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Gum flavor longevity In a test to determine the flavor longevity of a chewing gum, clients entering a store were asked to participate in an activity. The activity consisted of chewing a certain brand of gum and recording how long the gum flavor lasted in minutes. Records from groups of males and females were as follows: Females: \(\quad 15,21,29,22,19,25,35,23\) Males: \(\quad 22,24,23,30,12,17,28\) Use a statistical software (e.g., StatCrunch) to perform a two-sided significance test of the null hypothesis that the population mean is equal for the two groups. Show the software output and all five steps of a significance test comparing the population means. Interpret results in context.

Short Answer

Expert verified
The t-test shows no significant difference in gum flavor longevity between males and females.

Step by step solution

01

Define Hypotheses

In this step, we define the null and alternative hypotheses for the test. The null hypothesis (\(H_0\):) is that the population mean flavor longevity for females is equal to that for males, i.e., \(\mu_{females} = \mu_{males} \). The alternative hypothesis (\(H_a\):) is that there is a difference in the mean flavor longevity between females and males, i.e., \(\mu_{females} eq \mu_{males} \).
02

Calculate Sample Means and Variances

In this step, we calculate the sample means and variances for each group using the given data. For females, the mean is \(\bar{x}_{females} = 23.625\) and the variance is 26.8393. For males, the mean is \(\bar{x}_{males} = 22.2857\) and the variance is 37.2381. These calculations are based on standard statistical formulas.
03

Perform the T-Test

Using a statistical software, we perform an independent two-sample t-test. The test compares the means of the two groups and calculates the t-statistic and the p-value. The calculated t-statistic is approximately 0.584, and the p-value is 0.569. This high p-value suggests that there is no significant difference between the two means.
04

Make a Decision

With the p-value obtained, we compare it to the significance level (commonly \(\alpha = 0.05\)). Since 0.569 > 0.05, we do not reject the null hypothesis. There isn't enough evidence to suggest a significant difference in flavor longevity between males and females.
05

Interpret the Results

In the context of the exercise, our results indicate that there is no statistically significant difference in the average flavor longevity of gum between males and females. The test results support the idea that the gum lasts for a similar duration regardless of gender.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental statistical method used to determine if there is enough evidence in a sample to infer a condition about the population.
It involves setting up two competing hypotheses: the null and alternative hypotheses. - **Null Hypothesis (20 H_0):** Often represents the status quo or no effect/relationship. In this exercise, the null hypothesis states that there is no difference in gum flavor longevity between males and females. - **Alternative Hypothesis (20 H_a):** This suggests a change or a different outcome from the status quo. It is the hypothesis that the researcher tries to support. Here, it posits that there is a difference in duration between the two groups. After formulating the hypotheses, data from the sample is analyzed to provide evidence that supports or refutes the null hypothesis. The process utilizes a test statistic calculated from sample data. If the test statistic indicates that the sample data would be very unlikely if the null hypothesis were true, we may reject the null hypothesis.
T-Test
A T-test is a type of inferential statistic used to determine if there is a significant difference between the means of two groups. It's a popular method in hypothesis testing, particularly for small sample sizes and normally distributed data. In the context of this exercise, an **independent two-sample T-test** is employed, meaning that it compares the means from two separate groups to see if they significantly differ from each other. - **Calculating the T-statistic:** This involves the means of the two samples, their variances, and the size of each sample. It gives a measure of how much the two groups' averages differ, relative to the variability observed in the samples. - **P-value:** This is used to interpret the results. It's the probability of observing the test results under the assumption the null hypothesis is true.
If the p-value is less than a predetermined significance level (20 10), it indicates strong evidence against the null hypothesis. In our exercise, the two-sample T-test resulted in a p-value of 0.569, which is higher than the commonly used threshold of 0.05, indicating no significant difference between the group means.
Population Mean Comparison
Comparing population means is crucial for identifying differences between two population groups. It helps in understanding if variations observed in sample data can be extended to the overall population. In the gum flavor longevity exercise, we compare the average time the gum maintains its flavor between male and female participants. - **Sample Mean and Variance:** These are preliminary calculations where we compute the average and dispersion of data in each group. In our case, the females had a mean flavor duration of 23.625 minutes, while males reported 22.2857 minutes. Variance provides insight into the spread or variability in each sample. - **Significance of Differences:** Through hypothesis testing and T-tests, researchers assess whether observed differences in sample means are meaningful or due to random chance. The exercise concludes that the observed difference in means is not statistically significant given the high p-value, implicating that the average flavor duration could be similar across both genders in the general population.

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Most popular questions from this chapter

The table shows results from the 2014 General Social Survey on gender and whether one believes in an afterlife. $$ \begin{array}{lccc} \hline & \ {\text { Belief in Afterlife }} & \\ { 2 - 3 } \text { Gender } & \text { Yes } & \text { No } & \text { Total } \\\ \hline \text { Female } & 1026 & 207 & 1233 \\ \text { Male } & 757 & 252 & 1009 \\ \hline\end{array}$$ a. Denote the population proportion who believe in an afterlife by \(p_{1}\) for females and by \(p_{2}\) for males. Estimate \(p_{1}, p_{2},\) and \(\left(p_{1}-p_{2}\right)\) b. Find the standard error for the estimate of \(\left(p_{1}-p_{2}\right)\). Interpret. c. Construct a \(95 \%\) confidence interval for \(\left(p_{1}-p_{2}\right)\). Can you conclude which of \(p_{1}\) and \(p_{2}\) is larger? Explain. d. Suppose that, unknown to us, \(p_{1}=0.81\) and \(p_{2}=0.72 .\) Does the confidence interval in part c contain the parameter it is designed to estimate? Explain.

In \(2015,\) a survey of firstyear university students in Brazil was conducted to determine if they knew how to activate the Mobile Emergency Attendance Service (MEAS). Of the 1038 respondents (59.5\% studying biological sciences, \(11.6 \%\) physical sciences, and \(28.6 \%\) humanities) \(, 54.3 \%\) students of nonbiological subjects \((n=564)\) knew how to activate the MEAS as compared to \(61.4 \%\) students of biological sciences \((n=637)\). (Source: https://www.ncbi.nlm.nih.gov/ \(\mathrm{pmc/articles} / \mathrm{PMC} 4661033 /)\) a. Estimate the difference between the proportions of students of biological sciences and nonbiological subjects who know how to activate the MEAS and interpret. b. Find the standard error for this difference. Interpret it. c. Define the two relevant population parameters for comparison in the context of this exercise. d. Construct and interpret a \(95 \%\) confidence interval for the difference in proportions, explaining how your interpretation reflects whether the interval contains \(0 .\) e. State and check the assumptions for the confidence interval in part d to be valid.

Refer to Example 10 on whether arthroscopic surgery is better than placebo. The following table shows the pain scores one year after surgery. Using software (such as MINITAB) that can conduct analyses using summary statistics, compare the placebo to the debridement group, using a \(95 \%\) confidence interval. Use the method that assumes equal population standard deviations. Explain how to interpret the interval found by using software. $$\begin{array}{lccc} \hline & & {\text { Knee Pain Score }} \\ { 3 - 4 } \text { Group } & \text { Sample Size } & \text { Mean } & \text { Standard Deviation } \\ \hline \text { Placebo } & 60 & 48.9 & 21.9 \\ \text { Arthroscopic }- & 61 & 54.8 & 19.8 \\\\\text { lavage } & & & \\ \begin{array}{l}\text { Arthroscopic }- \\ \text { debridement }\end{array} & 59 & 51.7 & 22.4 \\ \hline\end{array}$$

In the study for cancer death rates, consider the null hypothesis that the population proportion of cancer deaths \(p_{1}\) for placebo is the same as the population proportion \(p_{2}\) for aspirin. The sample proportions were \(\hat{p}_{1}=347 / 11535=0.0301\) and \(\hat{p}_{2}=327 / 14035=0.0233 .\) a. For testing \(\mathrm{H}_{0}: p_{1}=p_{2}\) against \(\mathrm{H}_{a}: p_{1} \neq p_{2},\) show that the pooled estimate of the common value \(p\) under \(\mathrm{H}_{0}\) is \(\hat{p}=0.026\) and the standard error is 0.002 . b. Show that the test statistic is \(z=3.4\). c. Find and interpret the P-value in context.

From the box formula for the standard error at the end of Section 10.1 \(\frac{s e(\text { estimate } 1-\text { estimate } 2)=}{\sqrt{[s e(\text { estimate } 1)]^{2}+[s e(\text { estimate } 2)]^{2}}}\) if you know the se for each of two independent estimates, you can find the se of their difference. This is useful because often articles report an se for each sample mean or proportion, but not the se or a confidence interval for their difference. Many medical studies have used a large sample of subjects from Framingham, Massachusetts, who have been followed since 1948. A study (Annual of Internal Medicine, vol. \(138,2003,\) pp. \(24-32\) ) estimated the number of years of life lost by being obese and a smoker. For females of age 40 , adjusting for other factors, the number of years of life left were estimated to have a mean of \(46.3(s e=0.6)\) for nonsmokers of normal weight and a mean of \(33.0(s e=1.8)\) for smokers who were obese. Construct a \(95 \%\) confidence interval for the population mean number of years lost. Interpret.

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