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Four different paints are advertised as having the same drying time. To check the manufacturer's claims, five samples were tested for each of the paints. The time in minutes until the paint was dry enough for a second coat to be applied was recorded. The following data were obtained. \(\begin{array}{cccc}\text { Paint 1 } & \text { Paint 2 } & \text { Paint 3 } & \text { Paint 4 } \\ 128 & 144 & 133 & 150 \\ 137 & 133 & 143 & 142 \\\ 135 & 142 & 137 & 135 \\ 124 & 146 & 136 & 140 \\ 141 & 130 & 131 & 153\end{array}\) At the \(\alpha=.05\) level of significance, test to see whether the mean drying time is the same for each type of paint.

Short Answer

Expert verified
Perform an ANOVA test and compare the F-statistic with the critical value to test the null hypothesis.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that there is no difference in the mean drying times for the four paints, i.e., \(\mu_1 = \mu_2 = \mu_3 = \mu_4\). The alternative hypothesis (\(H_a\)) states that at least one paint has a different mean drying time, i.e., \(\mu_i eq \mu_j\) for some \(i eq j\).
02

Calculate Group Means and Overall Mean

Compute the mean drying time for each paint type and the overall mean of the drying times for all samples. - Paint 1 mean: \((128 + 137 + 135 + 124 + 141)/5 = 133\) - Paint 2 mean: \((144 + 133 + 142 + 146 + 130)/5 = 139\)- Paint 3 mean: \((133 + 143 + 137 + 136 + 131)/5 = 136\)- Paint 4 mean: \((150 + 142 + 135 + 140 + 153)/5 = 144\)- Overall mean: \((133 + 139 + 136 + 144)/4 = 138\)
03

Calculate Sum of Squares

Calculate the total sum of squares (SST), the sum of squares between groups (SSB), and the sum of squares within groups (SSW).- SST = \(\sum (X_{ij} - \text{Overall Mean})^2\)- SSB = \(n \times \sum (\text{Group mean} - \text{Overall mean})^2\)- SSW = \(\sum \sum(X_{ij} - \text{Group mean})^2\).Plug in numbers and calculate these sums.
04

Calculate Degrees of Freedom

Determine the degrees of freedom for between groups (k-1) and within groups (N-k), where \(k\) is the number of paints and \(N\) is the total number of observations, i.e., 4-1=3 and 20-4=16.
05

Compute Mean Square Values

Calculate mean square values for the between groups (MSB = SSB/df between) and within groups (MSW = SSW/df within).
06

Perform F-Test

Calculate the F-statistic by dividing the MSB by MSW. Compare this value with the critical F-value for df between and df within at \(\alpha=0.05\), using an F-distribution table.
07

Draw Conclusion

If the calculated F-statistic is greater than the critical F-value, reject the null hypothesis. If not, 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.

Hypothesis Testing
Hypothesis testing is a staple in statistics for determining if there is enough evidence to reject a theoretical assumption. It is primarily used to weigh the strength of evidence in a data set. In the context of this exercise, we are examining **whether different types of paint have the same drying time**.

Here's how it works:- **Null Hypothesis (\( H_0 \)):** This is the initial assumption that there is no effect or difference. For our paint drying times, the null hypothesis states that all paints have the same mean drying time.- **Alternative Hypothesis (\( H_a \)):** This opposite assertion claims that at least one paint has a different mean drying time. This is what you try to support with your data.
The goal is to use statistical calculations to determine whether the data you collected is consistent with the null hypothesis, or whether the data suggests the alternative hypothesis could be true. By selecting a significance level, such as \( \alpha=0.05 \), you set your threshold for deciding when to reject the null hypothesis. If your final test statistic is within the critical region, it means the result is statistically significant, leading to a rejection of the null hypothesis.
Mean Comparison
Comparing means across different groups is a crucial step in hypothesis testing, allowing us to explore whether different samples come from the same population. In the scenario with paints, **we calculate and examine the mean drying times** for each paint type.
To effectively compare these means, you do the following steps:- **Calculate the Mean:** Find the average drying time for each paint. For example, Paint 1's mean is calculated as shown, \( \text{Mean} = \frac{128 + 137 + 135 + 124 + 141}{5} = 133 \).- **Overall Mean:** This is the mean of all sample means, important for evaluating the overall variability. In our case, the overall mean for all paints is 138 as calculated.
By assessing how much each group's mean differs from the overall mean, you gather insights on whether the variances in means are large enough to reject the null hypothesis. Mean comparison is pivotal in determining whether observed differences are due to random chance or a true effect.
F-Test
An F-test is a statistical test used to compare variances and is most often used in the context of an Analysis of Variance (ANOVA). In our paint drying example, the **F-test helps determine if the means of several groups are equal**, supporting or refuting the hypothesis that paint type affects drying time.
Here's a breakdown:- **F-Statistic Calculation:** You calculate the F-statistic by dividing the mean square between the groups (MSB) by the mean square within the groups (MSW). This helps understand the degree of variance of the group means relative to the variance within each group.- **Critical F-Value:** Look up the critical F-value from the F-distribution table for your chosen significance level (\( \alpha = 0.05 \)) and degrees of freedom calculated from your data.
The key to evaluating the F-test is comparing the calculated F-statistic to the critical F-value. If the F-statistic is larger than the critical F-value, you have substantial evidence that the means are not equal and reject the null hypothesis. Through this **procedure**, the F-test is a robust method to infer the equality of group means based on sample data.

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Most popular questions from this chapter

A study reported in the Journal of Small Business Management concluded that selfemployed individuals do not experience higher job satisfaction than individuals who are not self-employed. In this study, job satisfaction is measured using 18 items, each of which is rated using a Likert-type scale with \(1-5\) response options ranging from strong agreement to strong disagreement. A higher score on this scale indicates a higher degree of job satisfaction. The sum of the ratings for the 18 items, ranging from \(18-90\), is used as the measure of job satisfaction. Suppose that this approach was used to measure the job satisfaction for lawyers, physical therapists, cabinetmakers, and systems analysts. The results obtained for a sample of 10 individuals from each profession follow. At the \(\alpha=.05\) level of significance, test for any difference in the job satisfaction among the four professions.

Bank of America's Consumer Spending Survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment (US Airways Attaché, December 2003 ). Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population 1 ) and the annual credit card charges for dining out (population 2 ). Using the difference data, the sample mean difference was \(d=\$ 850,\) and the sample standard deviation was \(s_{d}=\$ 1123\) a. Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out. b. Use a .05 level of significance. Can you conclude that the population means differ? What is the \(p\) -value? c. Which category, groceries or dining out, has a higher population mean annual credit card charge? What is the point estimate of the difference between the population means? What is the \(95 \%\) confidence interval estimate of the difference between the population means?

The following results are for independent random samples taken from two populations. \(\begin{array}{ll}\text { Sample } 1 & \text { Sample } 2 \\\ n_{1}=20 & n_{2}=30 \\ \bar{x}_{1}=22.5 & \bar{x}_{2}=20.1 \\ s_{1}=2.5 & s_{2}=4.8\end{array}\) a. What is the point estimate of the difference between the two population means? b. What is the degrees of freedom for the \(t\) distribution? c. \(\quad\) At \(95 \%\) confidence, what is the margin of error? d. What is the \(95 \%\) confidence interval for the difference between the two population means?

Consider the following hypothesis test. \(H_{0}: \mu_{1}-\mu_{2} \leq 0\) \(H_{\mathrm{a}}: \mu_{1}-\mu_{2}>0\) The following results are for two independent random samples taken from the two populations. \(\begin{array}{ll}\text { Sample } 1 & \text { Sample } 2 \\\ n_{1}=40 & n_{2}=50 \\ \bar{x}_{1}=25.2 & \bar{x}_{2}=22.8 \\ \sigma_{1}=5.2 & \sigma_{2}=6.0\end{array}\) a. What is the value of the test statistic? b. What is the \(p\) -value? c. With \(\alpha=.05,\) what is your hypothesis testing conclusion?

Commercial real estate prices and rental rates suffered substantial declines in the past year (Newsweek, July 27,2009 ). These declines were particularly severe in Asia; annual lease rates in Tokyo, Hong Kong, and singapore declined by \(40 \%\) or more. Even with such large declines, annual lease rates in Asia were still higher than those in many cities in Europe and the United States. Annual lease rates for a sample of 30 commercial properties in Hong Kong showed an average of \(\$ 1,114\) per square meter with a standard deviation of \(\$ 230\) Annual lease rates for a sample of 40 commercial properties in Paris showed an average lease rate of \(\$ 989\) per square meter with a standard deviation of \(\$ 195\) a. \(\quad\) On the basis of the sample results, can we conclude that the mean annual lease rate is higher in Hong Kong than in Paris? Develop appropriate null and alternative hypotheses. b. Use \(\alpha=.01 .\) What is your conclusion? Are rental rates higher in Hong Kong?

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