/*! 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 52 Antibiotics in Infancy Exercise ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Antibiotics in Infancy Exercise 2.19 describes a Canadian longitudinal study that examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later in life. The study included 616 children and found that 438 of the children had received antibiotics during the first year of life. Test to see if this provides evidence that more than \(70 \%\) of Canadian children receive antibiotics during the first year of life. Show all details of the hypothesis test, including hypotheses, the standardized test statistic, the p-value, the generic conclusion using a \(5 \%\) significance level, and a conclusion in context.

Short Answer

Expert verified
First the hypotheses are set up as \(H_0: p = 0.7\) and \(H_1: p > 0.7\). The test statistic is calculated using these and the sample data. The p-value is found using this test statistic and a standard normal distribution. If the p-value is less than or equal to the significance level of 0.05, then the null hypothesis is rejected in favor of the alternative. A contextual conclusion is then made.

Step by step solution

01

Formulating the Hypotheses

The null hypothesis is that 70% of Canadian children receive antibiotics during their first year of life: \(H_0: p = 0.7\). The alternative hypothesis is that more than 70% of Canadian children receive antibiotics during their first year of life: \(H_1: p > 0.7\)
02

Calculating Test Statistic

The test statistic for a population proportion is calculated using the formula: \[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\] where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion and n is the sample size. Here \(\hat{p} = \frac{438}{616} \approx 0.71\), \(p_0 = 0.7\), and \(n = 616\). This will give you the observed test statistic.
03

Calculating the P-value

To get the p-value, use a standard normal (Z) distribution table or a technology that can do the calculation to find the probability that a Z value is larger than the calculated test statistic. This is appropriate since the alternative hypothesis states \(H_1: p > 0.7\). The p-value represents the probability of obtaining these (or more extreme) results if the null hypothesis were true.
04

Comparing P-value with Significance Level

You should compare the calculated p-value with the given significance level. If the p-value is smaller than or equal to the significance level, then you will reject the null hypothesis.
05

Drawing Conclusion in Context

You should then conclude in context of the problem. The conclusion should relate to the original research question.

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.

Understanding Population Proportion
When conducting a hypothesis test involving proportions, it's essential to first define the concept of population proportion. A population proportion is a fraction of the population that possesses a particular characteristic. In our example, this refers to the proportion of Canadian children who receive antibiotics in their first year of life.
The symbol used for population proportion is often \(p\), representing the hypothesized value in the hypothesis tests. In the context of our exercise, the null hypothesis proposes that the population proportion is 0.7, which means 70% of children receive antibiotics during infancy. Meanwhile, in our sample, we observed that 438 out of 616 children received antibiotics, giving a sample proportion (\(\hat{p}\)) of approximately 0.71. This sample proportion is computed as follows:
  • \(\hat{p} = \frac{438}{616} \approx 0.71\).

By comparing these proportions, you'll go on to use hypothesis testing to decide if the observed proportion differs significantly from the hypothesized proportion.
Demystifying the Test Statistic
The test statistic in a hypothesis test provides a standardized measure of the difference between the observed sample proportion and the hypothesized population proportion. It essentially offers a mathematical way to see if the data collected could have happened by random chance.
In the case of testing a population proportion, the test statistic is calculated using the formula:
  • \[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]

where:
  • \(\hat{p}\) is the sample proportion (0.71 in our case),
  • \(p_0\) is the null hypothesis proportion (0.7 here),
  • \(n\) is the sample size (616 here).

Plugging in the values, the Z-score quantifies how far away our sample proportion is from the hypothesized value under the assumption that the null hypothesis is true.
This test statistic is crucial, as it forms the basis for finding the p-value in the next step.
Decoding the P-value
The p-value is a probability that helps us make informed decisions in hypothesis testing. It represents the likelihood of observing our data, or something more extreme, assuming the null hypothesis is true.
To find the p-value, you'll use the Z-score calculated in the previous step. Since we are dealing with a right-tailed test (our alternative hypothesis is that more than 70% of children receive antibiotics), the p-value indicates the area to the right of the calculated Z-score in the standard normal distribution.
A small p-value suggests that the observed data are unlikely under the null hypothesis. Conversely, a large p-value would imply that the data do not provide strong evidence against the null hypothesis.
Once you've determined the p-value using statistical software or a Z-distribution table, it will directly inform whether the null hypothesis should be rejected or not.
Exploring the Significance Level
The significance level, often denoted as \(\alpha\), is a threshold set before conducting a hypothesis test. It represents the probability of wrongly rejecting the null hypothesis if it is true.
In many studies, a common choice for the significance level is \(\alpha = 0.05\), which equates to a 5% risk of concluding that a difference exists when there is none. Simply put, if your p-value is less than or equal to this significance level, you'll reject the null hypothesis, concluding that there is enough evidence to suggest a significant difference or effect.
In the context of our antibiotic study, a significance level of 5% means if the p-value calculated from the Z-score is 0.05 or less, we will reject the null hypothesis. This would suggest that there is evidence that more than 70% of infants receive antibiotics in their first year.
Understanding and setting the appropriate significance level is crucial for interpreting results and supporting decisions made from hypothesis tests.

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

In Exercises 6.7 and 6.8 , compute the standard error for sample proportions from a population with the given proportion using three different sample sizes. What effect does increasing the sample size have on the standard error? Using this information about the effect on the standard error, discuss the effect of increasing the sample size on the accuracy of using a sample proportion to estimate a population proportion. A population with proportion \(p=0.75\) for sample sizes of \(n=40, n=300,\) and \(n=1000 .\)

Use a t-distribution to answer the question. Assume the sample is a random sample from a distribution that is reason ably normally distributed and we are doing inference for a sample mean. Find the area in a t-distribution above 2.3 if the sample has size \(n=6\).

In Exercises 6.156 to 6.161: (a) Find the relevant sample proportions in each group and the pooled proportion. (b) Complete the hypothesis test using the normal distribution and show all details. Test whether there is a difference between two groups in the proportion who voted, if 45 out of a random sample of 70 in Group 1 voted and 56 out of a random sample of 100 in Group 2 voted.

Use StatKey or other technology to generate a bootstrap distribution of sample means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviation as an estimate of the population standard deviation. Mean price of used Mustang cars online (in \$1000s) using the data in MustangPrice with \(n=25\), \(\bar{x}=15.98,\) and \(s=11.11\)

A survey of 1000 adults in the US conducted in March 2011 asked "Do you favor or oppose 'sin taxes' on soda and junk food?" The proportion in favor of taxing these foods was \(32 \% .10\) (a) Find a \(95 \%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food. (b) What is the margin of error? (c) If we want a margin of error of only \(1 \%\) (with \(95 \%\) confidence \()\), what sample size is needed?

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.