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Ten subjects are randomly drawn from a population. Scores in the population are normally distributed. For the sample \(\sum \mathrm{X}=1038\) and \(\sum \mathrm{X}^{2}=107888\). Test the hypothesis that \(\bar{\mu}=100 .\) Adopt a \(.05\) level of significance. Use a two-tailed test.

Short Answer

Expert verified
Based on the given sample data, we have calculated the sample mean as \(\bar{x}=103.8\), the sample standard deviation as \(s=12.42\), and the t-statistic as \(t=0.976\). By comparing the t-statistic to the critical t-values (\(t_{0.025} = \pm 2.262\)), we fail to reject the null hypothesis. Therefore, with a 0.05 level of significance, we do not have enough evidence to reject the hypothesis that the population mean is 100.

Step by step solution

01

State the hypotheses

The null hypothesis (\(H_0\)): The population mean is equal to 100. \[H_0: \mu = 100\] The alternative hypothesis (\(H_1\)): The population mean is not equal to 100. \[H_1: \mu \neq 100\]
02

Calculate sample mean and sample variance

Sample mean (\(\bar{x}\)) can be calculated as: \[\bar{x}=\frac{\sum \mathrm{X}}{n}\] Sample variance (\(s^2\)) can be calculated as: \[s^2 = \frac{\sum \mathrm{X}^{2}-n(\bar{x})^2}{n-1}\] Calculate sample mean: \[\bar{x}=\frac{1038}{10} = 103.8\] Calculate sample variance: \[s^2=\frac{107888-10(103.8)^2}{10-1}=154.4\] Sample standard deviation (\(s\)) can be calculated as the square root of the sample variance. \[s = \sqrt{154.4} = 12.42\]
03

Find critical values

Since we are adopting a \(0.05\) level of significance and using a two-tailed test, we need to find the critical values of \(t\) with \(n-1=9\) degrees of freedom and \(\alpha/2 = 0.025\) in each tail. Refer to the t-distribution table or use a calculator to find these values: \[t_{0.025} = \pm 2.262 \]
04

Calculate t-statistic

T-statistic can be calculated using the formula: \[t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}\] Here, \(\mu_0 = 100\), which is the value in the null hypothesis. Calculate t-statistic: \[t = \frac{103.8 - 100}{12.42 / \sqrt{10}} = 0.976\]
05

Make a decision

We can now compare the t-statistic to the critical t-values. Since \(-2.262 < 0.976 < 2.262\), we fail to reject the null hypothesis \(H_0\). In conclusion, with a 0.05 level of significance, we do not have enough evidence to reject the null hypothesis that the population mean is 100.

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Key Concepts

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

t-distribution
The t-distribution is a type of probability distribution that is symmetrical and bell-shaped, similar to the normal distribution but with heavier tails. This characteristic becomes relevant when working with smaller sample sizes because it accounts for the increased variability typically observed in such scenarios.
The t-distribution is especially useful in hypothesis testing when the sample size is small (typically less than 30) and the population standard deviation is unknown. It allows us to estimate the range within which the sample mean would likely fall if the null hypothesis is true.
  • The degrees of freedom (df) for the t-distribution depend on the sample size: it is calculated as the number of observations minus one (n-1).
  • As the sample size increases, the t-distribution approaches the normal distribution.
  • In our exercise, with 10 subjects, we have 9 degrees of freedom (10 - 1 = 9).
Understanding the t-distribution helps in making informed decisions about your hypothesis test, especially when it comes to identifying critical values.
sample mean
The sample mean, denoted as \( \bar{x} \), is a measure that represents the central tendency of the collected sample data. It provides an estimate of the population mean and is used extensively in hypothesis testing to test claims about the population mean.
To compute the sample mean, you calculate the sum of all sample observations and divide by the number of observations. The formula for the sample mean is: \[ \bar{x} = \frac{\sum X}{n} \]
  • In our exercise example, the sum of the sample observations is 1038, and there are 10 subjects. So, \( \bar{x} = \frac{1038}{10} = 103.8 \).
  • The sample mean of 103.8 is our best estimate for the true mean of the population in this context.
Using the sample mean in hypothesis testing helps determine if the observed data reflect the population mean hypothesized.
sample variance
Sample variance is an estimate of the variance of the population from which the sample was drawn. It provides information about the spread or variability of the sample data. It measures how much the data in the sample are spread out around the mean.
The formula for calculating sample variance \( s^2 \) is as follows: \[ s^2 = \frac{\sum X^2 - n(\bar{x})^2}{n-1} \]
  • "\( \sum X^2 \)" represents the sum of the squares of each observation in the sample.
  • "n" is the number of observations in the sample.
  • "\( n-1 \)" is known as degrees of freedom and is used here to eliminate bias in the estimation of population variance.
  • In our example, the sample variance was calculated as \( 154.4 \).
This measure is crucial as it helps in understanding the dispersion effects on the sample mean, which is especially important for reliable hypothesis testing.
two-tailed test
A two-tailed test is a type of hypothesis test used when we are interested in determining whether there is either a significant increase or decrease in a population parameter compared to a hypothesized value. Unlike one-tailed tests, which assess only one direction, two-tailed tests consider both directions, making them more conservative.
For our scenario, the null hypothesis is that the population mean is equal to 100 \( (H_0: \mu = 100) \), and the alternative hypothesis is that the population mean is not equal to 100 \( (H_1: \mu eq 100) \).
  • When using a two-tailed test, the level of significance \( \alpha \) is split between the two tails of the distribution. Typically, 0.05 is divided into 0.025 for each tail.
  • This kind of test seeks to find evidence that the sample mean is either significantly higher or lower than the population mean.
  • In our example, we found critical values using the t-distribution with 9 degrees of freedom: \( \pm 2.262 \).
Understanding when and how to use a two-tailed test is vital, as it affects how we interpret our test results and determine if our hypothesis stands.

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Most popular questions from this chapter

\(\mathrm{H}_{0}: \mu=\mu_{0}\) \(\mathrm{H}_{1}: \mu=\mu_{1}\) and \(\alpha\) and \(\beta\) are probabilities of making type I and type II errors respectively, show that for a two-tailed test the required sample size \(\mathrm{n}\) is given approximately by $$ \mathrm{n}=\left[\left\\{\left(\mathrm{Z}_{\alpha / 2}+\mathrm{Z}_{\mathrm{B}}\right)^{2} \sigma^{2}\right\\} /\left(\mu_{1}-\mu_{0}\right)^{2}\right] \text { , } $$ provided that $$ \operatorname{Pr}\left[Z<-Z_{\alpha / 2}-\left\\{\left(\sqrt{n}\left|\mu_{1}-\mu_{0}\right|\right) / \sigma\right\\}\right] $$ is small when \(\mu=\mu_{1}\).

Suppose we have a type of battery for which we take a sample of 10 batteries. The mean operating life of these batteries is \(18.0\) hours with a standard deviation of \(3.0\) hours. Suppose also that we have a new type of battery for which we take a sample of 17 batteries. The mean of this sample is \(22.0\) hours with a standard deviation of \(6.0\) hours. Determine for a \(1 \%\) level of significance whether there is a significant difference between the means of the two samples. Also determine at a \(1 \%\) level of significance whether we can conclude that the new batteries are superior to the old ones.

Suppose that you want to decide which of two equally-priced brands of light bulbs lasts longer. You choose a random sample of 100 bulbs of each brand and find that brand \(\mathrm{A}\) has sample mean of 1180 hours and sample standard deviation of 120 hours, and that brand \(\mathrm{B}\) has sample mean of 1160 hours and sample standard deviation of 40 hours. What decision should you make at the \(5 \%\) significance level?

Suppose we have a binomial distribution for which \(\mathrm{H}_{0}\) is \(p=(1 / 2)\) where \(p\) is the probability of success on a single trial. Suppose the type I error, \(\alpha=.05\) and \(\mathrm{n}=100\). Calculate the power of this test for each of the following alternate hypotheses, \(\mathrm{H}_{1}: \mathrm{p}=.55, \mathrm{p}=.60, \mathrm{p}=.65, \mathrm{p}=.70\), and \(\mathrm{p}=.75\). Do the same when \(\alpha=.01\)

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?

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