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State the conclusion of the test based on this p-value in terms of "Reject \(H_{0} "\) or "Do not reject \(H_{0} "\), if we use a \(5 \%\) significance level. \(\mathrm{p}\) -value \(=0.1145\)

Short Answer

Expert verified
Do not reject the null hypothesis \(H_{0}\) because the p-value is greater than the significance level.

Step by step solution

01

Understanding the concept of p-value and significance level

In hypothesis testing, p-value is the probability that explains how extreme the data are given that the null hypothesis \(H_{0}\) is true. The lower the p-value, the stronger is the evidence against \(H_{0}\). On the other hand, the significance level (often denoted by \(\alpha\)) represents the threshold below which the null hypothesis can be rejected. Common choices for \(\alpha\) are 5% and 1%.
02

Comparing the p-value to the significance level

Next, we compare our p-value (0.1145 here in this problem) with the given significance level, which is 5% or 0.05. If the p-value is less than or equal to the significance level, we reject the null hypothesis. If the p-value is greater than the significance level, we do not reject the null hypothesis.
03

Forming the conclusion

Since the p-value of 0.1145 is greater than the significance level of 0.05, we do not reject \(H_{0}\).

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Key Concepts

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

P-value
When performing hypothesis testing, the p-value is a critical component that helps you decide whether to support the null hypothesis (\(H_{0}\)) or not. It is defined as the probability of obtaining test results that are at least as extreme as the observed results, assuming that the null hypothesis is true. In simpler terms, a p-value tells you how likely it is to get your data or something more unusual if the null hypothesis is actually true.

To understand the p-value, imagine it as a gauge of surprising your data is. A very small p-value indicates that the observed data is highly unexpected under the assumption of the null hypothesis. When the p-value is low enough, typically below the significance level, it suggests that the data is inconsistent with the null hypothesis, prompting researchers to consider alternative hypotheses.

Going back to our original problem where the p-value is 0.1145, this means there’s an 11.45% chance of finding the test results we did if the null hypothesis is true. Since this is a relatively high probability, it does not provide strong evidence against the null hypothesis.
Significance Level
The significance level, denoted as \(\alpha\), is a threshold chosen by the researcher to decide whether a p-value is sufficiently low to reject the null hypothesis. It's often considered as the risk level of accepting an alternative hypothesis when the null hypothesis is actually true, which is known as Type I error. The most commonly used significance levels are 5% (\(\alpha = 0.05\)) and 1% (\(\alpha = 0.01\)).

Using the significance level as a cutoff, if a p-value falls below it, the result of the test is considered statistically significant. This means the evidence is strong enough to reject the null hypothesis in favor of the alternative hypothesis. Conversely, if the p-value is above the cutoff, you would not reject the null hypothesis, meaning the evidence is not strong enough to support the alternative.

In the given exercise where the p-value is 0.1145, and the significance level is set at 5%, the conclusion is that the p-value exceeds the significance level, hence we do not have sufficient evidence to reject the null hypothesis.
Null Hypothesis
The null hypothesis (\(H_{0}\)) is a statement being tested in a statistical hypothesis test that usually postulates no effect or no difference. It's the default position or the status quo that there is nothing unusual happening, and any observed effect is due to random chance. The null hypothesis is what you attempt to disprove or find evidence against through your test.

It is crucial to frame the null hypothesis correctly as it sets the stage for the entire hypothesis testing procedure. The outcome of the test will lead to either retaining the null hypothesis—implying there’s insufficient evidence to support an alternative claim—or rejecting it in favor of the alternative hypothesis, which suggests that the observed data are unlikely under the null hypothesis.

In the context of our exercise, with a p-value of 0.1145 against a significance level of 5%, the data do not provide strong enough evidence to reject the null hypothesis. This does not necessarily mean that the null hypothesis is true, but rather that there is no statistical basis for preferring the alternative hypothesis based on the provided data.

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Most popular questions from this chapter

Influencing Voters Exercise 4.39 on page 272 describes a possible study to see if there is evidence that a recorded phone call is more effective than a mailed flyer in getting voters to support a certain candidate. The study assumes a significance level of \(\alpha=0.05\) (a) What is the conclusion in the context of thisstudy if the p-value for the test is \(0.027 ?\) (b) In the conclusion in part (a), which type of error are we possibly making: Type I or Type II? Describe what that type of error means in this situation. (c) What is the conclusion if the p-value for the test is \(0.18 ?\)

Mating Choice and Offspring Fitness Does the ability to choose a mate improve offspring fitness in fruit flies? Researchers have studied this by taking female fruit flies and randomly dividing them into two groups; one group is put into a cage with a large number of males and able to freely choose who to mate with, while flies in the other group are each put into individual vials, each with only one male, giving no choice in who to mate with. Females are then put into egg laying chambers, and a certain number of larvae collected. Do the larvae from the mate choice group exhibit higher survival rates? A study \(^{44}\) published in Nature found that mate choice does increase offspring fitness in fruit flies (with p-value \(<0.02\) ), yet this result went against conventional wisdom in genetics and was quite controversial. Researchers attempted to replicate this result with a series of related experiments, \({ }^{45}\) with data provided in MateChoice. (a) In the first replication experiment, using the same species of fruit fly as the original Nature study, 6067 of the 10000 larvae from the mate choice group survived and 5976 of the 10000 larvae from the no mate choice group survived. Calculate the p-value. (b) Using a significance level of \(\alpha=0.05\) and \(\mathrm{p}\) -value from (a), state the conclusion in context. (c) Actually, the 10,000 larvae in each group came from a series of 50 different runs of the experiment, with 200 larvae in each group for each run. The researchers believe that conditions dif- fer from run to run, and thus it makes sense to treat each \(\mathrm{run}\) as a case (rather than each fly). In this analysis, we are looking at paired data, and the response variable would be the difference in the number of larvae surviving between the choice group and the no choice group, for each of the 50 runs. The counts (Choice and NoChoice and difference (Choice \(-\) NoChoice) in number of surviving larva are stored in MateChoice. Using the single variable of differences, calculate the p-value for testing whether the average difference is greater than \(0 .\) (Hint: this is a single quantitative variable, so the corresponding test would be for a single mean.) (d) Using a significance level of \(\alpha=0.05\) and the p-value from (c), state the conclusion in context. (e) The experiment being tested in parts (a)-(d) was designed to mimic the experiment from the original study, yet the original study yielded significant results while this study did not. If mate choice really does improve offspring fitness in fruit flies, did the follow-up study being analyzed in parts (a)-(d) make a Type I, Type II, or no error? (f) If mate choice really does not improve offspring fitness in fruit flies, did the original Nature study make a Type I, Type II, or no error?

Test \(\mathrm{A}\) is described in a journal article as being significant with " \(P<.01\) "; Test \(\mathrm{B}\) in the same article is described as being significant with " \(P<\).10." Using only this information, which test would you suspect provides stronger evidence for its alternative hypothesis?

Female primates visibly display their fertile window, often with red or pink coloration. Do humans also do this? A study \(^{18}\) looked at whether human females are more likely to wear red or pink during their fertile window (days \(6-14\) of their cycle \()\). They collected data on 24 female undergraduates at the University of British Columbia, and asked each how many days it had been since her last period, and observed the color of her shirt. Of the 10 females in their fertile window, 4 were wearing red or pink shirts. Of the 14 females not in their fertile window, only 1 was wearing a red or pink shirt. (a) State the null and alternative hypotheses. (b) Calculate the relevant sample statistic, \(\hat{p}_{f}-\hat{p}_{n f}\), for the difference in proportion wearing a pink or red shirt between the fertile and not fertile groups. (c) For the 1000 statistics obtained from the simulated randomization samples, only 6 different values of the statistic \(\hat{p}_{f}-\hat{p}_{n f}\) are possible. Table 4.7 shows the number of times each difference occurred among the 1000 randomizations. Calculate the p-value.

In Exercise 3.129 on page \(254,\) we see that the home team was victorious in 70 games out of a sample of 120 games in the FA premier league, a football (soccer) league in Great Britain. We wish to investigate the proportion \(p\) of all games won by the home team in this league. (a) Use StatKeyor other technology to find and interpret a \(90 \%\) confidence interval for the proportion of games won by the home team. (b) State the null and alternative hypotheses for a test to see if there is evidence that the proportion is different from 0.5 . (c) Use the confidence interval from part (a) to make a conclusion in the test from part (b). State the confidence level used. (d) Use StatKey or other technology to create a randomization distribution and find the p-value for the test in part (b). (e) Clearly interpret the result of the test using the p-value and using a \(10 \%\) significance level. Does your answer match your answer from part (c)? (f) What information does the confidence interval give that the p-value doesn't? What information does the p-value give that the confidence interval doesn't? (g) What's the main difference between the bootstrap distribution of part (a) and the randomization distribution of part (d)?

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