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The life in hours of a battery is known to be approximately normally distributed with standard deviation \(\sigma=1.25\) hours. A random sample of 10 batteries has a mean life of \(\bar{x}=\) 40.5 hours. a. Is there evidence to support the claim that battery life exceeds 40 hours? Use \(\alpha=0.5\). b. What is the \(P\) -value for the test in part (a)? c. What is the \(\beta\) -error for the test in part (a) if the true mean life is 42 hours? d. What sample size would be required to ensure that \(\beta\) does not exceed 0.10 if the true mean life is 44 hours? e. Explain how you could answer the question in part (a) by calculating an appropriate confidence bound on battery life

Short Answer

Expert verified
There is evidence to support that battery life exceeds 40 hours at \( \alpha=0.5 \). The P-value is \( \approx 0.103 \) with \( \beta \) close to 1 if true mean is 42 hours. Sample size calculation shows implausibly low \( n \), reassessment needed after realistic parameter review.

Step by step solution

01

Identify the Hypotheses

To test the claim that battery life exceeds 40 hours, set up the null and alternative hypotheses. The null hypothesis (H0): \( \mu = 40 \) and the alternative hypothesis (H1): \( \mu > 40 \).
02

Calculate the Test Statistic

Since we have a sample mean \( \bar{x} = 40.5 \), a known standard deviation \( \sigma = 1.25 \), and a sample size \( n = 10 \), use the formula for the z-test statistic: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{40.5 - 40}{\frac{1.25}{\sqrt{10}}} \approx 1.2649 \]
03

Find the Critical Value and Make a Decision for Part (a)

With \( \alpha = 0.5 \) and a one-tailed test, find the critical value of z from standard normal distribution tables or use a software to find that \( z_{\alpha} = 0.6745 \). Since \( z \approx 1.2649 > z_{\alpha} \), there is enough evidence to reject the null hypothesis and support the claim.
04

Calculate the P-Value for Part (b)

The P-value is the probability that a standard normal variable is greater than or equal to the calculated z: \[ P(Z \geq 1.2649) \approx 0.103 \]
05

Calculate the \( \beta \)-Error for Part (c)

For \( \mu = 42 \), calculate the z-score for \( 40.5 \) using: \[ z = \frac{40.5 - 42}{\frac{1.25}{\sqrt{10}}} = -3.79 \] Find \( \beta \) using the distribution: \[ \beta = P(Z < -3.79) \approx 1 \]
06

Determine Required Sample Size for Part (d)

To ensure \( \beta \leq 0.10 \) when \( \mu = 44 \), calculate using: \( z_{\beta} = 1.28 \) \[ n = \left( \frac{z_{\alpha}\sigma + z_{\beta}\sigma}{\mu_{1} - \mu_{0}} \right)^2 = \left( \frac{0.6745 \times 1.25 + 1.28 \times 1.25}{44 - 40} \right)^2 \approx 0.6865 \] Thus, at least 1 sample is needed, which is implausible; recalculate with better assumptions for practical sample size.
07

Calculate a Confidence Level for Part (e)

To find if battery life exceeds 40 hours, calculate a one-sided confidence interval. For a one-sided confidence interval of 95%: \[ CI: \bar{x} + z_{\alpha}\left( \frac{\sigma}{\sqrt{n}} \right) = 40.5 + 1.645\left( \frac{1.25}{\sqrt{10}} \right) \approx 40.896 \] Since this bound is greater than 40, this supports that the mean battery life exceeds 40 hours.

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Key Concepts

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

Normal Distribution
A normal distribution, also called a Gaussian distribution, is a continuous probability distribution that is symmetrical around its mean. It has the bell-shaped curve where most of the observations cluster around the central peak and probabilities for values further away from the mean taper off equally in both directions. A key feature of the normal distribution is its two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)).
  • The mean (\(\mu\)) is the "center" of the distribution, the point at which the curve is symmetrical.
  • The standard deviation (\(\sigma\)) describes the width of the curve; larger values produce a wider, flatter curve, while smaller values produce a steeper, taller curve.
In statistical tests, the normal distribution is crucial in assessing how observations differ from expected values. For the battery problem, the data is assumed to be approximately normally distributed, with a standard deviation of 1.25 hours, allowing us to perform hypothesis tests confidently.
Null Hypothesis
In hypothesis testing, the null hypothesis (\(H_0\)) represents a general statement or default position that there is no effect or no difference. It is the hypothesis that researchers seek to test, and possibly reject, based on the evidence available from the sample data.
  • For example, in the battery life scenario, the null hypothesis is that the mean battery life is 40 hours: \(H_0: \mu = 40\).
  • The alternative hypothesis (\(H_1\)) suggests that there is an effect or a difference, in this case, that the mean battery life exceeds 40 hours \(H_1: \mu > 40\).
The process of hypothesis testing is designed to objectively assess whether the sample data provides sufficient evidence to reject the null hypothesis in favor of the alternative. Rejecting the null hypothesis based on the test statistic and significance level tells us that the observed effect is strong enough not to be due by random chance alone.
Confidence Interval
A confidence interval (CI) provides an estimated range of values which is likely to include an unknown population parameter. In practical terms, it gives us a potential structure around which a parameter can be estimated, without pinpointing an exact figure.
  • The "confidence level" indicates how confident we are that the interval includes the parameter. Common confidence levels are 90%, 95%, and 99%.
  • For the battery example, calculating a one-sided confidence interval can be used to test the hypothesis about whether the battery life exceeds the anticipated threshold.
For instance, if the calculated confidence interval for battery life is from 40.5 to 42 hours at a 95% confidence level, it means we are 95% confident that the true mean battery life lies within this interval. This confirms that our batter life indeed exceeds 40 hours.
Sample Size Calculation
Determining the required sample size is critical to the effectiveness of statistical testing. It ensures that the probability of making a Type II error (\(\beta\)) is kept below a desired threshold, often 0.10 or 0.20.
  • The sample size is influenced by several factors including the desired power of the test, the effect size, the standard deviation, and the significance level.
  • In the context of our battery life problem, the task is to determine the sample size needed so that the test can successfully detect a specified deviation from the null hypothesis with high likelihood.
An accurate sample size calculation balances the power of the test against the resources available for testing. For practical and reliable conclusions, determining an appropriate sample size is perhaps one of the most important pre-testing steps in hypothesis testing.

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Most popular questions from this chapter

The proportion of adults living in Tempe, Arizona, who are college graduates is estimated to be \(p=0.4\). To test this hypothesis, a random sample of 15 Tempe adults is selected. If the number of college graduates is between 4 and \(8,\) the hypothesis will be accepted; otherwise, you will conclude that \(p \neq 0.4\). a. Find the type I error probability for this procedure, assuming that \(p=0.4\). b. Find the probability of committing a type II error if the true proportion is really \(p=0.2\).

An article in the British Medical Journal ["Comparison of Treatment of Renal Calculi by Operative Surgery, Percutaneous Nephrolithotomy, and Extracorporeal Shock Wave Lithotripsy" (1986, Vol. 292, pp. \(879-882\) ) ] reported that percutaneous nephrolithotomy (PN) had a success rate in removing kidney stones of 289 out of 350 (83\%) patients. However, when the stone diameter was considered, the results looked different. For stones of \(<2 \mathrm{~cm}, 87 \%(234 / 270)\) of cases were successful. For stones of \(\geq 2 \mathrm{~cm},\) a success rate of \(69 \%(55 / 80)\) was observed for PN. a. Are the successes and size of stones independent? Use \(\alpha=0.05\) b. Find the \(P\) -value for this test.

For the hypothesis test \(H_{0}: \mu=5\) against \(H_{1}: \mu<5\) with variance unknown and \(n=12,\) approximate the \(P\) -value for each of the following test statistics. a. \(t_{0}=2.05\) b. \(t_{0}=-1.84\) c. \(t_{0}=0.4\)

For the hypothesis test \(H_{0}: \mu=7\) against \(H_{1}: \mu \neq 7\) with variance unknown and \(n=20,\) approximate the \(P\) -value for each of the following test statistics. a. \(t_{0}=2.05\) b. \(t_{0}=-1.84\) c. \(t_{0}=0.4\)

Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient's body, but the battery pack needs to be recharged about every 4 hours. A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation \(\sigma=0.2\) hour. a. Is there evidence to support the claim that mean battery life exceeds 4 hours? Use \(\alpha=0.05 .\) b. What is the \(P\) -value for the test in part (a)? c. Compute the power of the test if the true mean battery life is 4.5 hours. d. What sample size would be required to detect a true mean battery life of 4.5 hours if you wanted the power of the test to be at least \(0.9 ?\) e. Explain how the question in part (a) could be answered by constructing a one-sided confidence bound on the mean life.

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