Question

A manufacturer of hard safety hats for construction workers is concerned about the mean and the variation of the forces helmets transmit to wearers when subjected to a standard external force. The manufacturer desires the mean force transmitted by helmets to be 800 pounds (or less), well under the legal 1000 -pound limit, and \(\sigma\) to be less than \(40 .\) A random sample of \(n=40\) helmets was tested, and the sample mean and variance were found to be equal to 825 pounds and 2350 pounds \(^{2}\), respectively. a. If \(\mu=800\) and \(\sigma=40,\) is it likely that any helmet, subjected to the standard external force, will transmit a force to a wearer in excess of 1000 pounds? Explain. b. Do the data provide sufficient evidence to indicate that when the helmets are subjected to the standard external force, the mean force transmitted by the helmets exceeds 800 pounds?

Short answer

Also, does the sample data provide sufficient evidence that the mean force transmitted by the helmets exceeds 800 pounds? Answer: No, it is highly unlikely that a helmet will transmit a force in excess of 1000 pounds if the population mean is 800 pounds and the standard deviation is 40 pounds. And yes, the sample data provides sufficient evidence that the mean force transmitted by the helmets exceeds 800 pounds.

Step-by-step solution

Find the probability of transmitting a force above 1000 pounds

First, let's calculate the z-score for transmitting a force of 1000 pounds or more. The formula for z-score is: $$ Z = \frac{X - \mu}{\sigma} $$ Where \(X\) is the value we're interested in, \(\mu\) is the population mean, and \(\sigma\) is the population standard deviation. Plugging in the values: $$ Z = \frac{1000 - 800}{40} $$ $$ Z = 5 $$ Now, we can find the probability of transmitting a force above 1000 pounds using a standard normal distribution table. We look up the value for Z = 5 in the table, which gives us a probability of close to 1. Therefore, the probability of transmitting a force below 1000 pounds is very close to 0. Hence, it is highly unlikely that any helmet will transmit a force in excess of 1000 pounds if the population mean is 800 pounds and standard deviation is 40 pounds.

Perform hypothesis testing

In this step, we want to test if the sample data provides sufficient evidence that the mean force transmitted by the helmets exceeds 800 pounds. First, let's write the null and alternative hypotheses: - Null hypothesis (H0): \(\mu = 800\) - Alternative hypothesis (H1): \(\mu > 800\) Now, we need to calculate the t-score. The formula for the t-score is: $$ t = \frac{\bar{X} - \mu}{\frac{S}{\sqrt{n}}} $$ Where \(\bar{X}\) is the sample mean, \(S\) is the sample standard deviation, and \(n\) is the sample size. Here, we're given the sample variance, so we first need to find the sample standard deviation by taking the square root of the sample variance: $$ S = \sqrt{2350} \approx 48.47 $$ Now, we can plug in the values into the t-score formula: $$ t = \frac{825 - 800}{\frac{48.47}{\sqrt{40}}} $$ $$ t \approx 2.58 $$ Next, we need to find the critical value for the t-distribution with 39 degrees of freedom and a one-tailed significance level of 0.05. Using a t-distribution table or an online calculator, we find that the critical value is approximately 1.685. Since the calculated t-score (2.58) is greater than the critical value (1.685), we reject the null hypothesis in favor of the alternative hypothesis. Therefore, the data provides sufficient evidence to indicate that when the helmets are subjected to the standard external force, the mean force transmitted by the helmets exceeds 800 pounds.

Key concepts

Z-score

The **Z-score** is a statistical measure that helps us understand where a data point lies in relation to the mean of a data set. In simpler terms, it tells us how far and in which direction a value deviates from the mean, measured in standard deviations.

For example, in the context of our safety helmet manufacturer exercise, we are interested in knowing how unusual it is for a helmet to transmit a force of 1000 pounds when the expected mean force is 800 pounds. To determine this, we calculate the Z-score using the formula: \[Z = \frac{X - \mu}{\sigma}\]

In this exercise:- \(X\) is the value of interest (1000 pounds).- \(\mu\) is the mean (800 pounds).- \(\sigma\) is the standard deviation (40 pounds).

This results in:\[Z = \frac{1000 - 800}{40} = 5\]

A Z-score of 5 indicates that 1000 pounds is 5 standard deviations above the mean, which suggests that it is extremely unlikely for a helmet to transmit such a force if the true mean is indeed 800 pounds.

T-score

The **T-score** is vital for determining whether we can conclude that the sample mean is significantly different from a known or hypothesized population mean, especially when the sample size is small or the population standard deviation is unknown.

In the helmet scenario, we perform hypothesis testing to see if the mean force transmitted is indeed greater than 800 pounds, as the sample mean calculated from the test helmets is 825 pounds. Since the population standard deviation is unknown, we use the T-score calculated as:\[t = \frac{\bar{X} - \mu}{\frac{S}{\sqrt{n}}}\]
Where:

• \(\bar{X}\) is the sample mean (825 pounds).
• \(\mu\) is the population mean under the null hypothesis (800 pounds).
• \(S\) is the sample standard deviation, derived from the sample variance.
• \(n\) is the sample size (40).

Calculated here:\[t = \frac{825 - 800}{\frac{48.47}{\sqrt{40}}} \approx 2.58\]

A T-score of 2.58 compared against the critical T-score threshold of 1.685 (for a significance level of 0.05 and 39 degrees of freedom) leads us to reject the null hypothesis, indicating the sample mean significantly exceeds 800 pounds.

Standard Deviation

The **Standard Deviation (\(\sigma\))** is a key concept in statistics that measures the amount of variation or dispersion in a set of values. It provides insight into how spread out the data is around the mean.

In this safety helmet exercise, standard deviation is essential for understanding how much the forces transmitted by these helmets might vary. A lower standard deviation signifies that the forces are closely clustered around the mean, which is desirable when consistency is crucial.

Given that the population standard deviation is stated as 40 pounds in the hypothetical scenario, it tells us about the expected range within which most of the helmet forces should fall. When we calculated the sample standard deviation from the provided variance of 2350 pounds squared:\[S = \sqrt{2350} \approx 48.47\]
This computed sample standard deviation of 48.47 pounds indicates an observation of more variability in the sample than in the population hypothetically expected. Such information helps in understanding the consistency and reliability of the helmets when subjected to the standard external force.

Null and Alternative Hypotheses

In hypothesis testing, the **Null and Alternative Hypotheses** help us determine if there’s enough evidence in a sample of data to infer that a certain condition is true for the entire population.

**Null Hypothesis (\(H_0\)):** This is the initial assumption that there is no effect or difference. For the helmet exercise, the null hypothesis states that the mean force transmitted by the helmets is 800 pounds.- \(H_0: \mu = 800\)

**Alternative Hypothesis (\(H_1\)):** This suggests that there is a statistically significant effect or difference. In our case, it proposes that the mean force transmitted exceeds 800 pounds.- \(H_1: \mu > 800\)

The hypothesis test, through the calculation of the T-score, allows us to decide between these two hypotheses. If the evidence (T-score) is strongly against the null hypothesis, we reject it in favor of the alternative hypothesis. In this scenario, the calculated T-score indicated strong evidence that the mean force is indeed greater than 800 pounds, hence rejecting the null hypothesis.