) degrees of freedom.
04

Make the conclusion

If the absolute value of \(r\) is found to be larger than the critical value, then one rejects the null hypothesis. Conclude that data strongly suggests there is a negative linear relationship between temperature and pH level of skim milk if the correlation coefficient found was negative and strong (close to -1). If not, conclude that the data doesn't suggest a strong negative linear relationship between temperature and pH.

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 91影视!

Key Concepts

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

Correlation Coefficient
The correlation coefficient, often represented by the symbol \( r \), measures the strength and direction of a linear relationship between two variables. It ranges from -1 to 1. When \( r \) is -1, it indicates a perfect negative linear relationship, while \( r = +1 \) signifies a perfect positive linear relationship. A value of 0 means no linear relationship exists.
Calculating \( r \) involves using the formula:
  • Numerator: \( n\sum xy - (\sum x)(\sum y) \)
  • Denominator: \( \sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]} \)
Once calculated, \( r \) will help us determine if temperature and milk pH have a significant linear relationship. Identifying whether \( r \) is strong or weak is crucial in understanding the relationship's impact.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It involves two hypotheses:
  • Null Hypothesis \( H_0 \): Assumes no effect or no difference. In this context, \( H_0: \rho = 0 \) means no linear correlation between temperature and pH.
  • Alternative Hypothesis \( H_1 \): Suggests a potential effect or difference. Here, \( H_1: \rho eq 0 \), implies a linear correlation exists.
The hypothesis test assesses if there is enough evidence in the sample to reject \( H_0 \). If we can refute \( H_0 \), it supports the idea that a linear relationship may exist. The test compares the calculated correlation coefficient to a critical value determined for a specific significance level.
Significance Level
The significance level, denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It helps control the chances of a Type I error, which occurs when we wrongly reject \( H_0 \). Commonly, in hypothesis testing, \( \alpha \) is set at widely accepted values like 0.05 or 0.01.
In our situation, \( \alpha = 0.01 \) is used. This means we want a 99% confidence level in our decision, reducing the likelihood of concluding a relationship falsely.
If the calculated \( r \) exceeds the critical threshold based on \( \alpha \) and degrees of freedom \( (n - 2) \), we reject \( H_0 \). A small \( \alpha \) aids in making a robust decision, ensuring that any determined relationship is statistically significant.

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

Hormone replacement therapy (HRT) is thought to increase the risk of breast cancer. The accompanying data on \(x=\) percent of women using HRT and \(y=\) breast cancer incidence (cases per 100,000 women) for a region in Germany for 5 years appeared in the paper. The authors of the paper used a simple linear regression model to describe the relationship between HRT use and breast cancer incidence. \begin{tabular}{cc} & Breast Cancer \\ HRT Use & Incidence \\ \hline \(46.30\) & \(103.30\) \\ \(40.60\) & \(105.00\) \\ \(39.50\) & \(100.00\) \\ \(36.60\) & \(93.80\) \\ \(30.00\) & \(83.50\) \\ \hline \end{tabular} a. What is the equation of the estimated regression line? b. What is the estimated average change in breast cancer incidence associated with a 1 percentage point increase in HRT use? c. What would you predict the breast cancer incidence to be in a year when HRT use was \(40 \%\) ? d. Should you use this regression model to predict breast cancer incidence for a year when HRT use was \(20 \%\) ? Explain. e. Calculate and interpret the value of \(r^{2}\). f. Calculate and interpret the value of \(s_{e}\).

If the sample correlation coefficient is equal to 1, is it necessarily true that \(\rho=1\) ? If \(\rho=1\), is it necessarily true that \(r=1 ?\)

The authors of the paper studied a number of variables they thought might be related to bone mineral density (BMD). The accompanying data on \(x=\) weight at age 13 and \(y=\) bone mineral density at age 27 are consistent with summary quantities for women given in the paper. A simple linear regression model was used to describe the relationship between weight at age 13 and \(\mathrm{BMD}\) at age 27\. For this data: $$ \begin{array}{lll} a=0.558 & b=0.009 & n=15 \\ \mathrm{SSTo}=0.356 & \text { SSResid }=0.313 & \end{array} $$ a. What percentage of observed variation in \(\mathrm{BMD}\) at age 27 can be explained by the simple linear regression model? b. Give a point estimate of \(\sigma\) and interpret this estimate. c. Give an estimate of the average change in BMD associated with a \(1 \mathrm{~kg}\) increase in weight at age 13 . d. Compute a point estimate of the mean BMD at age 27 for women whose age 13 weight was \(60 \mathrm{~kg}\).

A subset of data read from a graph that appeared in the paper "Decreased Brain Volume in Adults with Childhood Lead Exposure" (Public Library of Science Medicine [May 27, 2008]: ell2) was used to produce the following Minitab output, where \(x=\) mean childhood blood lead level \((\mu \mathrm{g} / \mathrm{dL})\) and \(y=\) brain volume change (percentage). (See Exercise \(13.19\) for a more complete description of the study described in this paper) Regression Analysis: Response versus Mean Blood Lead Level The regression equation is Response \(=-0.00179-0.00210\) Mean Blood Lead Level \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & -0.001790 & 0.008303 & -0.22 & 0.830 \\ \text { Mean Blood Lead Level } & -0.0021007 & 0.0005743 & -3.66 & 0.000\end{array}\) a. What is the equation of the estimated regression line? b. For this dataset, \(n=100, \bar{x}=11.5, s_{e}=0.032\), and \(S_{x x}=1764 .\) Estimate the mean brain volume change for people with a childhood blood lead level of \(20 \mu \mathrm{g} / \mathrm{dL}\), using a \(90 \%\) confidence interval. c. Construct a \(90 \%\) prediction interval for brain volume change for a person with a childhood blood lead level of \(20 \mu \mathrm{g} / \mathrm{dL}\). d. Explain the difference in interpretation of the intervals computed in Parts (b) and (c).

A simple linear regression model was used to describe the relationship between sales revenue \(y\) (in thousands of dollars) and advertising expenditure \(x\) (also in thousands of dollars) for fast-food outlets during a 3 -month period. A sample of 15 outlets yielded the accompanying summary quantities. $$ \begin{aligned} &\sum x=14.10 \quad \sum y=1438.50 \quad \sum x^{2}=13.92 \\ &\sum y^{2}=140,354 \quad \sum x y=1387.20 \\ &\sum(y-\vec{y})^{2}=2401.85 \quad \sum(y-\hat{y})^{2}=561.46 \end{aligned} $$ a. What proportion of observed variation in sales revenue can be attributed to the linear relationship between revenue and advertising expenditure? b. Calculate \(s_{e}\) and \(s_{b}\) c. Obtain a \(90 \%\) confidence interval for \(\beta\), the average change in revenue associated with a \(\$ 1000\) (that is, 1 -unit) increase in advertising expenditure.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation
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.