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In a test to see whether males, on average, have bigger noses than females, the study indicates that " \(p<0.01\)."

Short Answer

Expert verified
The p-value of less than 0.01 indicates a significant statistical difference in the average nose sizes between males and females, with males having larger noses on average.

Step by step solution

01

Understanding P-value

The P-value is a statistical measure that helps scientists determine whether or not their hypotheses are correct. It is the estimated probability of rejecting the null hypothesis (the hypothesis that there is no difference or relationship) in a particular experiment or study. P-values are used in hypothesis testing to help you support or disprove your null hypothesis.
02

Interpreting the p-value

Here, we have \(p<0.01\). This is a very small p-value. In statistical language, a p-value of less than 0.01 will often mean 'strong evidence against the null hypothesis', or 'very statistically significant', often taken as certainty that the null hypothesis must be rejected.
03

Apply the meaning of the p-value to our problem

In this case, the null hypothesis would be that there is no difference in average nose sizes between males and females. With \(p<0.01\), we reject the null hypothesis. Therefore, it's very likely that a significant difference in nose sizes does exist between males and females, with males having larger noses on average.

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Key Concepts

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

Statistical Significance

The term 'statistical significance' plays a pivotal role in the realm of hypothesis testing. It helps us to decide whether the observed effects in our data are genuine or just a result of random chance. When the p-value is low, as in the case of p<0.01, it suggests that the observed data is highly unlikely under the assumption that the null hypothesis is true. To put this into context, if the null hypothesis purported that there is no difference between male and female nose sizes, a p-value less than 0.01 would indicate that such a no-difference scenario occurring by chance is less than 1% probable. Therefore, we would conclude that the findings are statistically significant and that the alternate hypothesis – that there is a difference – has stronger support.

It's crucial to understand that 'statistical significance' does not necessarily equate to 'practical significance.' A result can be statistically significant but still be of negligible real-world importance if the observed difference is too small to matter in practical scenarios. Always ensure the effect size – the magnitude of the observed effect – is considered alongside the p-value for a comprehensive understanding of the results' significance.
In the study context, a statistical significance of the difference in nose sizes should also be understood keeping in mind the biological variability and the possible impact of such a difference.

Null Hypothesis

At the heart of hypothesis testing lies the null hypothesis, denoted as H0. This is a default assumption that there is no effect or no difference in the context of an experiment or study. It serves as the starting point for statistical testing and is a statement meant to be tested against the alternative hypothesis, Ha, which represents what the researcher aims to prove – in the example, that males have larger noses than females.

  • The null hypothesis is crucial because it provides a clear and testable statement that can be challenged with data.
  • It must be specific and stated before the data is collected, ensuring that the test's conclusions are unbiased.
  • The process of hypothesis testing is focused on discrediting the null hypothesis in favor of the alternative hypothesis.

In the scenario provided, the null hypothesis (H0) would suggest no difference in nose sizes between genders. The very small p-value associated with the test results leads us to reject this null hypothesis. However, it's important to remember that rejecting the null does not prove the alternative hypothesis to be true; it just indicates that there is strong evidence against the null and in favor of the alternative hypothesis within a set confidence level.

Hypothesis Testing

Hypothesis testing is a systematic method used to evaluate statistical evidence in data. It allows us to decide whether to accept or reject our null hypothesis and is fundamental to making inferences in statistics.

Steps in Hypothesis Testing

  • Formulate the null and alternative hypotheses.
  • Choose a significance level (alpha), such as 0.05 or 0.01, which determines how unlikely the observed data should be if the null hypothesis were true.
  • Conduct the appropriate statistical test which will yield a p-value.
  • Compare the p-value to the chosen significance level.
  • Draw conclusions about the hypotheses.

In the test regarding nose sizes, the p-value was found to be less than 0.01, which, when compared to the commonly used significance levels, leads to the rejection of the null hypothesis. This test informs us that there's strong statistical evidence suggesting a significant difference between the average nose sizes of males and females, with males having the larger average nose size, under the conditions of the study.

Hypothesis testing is not just about crunching numbers; it's also about contextual understanding. One must also ensure that the test used is appropriate for the data type and experiment, and confirm that the assumptions behind the statistical test are not violated. These considerations are vital to drawing reliable inferences from hypothesis testing.

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Most popular questions from this chapter

Data 4.3 on page 265 introduces a situation in which a restaurant chain is measuring the levels of arsenic in chicken from its suppliers. The question is whether there is evidence that the mean level of arsenic is greater than \(80 \mathrm{ppb},\) so we are testing \(H_{0}: \mu=80\) vs \(H_{a}:\) \(\mu>80,\) where \(\mu\) represents the average level of arsenic in all chicken from a certain supplier. It takes money and time to test for arsenic, so samples are often small. Suppose \(n=6\) chickens from one supplier are tested, and the levels of arsenic (in ppb) are: \(\begin{array}{llllll}68, & 75, & 81, & 93, & 95, & 134\end{array}\) (a) What is the sample mean for the data? (b) Translate the original sample data by the appropriate amount to create a new dataset in which the null hypothesis is true. How do the sample size and standard deviation of this new dataset compare to the sample size and standard deviation of the original dataset? (c) Write the six new data values from part (b) on six cards. Sample from these cards with replacement to generate one randomization sample. (Select a card at random, record the value, put it back, select another at random, until you have a sample of size \(6,\) to match the original sample size.) List the values in the sample and give the sample mean. (d) Generate 9 more simulated samples, for a total of 10 samples for a randomization distribution. Give the sample mean in each case and create a small dotplot. Use an arrow to locate the original sample mean on your dotplot.

Give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.7\) vs \(H_{a}: p<0.7\) Sample data: \(\hat{p}=125 / 200=0.625\) with \(n=200\)

Flying Home for the Holidays, On Time In Exercise 4.115 on page \(302,\) we compared the average difference between actual and scheduled arrival times for December flights on two major airlines: Delta and United. Suppose now that we are only interested in the proportion of flights arriving more than 30 minutes after the scheduled time. Of the 1,000 Delta flights, 67 arrived more than 30 minutes late, and of the 1,000 United flights, 160 arrived more than 30 minutes late. We are testing to see if this provides evidence to conclude that the proportion of flights that are over 30 minutes late is different between flying United or Delta. (a) State the null and alternative hypothesis. (b) What statistic will be recorded for each of the simulated samples to create the randomization distribution? What is the value of that statistic for the observed sample? (c) Use StatKey or other technology to create a randomization distribution. Estimate the p-value for the observed statistic found in part (b). (d) At a significance level of \(\alpha=0.01\), what is the conclusion of the test? Interpret in context. (e) Now assume we had only collected samples of size \(75,\) but got essentially the same proportions (5/75 late flights for Delta and \(12 / 75\) late flights for United). Repeating steps (b) through (d) on these smaller samples, do you come to the same conclusion?

Suppose you want to find out if reading speed is any different between a print book and an e-book. (a) Clearly describe how you might set up an experiment to test this. Give details. (b) Why is a hypothesis test valuable here? What additional information does a hypothesis test give us beyond the descriptive statistics we discuss in Chapter \(2 ?\) (c) Why is a confidence interval valuable here? What additional information does a confidence interval give us beyond the descriptive statistics of Chapter 2 and the results of a hypothesis test described in part (b)? (d) A similar study \(^{53}\) has been conducted, and reports that "the difference between Kindle and the book was significant at the \(p<.01\) level, and the difference between the iPad and the book was marginally significant at \(p=.06 . "\) The report also stated that "the iPad measured at \(6.2 \%\) slower reading speed than the printed book, whereas the Kindle measured at \(10.7 \%\) slower than print. However, the difference between the two devices [iPad and Kindle] was not statistically significant because of the data's fairly high variability." Can you tell from the first quotation which method of reading (print or e-book) was faster in the sample or do you need the second quotation for that? Explain the results in your own words.

In this exercise, we see that it is possible to use counts instead of proportions in testing a categorical variable. Data 4.7 describes an experiment to investigate the effectiveness of the two drugs desipramine and lithium in the treatment of cocaine addiction. The results of the study are summarized in Table 4.14 on page \(323 .\) The comparison of lithium to the placebo is the subject of Example 4.34 . In this exercise, we test the success of desipramine against a placebo using a different statistic than that used in Example 4.34. Let \(p_{d}\) and \(p_{c}\) be the proportion of patients who relapse in the desipramine group and the control group, respectively. We are testing whether desipramine has a lower relapse rate then a placebo. (a) What are the null and alternative hypotheses? (b) From Table 4.14 we see that 20 of the 24 placebo patients relapsed, while 10 of the 24 desipramine patients relapsed. The observed difference in relapses for our sample is $$\begin{aligned}D &=\text { desipramine relapses }-\text { placebo relapses } \\\&=10-20=-10\end{aligned}$$ If we use this difference in number of relapses as our sample statistic, where will the randomization distribution be centered? Why? (c) If the null hypothesis is true (and desipramine has no effect beyond a placebo), we imagine that the 48 patients have the same relapse behavior regardless of which group they are in. We create the randomization distribution by simulating lots of random assignments of patients to the two groups and computing the difference in number of desipramine minus placebo relapses for each assignment. Describe how you could use index cards to create one simulated sample. How many cards do you need? What will you put on them? What will you do with them?

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