/*! 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 40 Use the data from exercise \(9.3... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the data from exercise \(9.36\). a. Using the four-step procedure with a two-sided alternative hypothesis, should you be able to reject the hypothesis that the population mean is 5 pounds using a significance level of \(0.05\) ? Why or why not? The confidence interval is reported here: I am \(95 \%\) confident the population mean is between \(4.9\) and \(5.3\) pounds. b. Now test the hypothesis that the population mean is not 5 pounds using the four-step procedure. Use a significance level of \(0.05\) and number your steps.

Short Answer

Expert verified
For part a, the null hypothesis that the population mean is 5 pounds cannot be rejected at a significance level of 0.05 as the confidence interval includes the value of 5 pounds. For part b, the conclusion would depend on the p-value calculated based on the available data.

Step by step solution

01

State the hypotheses

The null hypothesis is that the population mean is 5 pounds \(H_0: \mu = 5\). The alternative hypothesis is the population mean is not 5 pounds \(H_1: \mu ≠ 5\).
02

Analyze the confidence interval

If the hypothesized population mean is within the \(95\%\) confidence interval, then you would not be able to reject the null hypothesis at a significance level of \(0.05\). The confidence interval given is between \(4.9\) and \(5.3\) pounds, which includes the hypothesized population mean of \(5\) pounds.
03

Conclusion for Question a

Since the hypothesized population mean falls within the \(95\%\) confidence interval, you cannot reject the null hypothesis at the .05 significance level. So, the answer to question a is no, based on the confidence interval, you should not be able to reject the null hypothesis.
04

Derive the test statistic and p-value

To answer question b, we would need additional information like the sample mean, sample standard deviation, and sample size to compute the test statistic (z or t depending on the available info) and the corresponding p-value.
05

Conclusion for Question b

If provided with the relevant data, the p-value will be compared with the significance level (0.05 in this case). If the p-value is less than the significance level, we will reject the null hypothesis and conclude that the population mean is not 5 pounds. If the p-value is greater than or equal to the significance level, we will fail to reject the null hypothesis.

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Key Concepts

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

Confidence Interval
Confidence intervals offer a range of values, derived from the statistical data at hand, that are believed to capture the true population parameter with a specified probability.
In the context of hypothesis testing, especially two-sided tests like the one discussed, if the null hypothesized parameter (like the population mean here) falls within the confidence interval, it suggests that there isn't strong enough evidence to reject the null hypothesis.
In simpler terms:
  • The confidence interval provides a range within which the true mean is likely to lie.
  • If this range includes our null hypothesis value (for example, 5 pounds), we don't have enough evidence against the null hypothesis.
The confidence interval in this exercise is from 4.9 to 5.3 pounds, which includes the hypothesized mean of 5 pounds, thus failing to reject the null hypothesis at the stated significance level.
Significance Level
The significance level, commonly denoted as \( \alpha \), is the threshold used in hypothesis testing to determine whether a result is statistically significant.
In this exercise, a significance level of 0.05 is used.
This threshold dictates the probability of rejecting the null hypothesis when it is actually true, which is known as a Type I error.
  • A significance level of 0.05 means there is a 5% risk of concluding that something is visible when in fact it is not.
  • This is typically chosen to establish a balance between being overly cautious and being too quick to infer significance.
A lower significance level means more evidence is needed against the null hypothesis to reject it, while a higher significance level increases the chance of mistakenly rejecting a true null hypothesis.
Null Hypothesis
The null hypothesis, represented as \( H_0 \), is a statement of no effect or no difference.
It is the assumption that any kind of effect or difference observed in the data is due to sampling or random variation.
In hypothesis tests, the null hypothesis is tested against an alternative hypothesis, \( H_1 \).
  • For this exercise, the null hypothesis is that the population mean is 5 pounds.
  • The alternative hypothesis states that the population mean is not 5 pounds.
By default, the null hypothesis is assumed to be true until evidence suggests otherwise.
This evidence is typically in the form of sample data that counters the null hypothesis, leading to its rejection. If the data falls within a statistically significant range (outside of the acceptable confidence interval or below the chosen significance threshold), the null hypothesis is rejected.

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Most popular questions from this chapter

In the 2015 AFC Championship game, there was a charge the New England Patriots deflated their footballs for an advantage. The balls should be inflated to between \(12.5\) and \(13.5\) pounds per square inch. The measurements were \(11.50,10.85,11.15,10.70,11.10,11.60,11.85,11.10,10.95\), \(10.50\), and \(10.90\) psi (pounds per square inch). (Source: http:// online.wsj.com/public/resources/documents/Deflategate.pdf) a. Test the hypothesis that the population mean is less than \(12.5\) psi using a significance level of \(0.05 .\) State clearly whether the Patriots' balls are deflated or not. Assume the conditions for a hypothesis test are satisfied. b. Use the data to construct a \(95 \%\) confidence interval for the mean psi for the Patriots' footballs. How does this confidence interval support your conclusion in part a?

According to a 2017 report by ComScore .com, the mean time spent on smartphones daily by the American adults is \(2.85\) hours. Assume this is correct and assume the standard deviation is \(1.4\) hours. a. Suppose 150 American adults are randomly surveyed and asked how long they spend on their smartphones daily. The mean of the sample is recorded. Then we repeat this process, taking 1000 surveys of 150 American adults and recording the sample means. What will be the shape of the distribution of these sample means? b. Refer to part (a). What will be the mean and the standard deviation of the distribution of these sample means?

In the United States, the population mean height for 3 -year-old boys is 38 inches (http://www.kidsgrowth .com). Suppose a random sample of 15 non-U.S. 3 -year-old boys showed a sample mean of \(37.2\) inches with a standard deviation of 3 inches. The boys were independently sampled. Assume that heights are Normally distributed in the population. a. Determine whether the population mean for non-U.S. boys is significantly different from the U.S. population mean. Use a significance level of \(0.05\) b. Now suppose the sample consists of 30 boys instead of 15, and repeat the test. c. Explain why the \(t\) -values and \(\mathrm{p}\) -values for parts a and \(\mathrm{b}\) are different.

Some sources report that the weights of full-term newborn babies have a mean of 7 pounds and a standard deviation of \(0.6\) pound and are Normally distributed. a. What is the probability that one newborn baby will have a weight within \(0.6\) pound of the mean - that is, between \(6.4\) and \(7.6\) pounds, or within one standard deviation of the mean? b. What is the probability the average of four babies' weights will be within \(0.6\) pound of the mean - that is, between \(6.4\) and \(7.6\) pounds? c. Explain the difference between \(\mathrm{a}\) and \(\mathrm{b}\).

A fast-food chain advertises that the mediumsize serving of French fries weighs 135 grams. A reporter took a random sample of 10 medium orders of French fries and weighed each order. The weights (in grams) were \(111,124,125,156,127,134\), \(135,136,139,141\). Assume the population distribution is Normal. (Source: soranews24.com) a. Test the hypothesis that the medium servings have a population mean different from 135 grams. Use a significance level of \(0.05\). b. Construct a \(95 \%\) confidence interval for the population mean. How does your confidence interval support your conclusion in part a? Do you think the consumers are being misled about the serving size? Explain.

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