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Suppose it is required that the mean operating life of size "D" batteries be 22 hours. Suppose also that the operating life of the batteries is normally distributed. It is known that the standard deviation of the operating life of all such batteries produced is \(3.0\) hours. If a sample of 9 batteries has a mean operating life of 20 hours, can we conclude that the mean operating life of size "D" batteries is not 22 hours? Then suppose the standard deviation of the operating life of all such batteries is not known but that for the sample of 9 batteries the standard deviation is \(3.0 .\) What conclusion would we then reach?

Short Answer

Expert verified
In the first part, using a z-test with a known population standard deviation, we find a p-value of 0.0456, which is less than the 0.05 significance level. Therefore, we reject the null hypothesis and conclude that the mean operating life of size "D" batteries is not 22 hours. In the second part, using a t-test with the given sample standard deviation, we find a p-value of 0.0818, which is greater than the 0.05 significance level. Therefore, we fail to reject the null hypothesis, and we do not have enough evidence to conclude that the mean operating life of size "D" batteries is not 22 hours.

Step by step solution

01

State the null and alternative hypotheses

We need to set up null and alternative hypotheses for the problem. The null hypothesis is that the mean operating life of size "D" batteries is 22 hours, and the alternative hypothesis is that it is not 22 hours. \(H_0: \mu = 22\) \(H_1: \mu \neq 22\) #Step 2: Calculate the z-score of the sample#
02

Calculate the z-score of the sample

Next, we will calculate the z-score of the sample using the formula: \(z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\) Where: - \(z\) is the z-score - \(\bar{x}\) is the sample mean (20 hours) - \(\mu\) is the population mean (22 hours) - \(\sigma\) is the population standard deviation (3.0 hours) - \(n\) is the sample size (9 batteries) \(z = \frac{20 - 22}{\frac{3}{\sqrt{9}}} = -2\) #Step 3: Find the p-value associated with the z-score#
03

Find the p-value associated with the z-score

Let's determine the p-value associated with the z-score. Since our alternative hypothesis is two-tailed (i.e., we are testing for a not equal to relationship), we need to find the probability that a z-score is more extreme than -2 in both the left and right tails of the normal distribution. P(Z < -2) = 0.0228 (from the standard normal probability table or using a calculator) P(Z > 2) = 0.0228 Two-tailed P(Z < -2 ∪ Z > 2) = 2 × 0.0228 = 0.0456 #Step 4: Compare the p-value with the significance level#
04

Compare the p-value with the significance level

Let's assume a significance level of 0.05. Since the p-value (0.0456) is less than the significance level (0.05), we reject the null hypothesis. Thus, there is enough evidence to conclude that the mean operating life of size "D" batteries is not 22 hours. #Part 2: Perform a t-test with the given sample standard deviation# #Step 5: Calculate the t-score of the sample#
05

Calculate the t-score of the sample

Now, we will calculate the t-score of the sample using the formula: \(t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\) Where: - \(t\) is the t-score - \(\bar{x}\) is the sample mean (20 hours) - \(\mu\) is the population mean (22 hours) - \(s\) is the sample standard deviation (3.0 hours) - \(n\) is the sample size (9 batteries) \(t = \frac{20 - 22}{\frac{3}{\sqrt{9}}} = -2\) #Step 6: Find the p-value associated with the t-score#
06

Find the p-value associated with the t-score

Since our alternative hypothesis is two-tailed, we need to find the probability that a t-score is more extreme than -2 in both the left and right tails of the Student's t-distribution with 8 degrees of freedom (df = 9 - 1 = 8). P(T < -2) = 0.0409 (from the t-probability table or using a calculator) P(T > 2) = 0.0409 Two-tailed P(T < -2 ∪ T > 2) = 2 × 0.0409 = 0.0818 #Step 7: Compare the p-value with the significance level#
07

Compare the p-value with the significance level

Let's assume the same significance level of 0.05. Since the p-value (0.0818) is greater than the significance level (0.05), we fail to reject the null hypothesis. With the given sample standard deviation, we do not have enough evidence to conclude that the mean operating life of size "D" batteries is not 22 hours.

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Key Concepts

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

z-score
In statistics, a z-score, also known as a standard score, is a way to describe the position of a raw score in relation to the mean of a group of scores. It is a measure of how many standard deviations an element is from the mean. For this reason, it's a powerful tool for comparing results across different data sets.
The formula for calculating a z-score is:
  • \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \)
Where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu \) is the population mean,
  • \( \sigma \) is the population standard deviation,
  • \( n \) is the sample size.
In our exercise, we calculate a z-score to assess whether the sample mean significantly differs from the population mean. A z-score of \(-2\) indicates the sample mean is 2 standard deviations away from the population mean, suggesting a potential difference.
t-test
The t-test is a statistical test that is used to compare the means of two groups. It's particularly useful when the population standard deviation is unknown, which often happens in real-life scenarios. Instead of using the population standard deviation, it uses the sample standard deviation and is applied using the t-distribution, which accounts for the smaller sample size by having 'fatter' tails than the normal distribution.
The formula for calculating the t-score is:
  • \( t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \)
Where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu \) is the population mean,
  • \( s \) is the sample standard deviation,
  • \( n \) is the sample size.
In our battery life example, a t-score of \(-2\) was calculated. However, the evidence wasn't strong enough to reject the null hypothesis when the sample standard deviation was used. This showcases the importance of considering sample size and variance when drawing conclusions from data.
standard deviation
Standard deviation is a measure that indicates the amount of variation or dispersion in a set of values. Simply put, it tells us how much the individual data points differ from the mean of the data set. A small standard deviation means the data points are close to the mean, while a large one indicates they are spread out over a wider range of values.
In statistical testing, like in our battery life test, standard deviation plays a crucial role. It allows us to understand how much our sample results can vary. In cases where the population standard deviation is known, we use it directly in z-score calculations. When unknown, we rely on the sample standard deviation for t-tests. This transformation ensures our tests accommodate variation in data, providing more accurate results.

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Most popular questions from this chapter

A manufacturer of transistors claims that its transistors will last an average of 1000 hours. To maintain this average, 25 transistors are tested each month. If the computed value of \(\mathrm{t}\) lies between \(-t_{.025}\) and \(t_{025}\), the manufacturer is satisfied with his claim. What conclusions should be drawn from a sample that has a mean \(\overline{\mathrm{x}}=1,010\) and a standard deviation \(\mathrm{s}=60\) ? Assume the distribution of the lifetime of the transistors is normal.

Two separate groups of subjects were tested. The experimental group (Group E) had 10 subjects; the control group (Group C) had 9 subjects. The data are given below; the scores are assumed to be normally distributed. Group E: \(12,13,16,14,15,12,15,14,13\), and 16 . Group C: \(10,13,14,12,15,16,12,14\), and 11 . Determine whether the means of the two groups differ significantly at the \(.05\) level of sionificance.

A certain printing press is known to turn out an average of 45 copies a minute. In an attempt to increase its output, an alteration is made to the machine, and then in 3 short test runs it turns out 46,47, and 48 copies a minute. Is this increase statistically significant, or is it likely to be simply the result of chance variation? Use a significance level of \(.05\).

For a shipment of cable, suppose that the specifications call for a mean breaking strength of 2,000 pounds. A sampling of the breaking strength of a number of segments of the cable has a mean breaking strength of 1955 pounds with an associated standard error of the mean of 25 pounds. Using the 5 percent level, test the significance of the difference found.

A manufacturer of car batteries guarantees that his batteries will last, on the average, 3 years with a standard deviation of 1 year. If 5 of these batteries have lifetimes of \(1.9,2.4,3.0\), \(3.5\), and \(4.2\) years, is the manufacturer still convinced that his batteries have a standard deviation of 1 year?

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