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Chapter 6: Problem 136

Use the t-distribution and the sample results to complete the test of the hypotheses. Use a \(5 \%\) significance level. Assume the results come from a random sample, and if the sample size is small, assume the underlying distribution is relatively normal. Test \(H_{0}: \mu=100\) vs \(H_{a}: \mu<100\) using the sample results \(\bar{x}=91.7, s=12.5,\) with \(n=30\).

Short Answer

Expert verified
The solution requires the application of the formula for test statistic \(t\), finding the critical value from the t-table, and making a decision based on a comparison of the two values. The actual decision (whether to reject or fail to reject the null hypothesis) will depend on the calculated values.

Step by step solution

01

Find the Test Statistic

The first step in performing a hypothesis test is to calculate the test statistic. For a t-distribution, the test statistic \(t\) is calculated using the formula: \(t = \frac{\bar{x} - \mu}{s/ \sqrt{n}}\)We substitute the given values into the formula:\(t = \frac{91.7 - 100}{12.5/ \sqrt{30}}\),then compute the value of \(t\).
02

Determine the Critical Value

To define the potential rejection region(s), we need to determine the critical value for the given significance level (\(5\%\)). Since the alternative hypothesis is \( \mu<100\), this is a left-tailed test. With \(n-1=29\) degrees of freedom and the given significance level (\(5\%\)), we use the t-distribution table to find the critical value, \(t_{cv}\).
03

Compare and Make a Decision

This step involves comparing the calculated test statistic (\(t\)) with the critical value (\(t_{cv}\)) found in step 2. If \(t<t_{cv}\), we reject the null hypothesis. We will draw a conclusion based on the comparison. If the null hypothesis is rejected, we say there is enough evidence to suggest that \(\mu<100\). Alternatively, if we failed to reject the null hypothesis, we say there's insufficient evidence to suggest that \(\mu<100\).

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Key Concepts

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

T-Distribution
Understanding the basics of t-distribution is vital for hypothesis testing in statistics, especially when working with small sample sizes and when the population standard deviation is not known. Think of it as a type of bell-shaped curve, similar to the normal distribution, but with fatter tails. It allows for more variability which is common in small samples.

The formula for the t-distribution test statistic is: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( \mu \) is the hypothesized population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.
  • The degrees of freedom, which are calculated as \( n-1 \), play a key role in shaping the t-distribution.
  • As the sample size increases, the t-distribution approaches the normal distribution.
  • For the given exercise, with a sample size of 30, the t-distribution is more appropriate than the normal distribution due to the small sample size.
When performing the test, the computed t-value is compared to a critical value from the t-distribution with \( n-1 \) degrees of freedom to determine the result of the hypothesis test.
Significance Level
The significance level, often denoted by \( \alpha \), represents the threshold for determining whether a statistical result is not due to random chance.

Common levels include 5%, 1%, and 0.1% - they quantify how 'significant' your results need to be to reject the null hypothesis. In the exercise, a significance level of \( 5\text{%} \) was chosen.
  • Choosing a smaller \( \alpha \) means being more conservative and requiring stronger evidence to reject the null hypothesis.
  • A significance level of 5% means that there is a 5% risk of concluding that there is an effect when there is no actual effect.
  • At a 5% significance level, we are saying that we are willing to accept a 5% chance of making a Type I error 鈥� wrongly rejecting the null hypothesis.
Deciding upon the significance level is a crucial step, as it directly impacts the probability of making an error in your hypothesis test.
Test Statistic
The test statistic is a standardized value that results from performing a statistical test. It鈥檚 essentially the star of the show in hypothesis testing as it helps to determine whether to reject the null hypothesis.

The concept revolves around how unusual or extreme the sample data is, considering the null hypothesis is true.
  • In the context of a t-test, the test statistic is calculated using sample data and reflects how far the sample mean deviates from the population mean proposed by the null hypothesis (\( \mu_{0} \)).
  • The exercise provided uses the formula for the t-test statistic: \[ t = \frac{\bar{x} - \mu}{s/ \sqrt{n}} \] which helps compare the observed sample mean (\( \bar{x} \)) to the hypothesized population mean (\( \mu \)) relative to the spread of the sample data.
  • We then compare this calculated test statistic to a critical value, and if our test statistic is more extreme (i.e., falls in the rejection region), we reject the null hypothesis.
Therefore, the test statistic plays a pivotal role in determining the outcome of a hypothesis test.

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Most popular questions from this chapter

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=8\) and \(n_{2}=10\)

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The distribution of sample means \(\bar{x}_{m}-\bar{x}_{f},\) where \(\bar{x}_{m}\) represents the mean Critical Reading score for a sample of 50 males and \(\bar{x}_{f}\) represents the mean Critical Reading score for a sample of 50 females, is centered at 5 with a standard deviation of \(22.5 .\) Give notation and define the quantity we are estimating with these sample differences. In the population of all students taking the test, who scored higher on average, males or females?

Table 6.29 gives a sample of grades on the first two quizzes in an introductory statistics course. We are interested in testing whether the mean grade on the second quiz is significantly higher than the mean grade on the first quiz. (a) Complete the test if we assume that the grades from the first quiz come from a random sample of 10 students in the course and the grades on the second quiz come from a different separate random sample of 10 students in the course. Clearly state the conclusion. (b) Now conduct the test if we assume that the grades recorded for the first quiz and the second quiz are from the same 10 students in the same order. (So the first student got a 72 on the first quiz and a 78 on the second quiz.) (c) Why are the results so different? Which is a better way to collect the data to answer the question of whether grades are higher on the second quiz? $$ \begin{array}{llllllllll} \hline \text { First Quiz } & 72 & 95 & 56 & 87 & 80 & 98 & 74 & 85 & 77 & 62 \\\ \text { Second Quiz } & 78 & 96 & 72 & 89 & 80 & 95 & 86 & 87 & 82 & 75 \\ \hline \end{array} $$

Describes scores on the Critical Reading portion of the Scholastic Aptitude Test (SAT) for college-bound students in the class of 2010. Critical Reading scores are approximately normally distributed with mean \(\mu=501\) and standard deviation \(\sigma=112\) (a) For each sample size below, use a normal distribution to find the percentage of sample means that will be greater than or equal to \(525 .\) Assume the samples are random samples. i. \(n=1\) ii. \(n=10\) iii. \(n=100\) iv. \(n=1000\) (b) Considering your answers from part (a), discuss the effect of the sample size on the likelihood of a sample mean being as far from the population mean as \(\bar{x}=525\) is from \(\mu=501\).

The All Countries dataset includes land area, in square kilometers, for all 213 countries in the world. The mean land area for all the countries is \(608,120 \mathrm{sq} \mathrm{km}\) with standard deviation 1,766,860 . For samples of size \(50,\) what percentage of sample means will be less than 400,000 sq km? What percentage will be greater than \(900,000 ?\)
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