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The accompanying observations on residual flame time (sec) for strips of treated children's nightwear were given in the article "An Introduction to Some Precision and Accuracy of Measurement Problems" (J. Test. Eval., 1982: 132-140). Suppose a true average flame time of at most \(9.75\) had been mandated. Does the data suggest that this condition has not been met? Carry out an appropriate test after first investigating the plausibility of assumptions that underlie your method of inference. \(\begin{array}{lllllll}9.85 & 9.93 & 9.75 & 9.77 & 9.67 & 9.87 & 9.67 \\ 9.94 & 9.85 & 9.75 & 9.83 & 9.92 & 9.74 & 9.99 \\ 9.88 & 9.95 & 9.95 & 9.93 & 9.92 & 9.89 & \end{array}\)

Short Answer

Expert verified
The data suggests the condition has not been met; the true mean flame time is likely greater than 9.75 seconds.

Step by step solution

01

Check Assumptions

Before conducting a hypothesis test, we need to check the assumptions of normality and independence for the sample data. The data consists of measurements of flame times from nightwear strips, which can typically be assumed to be independent. To check normality, we can create a histogram or a Q-Q plot of the data.
02

State the Hypotheses

We need to establish our null and alternative hypotheses. The null hypothesis (H_0) states that the true mean flame time is equal to or less than 9.75 seconds, which the mandate requires: \(H_0: \mu \leq 9.75\). The alternative hypothesis (H_a) states that the true mean flame time is greater than 9.75 seconds: \(H_a: \mu > 9.75\).
03

Calculate the Sample Mean and Standard Deviation

First, we find the sample mean \(\bar{x}\) and standard deviation \(s\) from the data. The sample mean is calculated as the sum of all observations divided by the number of observations, and the standard deviation measures the dispersion of the data around the mean.
04

Conduct a One-Sample T-Test

Since the population standard deviation is unknown and the sample size is small, we use a one-sample t-test to compare the sample mean to the hypothesized mean of 9.75. Calculate the test statistic \(t\) using \(t = \frac{\bar{x} - 9.75}{s/\sqrt{n}}\), where \(n\) is the sample size. Then, compare the calculated t-value to the critical t-value from the t-distribution table at an appropriate significance level (e.g., 0.05 for a 95% confidence level) and with \(n-1\) degrees of freedom.
05

Make a Decision

If the calculated t-value is greater than the critical t-value, we reject the null hypothesis and conclude that the true mean is greater than 9.75 seconds. Otherwise, we do not have enough evidence 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.

T-Test
The T-Test is a statistical test used to compare the sample mean of a dataset with a known value, especially when the population standard deviation is unknown. In this case, we are interested in determining if the average flame time of treated children's nightwear exceeds the mandated 9.75 seconds. When dealing with small sample sizes, the one-sample T-test is particularly useful.

A one-sample T-test involves calculating the T-statistic, which evaluates whether there is a significant difference between the sample mean and a known mean. The formula for the T-statistic is:
  • \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \)
where:
  • \(\bar{x}\) is the sample mean.
  • \(\mu_0\) is the hypothesized mean (in this case, 9.75 seconds).
  • \(s\) is the sample standard deviation.
  • \(n\) is the sample size.
The calculated T-statistic is compared to a critical value from the T-distribution table. If the T-statistic exceeds this critical value, we reject the null hypothesis. This would suggest that the true average flame time is indeed greater than 9.75 seconds.
Normality Assumption
Before performing a T-test, it is crucial to check if the data follows a normal distribution—this is known as the normality assumption. When the sample size is small (typically less than 30), non-normal data can lead to inaccurate results.

There are several ways to evaluate normality:
  • Graphical methods such as histograms or Q-Q plots help visualize if the data is approximately normal.
  • Numerical tests, like the Shapiro-Wilk test, provide a statistical basis for assessing normality.
In this situation, since the dataset consists of flame time measurements, it is generally assumed to be normally distributed, especially if the data was collected under similar conditions.
Sample Mean
The sample mean is a measure of central tendency, representing the average value of a dataset. It is calculated by summing all observed values and dividing by the number of observations.

For the flame time data, the sample mean is expressed as:
  • \(\bar{x} = \frac{\text{sum of all flame times}}{\text{number of observations}}\)
This value gives us an estimate of the average flame time of the treated children's nightwear. By comparing the sample mean to the mandated mean (9.75 seconds), we can infer whether the mean flame time differs significantly from the standard. The sample mean is critical in the T-test calculation, serving as a reference point in hypothesis testing.
Standard Deviation
Standard deviation is a measure of data dispersion around the mean. It quantifies how much individual data points deviate from the average value. A high standard deviation indicates widespread values, while a low standard deviation suggests that values are clustered closely around the mean.

In the context of our hypothesis test, the standard deviation plays a crucial role. It is used in the calculation of the T-statistic:
  • \( t = \frac{\bar{x} - 9.75}{s/\sqrt{n}} \)
where \(s\) represents the sample standard deviation. Using the sample standard deviation helps in understanding the variability in flame times and strengthens the basis of our statistical conclusions. By knowing the extent of variability, we can better assess whether observed differences between the sample mean and hypothesized mean are statistically significant.

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Most popular questions from this chapter

Let the test statistic \(Z\) have a standard normal distribution when \(H_{0}\) is true. Give the significance level for each of the following situations: a. \(H_{\mathrm{a}}: \mu>\mu_{0}\), rejection region \(z \geq 1.88\) b. \(H_{\mathrm{a}}: \mu<\mu_{0}\), rejection region \(z \leq-2.75\) c. \(H_{\mathrm{a}}: \mu \neq \mu_{0}\), rejection region \(z \geq 2.88\) or \(z \leq\) \(-2.88\)

Let \(\mu\) denote true average serum receptor concentration for all pregnant women. The average for all women is known to be \(5.63\). The article "Serum Transferrin Receptor for the Detection of Iron Deficiency in Pregnancy" (Amer. J. Clin. Nutrit., 1991: 1077-1081) reports that \(P\)-value \(>.10\) for a test of \(H_{0}: \mu=5.63\) versus \(H_{\mathrm{a}}: \mu \neq 5.63\) based on \(n=176\) pregnant women. Using a significance level of \(.01\), what would you conclude?

A new method for measuring phosphorus levels in soil is described in the article "A Rapid Method to Determine Total Phosphorus in Soils" (Soil Sci. Amer. \(J ., 1988: 1301-1304)\). Suppose a sample of 11 soil specimens, each with a true phosphorus content of \(548 \mathrm{mg} / \mathrm{kg}\), is analyzed using the new method. The resulting sample mean and standard deviation for phosphorus level are 587 and 10 , respectively. a. Is there evidence that the mean phosphorus level reported by the new method differs significantly from the true value of \(548 \mathrm{mg} / \mathrm{kg}\) ? Use \(\alpha=.05\). b. What assumptions must you make for the test in part (a) to be appropriate?

A regular type of laminate is currently being used by a manufacturer of circuit boards. A special laminate has been developed to reduce warpage. The regular laminate will be used on one sample of specimens and the special laminate on another sample, and the amount of warpage will then be determined for each specimen. The manufacturer will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate. State the relevant hypotheses, and describe the type I and type II errors in the context of this situation.

The recommended daily dietary allowance for zinc among males older than age 50 years is \(15 \mathrm{mg} /\) day. The article "Nutrient Intakes and Dietary Pattems of Older Americans: A National Study" (J. Gerontol., 1992: M145-150) reports the following summary data on intake for a sample of males age 65-74 years: \(n=115, \bar{x}=11.3\), and \(s=6.43\). Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance?

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