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A primer paint can be used on aluminum panels. The primer's drying time is an important consideration in the manufacturing process. Twenty panels are selected, and the drying times are as follows: \(1.6,1.3,1.5,1.6,1.7,1.9,1.8,1.6,\) \(1.4,1.8,1.9,1.8,1.7,1.5,1.6,1.4,1.3,1.6,1.5,\) and \(1.8 .\) Is there evidence that the mean drying time of the primer exceeds \(1.5 \mathrm{hr} ?\)

Short Answer

Expert verified
Yes, there is evidence that the mean drying time exceeds 1.5 hours.

Step by step solution

01

Formulate Hypotheses

To determine if the mean drying time exceeds 1.5 hours, we can use a hypothesis test. The null hypothesis \(H_0\) is that the mean drying time \(\mu\) is equal to 1.5 hours, and the alternative hypothesis \(H_a\) is that the mean drying time exceeds 1.5 hours. Mathematically, \( H_0: \mu = 1.5 \) and \( H_a: \mu > 1.5 \).
02

Calculate the Sample Mean

To compute the sample mean \( \bar{x} \), sum all the drying times and divide by the number of observations, which is 20. Calculating, we have \( \bar{x} = \frac{1.6 + 1.3 + 1.5 + 1.6 + 1.7 + 1.9 + 1.8 + 1.6 + 1.4 + 1.8 + 1.9 + 1.8 + 1.7 + 1.5 + 1.6 + 1.4 + 1.3 + 1.6 + 1.5 + 1.8}{20} = 1.63 \).
03

Compute the Sample Standard Deviation

First, find the sum of squared deviations from the sample mean. Then divide by the number of observations minus one to find the variance, and take the square root to get the sample standard deviation \( s \). Calculating, the standard deviation \( s \approx 0.207 \).
04

Perform the t-Test

Using the sample mean \( \bar{x} = 1.63 \), hypothesized mean \( \mu_0 = 1.5 \), standard deviation \( s \approx 0.207 \), and sample size \( n = 20 \), compute the t-statistic: \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \approx 2.8 \).
05

Determine the Critical Value and Conclusion

For a one-tailed t-test with \( n - 1 = 19 \) degrees of freedom, check a t-distribution table for a critical value at a significance level (commonly \( \alpha = 0.05 \)). The critical value is approximately 1.729. Since the calculated t-value (2.8) is greater than the critical value (1.729), we reject the null hypothesis.

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Key Concepts

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

t-test
A t-test is a statistical method used to compare a sample mean to a hypothetical or known population mean. In this case, we're interested in testing whether the mean drying time of a primer on aluminum panels is greater than a certain threshold, 1.5 hours. This situation is ideal for a one-sample t-test.

The t-test helps us decide whether any observed difference between the sample mean and the hypothesized population mean is statistically significant. It accounts for both the variability in the sample data and the size of the sample. Through the t-test, the null hypothesis postulates no difference or effect, stating that the mean drying time is 1.5 hours. The alternative hypothesis requires strong evidence to conclude that the mean drying time is actually greater than 1.5 hours.

Performing a t-test involves calculating a t-statistic, which is then compared against a critical value from the t-distribution table. This tells us if the null hypothesis should be rejected in favor of the alternative hypothesis. If the calculated t-value exceeds the critical value, it suggests that the observed sample mean is significantly different from the hypothesized mean.
sample mean
The sample mean is an important statistic that acts as an estimate of the population mean. It's calculated by summing all the individual data points and dividing by the number of observations in the dataset. In our example, the drying times of 20 aluminum panels were averaged.

Mathematically, the sample mean \( \bar{x} \) is calculated as follows: \( \bar{x} = \frac{\sum x_i}{n} \). The sum symbol represents adding together all the drying times, and \( n \) is the total number of panels tested, which is 20 in this instance. After performing the arithmetic, the sample mean was determined to be 1.63 hours, indicating that the drying process took an average of this time across all tested samples.

Understanding the sample mean helps in forming a valid hypothesis test as it serves as a direct comparison to the hypothesized population mean, guiding us in decision-making about whether the process exceeds known benchmarks.
standard deviation
Standard deviation is crucial in statistics as it measures the amount of variation or dispersion in a set of data points. It tells us how much the individual data points deviate from the sample mean on average.

To compute the sample standard deviation, first calculate the squared differences between each data point and the sample mean. These squared differences are then averaged by dividing by \( n - 1 \) (where \( n \) is the number of observations) to find the variance, before taking the square root to get the standard deviation. This two-step process is:
  • Calculate each data point’s deviation from the mean, square each, and sum them up.
  • Divide by \( n - 1 \) (for a sample rather than a whole population) and take the square root.
In the primer example, the calculated standard deviation is approximately 0.207. This indicates that the drying times for the panels varied slightly around the mean of 1.63 hours, helping us understand the consistency of the drying process. A smaller standard deviation points to more consistency.
critical value
The critical value in hypothesis testing acts as a threshold for deciding whether to reject the null hypothesis. It corresponds to a certain confidence level and is derived from a statistical distribution appropriate for the data and hypothesis type—in this case, the t-distribution.

When conducting a t-test, the critical value is determined based on the degrees of freedom (which is one less than the sample size, \( n - 1 \)) and the significance level (\( \alpha \)), which is often set at 0.05 for a 95% confidence. This critical value signifies the cut-off point, beyond which we consider the observed result (t-statistic) statistically significant enough to reject the null hypothesis.

In our example, with 19 degrees of freedom, the critical value at the 0.05 level was approximately 1.729. Since the calculated t-statistic of 2.8 exceeded this critical value, it provided sufficient evidence to reject the null hypothesis, concluding that the mean drying time likely exceeds 1.5 hours. Recognizing the correct critical value is key in determining the outcome of hypothesis testing.

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Most popular questions from this chapter

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