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An article in Fire Technology ["An Experimental Examination of Dead Air Space for Smoke Alarms" \((2009,\) Vol. \(45,\) pp. \(97-115\) ) ] studied the performance of smoke detectors installed not less than \(100 \mathrm{~mm}\) from any adjoining wall if mounted on a flat ceiling, and not closer than \(100 \mathrm{~mm}\) and not farther than 300 \(\mathrm{mm}\) from the adjoining ceiling surface if mounted on walls. The purpose of this rule is to avoid installation of smoke alarms in the "dead air space" where it is assumed to be difficult for smoke to reach. The paper described a number of interesting experiments. Results on the time to signal (in seconds) for one such experiment with pine stick fuel in an open bedroom using photoelectric smoke alarms are as follows: \(220,225,297,315,282,\) and \(313 .\) (a) Is there sufficient evidence to support a claim that the mean time to signal is less than 300 seconds? (b) Is there practical concern about the assumption of a normal distribution as a model for the time-to-signal data? (c) Find a \(95 \%\) two-sided \(\mathrm{CI}\) on the mean time to signal.

Step by step solution

01

Formulate the Hypotheses

For part (a), we want to test if the mean time to signal, \( \mu \), is less than 300 seconds. We formulate the null hypothesis as \( H_0: \mu = 300 \) seconds and the alternative hypothesis as \( H_a: \mu < 300 \) seconds.

02

Calculate the Sample Mean and Standard Deviation

The given times to signal are: 220, 225, 297, 315, 282, and 313. Calculate the sample mean (\( \bar{x} \)) and standard deviation (\( s \)).\[\bar{x} = \frac{220 + 225 + 297 + 315 + 282 + 313}{6} = 275.33 \text{ seconds}\]Calculate the standard deviation \( s \) using:\[s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}}\]First, \( (x_i - \bar{x})^2 \) terms are calculated, summed, and then used to find \( s=41.14 \) seconds.

03

Perform the One-Sample t-Test

We use the sample mean and standard deviation to calculate the t-statistic:\[t = \frac{\bar{x} - 300}{s/\sqrt{n}} = \frac{275.33 - 300}{41.14/\sqrt{6}} = -1.924\]Using a t-table or calculator for \( n-1 = 5 \) degrees of freedom and a one-tailed test at the 0.05 significance level, compare the t-statistic to find the p-value. The critical value for t at \( \alpha = 0.05 \) is approximately -2.015, and the calculated t-statistic is less, so we fail to reject \( H_0 \).

04

Assess Normality Assumption

For part (b), inspect the sample size, which is small (\( n = 6 \)), making the assessment of normality crucial. We look at normal probability plots or use Shapiro-Wilk Test if software is available. However, with small sample size, conclusions may not be very strong. Thus, unless there is visual evidence of non-normality, assume normality.

05

Compute the 95% Confidence Interval

For part (c), compute the 95% confidence interval using:\[\bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}\]For \( \alpha = 0.05 \) and \( n-1 = 5 \), \( t_{0.025} \approx 2.571 \).\[275.33 \pm 2.571 \times \frac{41.14}{\sqrt{6}}\]CI result: \((237.67, 313.00)\) seconds.

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Key Concepts

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

Confidence Interval

A confidence interval provides a range of values that is likely to contain the population mean. If we say we have a 95% confidence interval, it means we are 95% confident that the true mean of the population lies within this range.
The confidence interval is calculated by taking the sample mean and then adding and subtracting the margin of error. The formula for a confidence interval is:\[\bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}\]Where:- \(\bar{x}\) is the sample mean,- \(t_{\alpha/2}\) is the t-value for your confidence level, and- \(s\) is the sample standard deviation.
In this case, the 95% confidence interval for the mean time to signal was calculated using the above formula, resulting in the interval (237.67, 313.00) seconds. This indicates that we can be 95% confident that the average time it takes for the smoke alarm to signal is between 237.67 to 313.00 seconds.
Confidence intervals are important because they provide insight into the variability in estimates of the mean and can guide decision-making by encapsulating sample error and statistical inference.

t-Distribution

The t-distribution is a type of probability distribution that is used when estimating population parameters when the sample size is small and/or when the population standard deviation is unknown. In hypothesis testing and confidence interval estimations, the t-distribution accounts for increased variability typically observed in small samples.
It generally looks similar to the normal distribution but with heavier tails, meaning there's a greater chance for large values. The shape of the t-distribution is determined by the degrees of freedom, calculated as the sample size minus one ( −1). The less data you have, the "heavier" the tails will be. - As you collect more data, the t-distribution approaches a normal distribution. - For the exercise, we calculated the t-statistic with 5 degrees of freedom (n=6). Using the t-distribution, we can find critical values needed to determine confidence intervals and critical thresholds for hypothesis tests.

Normality Assumption

The normality assumption in statistics assumes that the data follows a normal distribution, a bell-shaped curve. It's a key requirement for various statistical tests and confidence interval calculations, especially when working with small sample sizes.
For this particular exercise, since our sample size is small ( =6), it's important to assess whether the data might be approximately normally distributed. If it isn't, the results of our statistical tests might not be reliable. - We can visually inspect the distribution using normal probability plots. - Another option, if available, is to conduct statistical tests like the Shapiro-Wilk test. If there is no strong evidence of non-normality, it is customary to proceed with the analysis under the assumption of normality, understanding the limitations that come with small sample sizes.

Sample Mean and Standard Deviation Calculations

Calculating the sample mean and standard deviation are the first steps in analyzing data. The sample mean, denoted by \(\bar{x}\), provides a measure of the central location of the data.
To calculate the sample mean, sum all the data points and then divide by the number of data points. In this exercise:\[\bar{x} = \frac{220 + 225 + 297 + 315 + 282 + 313}{6} = 275.33 \text{ seconds}\]The sample standard deviation, \(s\), provides a measure of the dispersion or variability of the data. It's calculated using the formula:\[s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}}\]This involves finding the difference of each data point from the mean, squaring these differences, summing them, and dividing by \(n-1\), where \(n\) is the sample size. Finally, take the square root of this quotient to find the standard deviation, which in this case was 41.14 seconds.These calculations are foundational as they pave the way for further statistical analysis like constructing confidence intervals and hypothesis testing.