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Tomatoes 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: \(\mathrm{I}\) 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
From the analysis, we fail to reject the null hypothesis at the 0.05 significance level. Therefore, we can't conclude that the population mean is not 5 pounds.

Step by step solution

01

State the Hypotheses

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

Formulate an Analysis Plan

The analysis would be based on the confidence interval. If the hypothesized population mean (5 pounds) is within the confidence interval, we would fail to reject the null hypothesis. Otherwise, we reject the null hypothesis. The significance level is 0.05.
03

Analyze Sample Data

The confidence interval is between 4.9 pounds to 5.3 pounds. We find that our hypothesized population mean (5 pounds) is within this confidence interval.
04

Interpret the Results

Since the hypothesized mean is within the confidence interval, we fail to reject the null hypothesis at the 0.05 significance level. Thus, there isn't strong enough evidence to conclude that the population mean is not 5 pounds.

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Key Concepts

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

Significance Level
In statistical hypothesis testing, the significance level is a critical concept that helps determine the likelihood of rejecting the null hypothesis when it is actually true. It's often denoted by the Greek letter alpha (\( \alpha \)). The significance level represents a threshold for deciding whether an observed effect is statistically significant. For example, a significance level of 0.05 means there is a 5% chance of rejecting the null hypothesis when it is true, also known as making a Type I error.

The choice of significance level can impact the conclusion of the hypothesis test. A lower significance level, like 0.01, would require stronger evidence to reject the null hypothesis compared to a higher level, such as 0.10. In the given exercise, the significance level is set at 0.05, so our decision hinges on whether the evidence from the data falls within this acceptable level of error.

When conducting a hypothesis test, always consider whether the chosen significance level suits the context of the study. This choice can affect the balance between risking Type I errors (false positives) and Type II errors (false negatives).
Confidence Interval
A confidence interval offers a range where we believe the true population parameter, like a mean, lies with a certain degree of confidence. Here, a 95% confidence interval suggests that we can be 95% confident the population mean falls within this range.

In our exercise, the confidence interval is between 4.9 pounds and 5.3 pounds. This range provides bounds for the true mean weight of the tomatoes. To link confidence intervals with hypothesis testing, look at whether the hypothesized mean value (5 pounds in this case) lies within the interval. If it does, like in our scenario, it suggests that the sample data does not provide enough evidence against the null hypothesis.

Confidence intervals are incredibly useful because they provide not just a single estimation (like a point estimate) but a range that reflects the possible variability of the estimate. It also inherently reflects the sample size, with larger samples generally narrowing the interval, offering a more precise estimate of the parameter.
Null Hypothesis
The null hypothesis is a key player in statistical hypothesis testing. It generally asserts there is no effect or no difference in the context being tested. The null hypothesis serves as a starting point or default assumption that there is no association or change, which we test against the evidence provided by data.

In the tomatoes exercise, the null hypothesis (\(H_0: \mu = 5\) suggests that the mean weight of the tomatoes is 5 pounds. When comparing to the alternative hypothesis, (\(H_1: \mu eq 5\)), we are testing whether there is statistically significant evidence to reject the idea that the mean is indeed 5 pounds.

In hypothesis testing, we either reject or fail to reject the null hypothesis based on statistical evidence and predefined significance criteria. Importantly, 'failing to reject' the null hypothesis doesn't prove it's true; it simply indicates insufficient evidence against it according to the study's design and data analysis. Making decisions around the null hypothesis helps guide research conclusions and the understanding of data patterns properly.

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Most popular questions from this chapter

Babies Weights (Example 2) 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; will be between \(6.4\) and \(7.6\) pounds? c. Explain the difference between a and \(\mathrm{b}\).

Vegetarians' Weights The mean weight of all 20-yearold women is 128 pounds (http://www.kidsgrowth.com). A random sample of 40 vegetarian women who are 20 years old showed a sample mean of 122 pounds with a standard deviation of 15 pounds. The women's measurements were independent of each other. a. Determine whether the mean weight for 20 -year old vegetarian women is significantly less than 128 , using a significance level of \(0.05\). b. Now suppose the sample consists of 100 vegetarian women who are 20 years old, and repeat the test. c. Explain what causes the difference between the \(\mathrm{p}\) -values for parts a and \(\mathrm{b}\).

Exam Scores The distribution of the scores on a certain exam is \(N(70,10)\), which means that the exam scores are Normally distributed with a mean of 70 and standard deviation of \(10 .\) a. Sketch the curve and label, on the \(x\) -axis, the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be bigger than 80\. Shade the region under the Normal curve whose area corresponds to this probability.

Oranges A statistics instructor randomly selected four bags of oranges, each bag labeled 10 pounds, and weighed the bags. They weighed \(10.2,10.5,10.3\), and \(10.3\) pounds. Assume that the distribution of weights is Normal. Find a \(95 \%\) confidence interval for the mean weight of all bags of oranges. Use technology for your calculations. a. Decide whether each of the following three statements is a correctly worded interpretation of the confidence interval, and fill in the blanks for the correct option(s). i. I am \(95 \%\) confident that the population mean is between ii. There is a \(95 \%\) chance that all intervals will be between iii. I am \(95 \%\) confident that the sample mean is between b. Does the interval capture 10 pounds? Is there enough evidence to reject the null hypothesis that the population mean weight is 10 pounds? Explain your answer.

Exam Scores The distribution of the scores on a certain exam is \(N(70,10)\), which means that the exam scores are Normally distributed with a mean of 70 and standard deviation of \(10 .\) a, Sketch the curve and label, on the \(x\) -axis, the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be between 50 and 90 . Shade the region under the Normal curve whose area corresponds to this probability.

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