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Chapter 6: Problem 135

In Exercises 6.135 to \(6.140,\) 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 \(H_{0}: \mu=15\) vs \(H_{a}: \mu>15\) using the sample results \(\bar{x}=17.2, s=6.4,\) with \(n=40\),

Short Answer

Expert verified
If the calculated test statistic (\(t\)) from Step 1 is greater than the critical value (\(t_{critical} = 1.685\)) from Step 2, we reject the null hypothesis \(H_{0}: \mu=15\), supporting the alternative hypothesis \(H_{a}: \mu>15\). If not, we do not reject the null hypothesis.

Step by step solution

01

Calculate the test statistic

The test statistic is calculated using the formula: \(t = \frac{\bar{x} - \mu_0}{s/ \sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu_0\) is the value of the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. Plugging in given values, we get \(t = \frac{17.2 - 15}{6.4/\sqrt{40}}\)
02

Find the critical value

From the t-table for a 5% significance level and degrees of freedom \(df = n - 1 = 39\), the critical value \(t_{critical} = 1.685\). Our test being a one-tailed test, we're only concerned with the extreme of the distribution in one direction.
03

Decision making

Compare the test statistic, \(t\), with the critical value, \(t_{critical}\). If \(t > t_{critical}\), then reject the null hypothesis. If \(t < t_{critical}\), then do not reject the null hypothesis. Compute the value of \(t\) from Step 1 and make the decision.

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Key Concepts

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

Understanding Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics used to make decisions based on data. It involves setting up two opposing statements, called the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\).
\[ \begin{align*}H_0 & : \mu = 15 \H_a & : \mu > 15\end{align*}\]
This indicates that the null hypothesis assumes the population mean \(\mu\) is 15, while the alternative hypothesis proposes it is greater than 15. The main goal is to determine if there is enough evidence in a sample to reject the null hypothesis.

**Key Steps in Hypothesis Testing**
  • Set up the hypotheses: Identify \(H_0\) and \(H_a\).
  • Choose a significance level: Often 5%, denoted as \(|\alpha| = 0.05\).
  • Compute the test statistic: Compare it to a critical value to make a decision.
By using these steps, we can evaluate the evidence and decide whether it supports \(H_a\) over \(H_0\).
Role of the Sample Mean
The sample mean, denoted as \(\bar{x}\), represents an average value calculated from the sample and is crucial in hypothesis testing. It helps estimate the population mean \(\mu\).

**Calculating the Sample Mean**
  • The sample mean \(\bar{x}\) provides an estimate of the true population mean.
  • In the given problem, \(\bar{x} = 17.2\), indicating a higher average than the hypothesized mean \(\mu = 15\).
  • This difference suggests evidence against the null hypothesis, which needs further testing.
The sample mean is central in calculating the test statistic, which is necessary for drawing conclusions in hypothesis testing.
The Significance Level
The significance level, denoted as \(\alpha\), helps in deciding how strong the evidence must be before rejecting the null hypothesis. In our example, a significance level of 5% is used.

**Understanding Significance Level**
  • The significance level \(\alpha = 0.05\) defines the probability of rejecting the null hypothesis when it is true.
  • A lower \(\alpha\) means we require stronger evidence to reject \(H_0\).
  • In practical terms, 5% signifies that there is a 5% risk of concluding that there is an effect when there is none.
This threshold helps manage the balance between Type I errors (false positives) and ensuring sensible conclusions based on statistical evidence.

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Most popular questions from this chapter

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 50 from Population 1 with mean 3.2 and standard deviation 1.7 and samples of size 50 from Population 2 with mean 2.8 and standard deviation 1.3

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 30 from a population with proportion 0.27

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? $$ \begin{array}{llllllllll} \hline \text { First Quiz } & 72 & 95 & 56 & 87 & 80 & 98 & 74 & 85 & 77 & 62 \\\ \text { Second Quiz } & 78 & 96 & 72 & 89 & 80 & 95 & 86 & 87 & 82 & 75 \\ \hline \end{array} $$

The dataset ICUAdmissions contains information on a sample of 200 patients being admitted to the Intensive Care Unit (ICU) at a hospital. One of the variables is HeartRate and another is Status which indicates whether the patient lived (Status \(=0\) ) or died (Status \(=1\) ). Use the computer output to give the details of a test to determine whether mean heart rate is different between patients who lived and died. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) $$ \begin{aligned} &\text { Two-sample } \mathrm{T} \text { for HeartRate }\\\ &\begin{array}{lrrrr} \text { Status } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 0 & 160 & 98.5 & 27.0 & 2.1 \\ 1 & 40 & 100.6 & 26.5 & 4.2 \end{array} \end{aligned} $$ Difference \(=m u(0)-m u(1)\) Estimate for difference: -2.12 \(95 \% \mathrm{Cl}\) for difference: (-11.53,7.28) T-Test of difference \(=0(\) vs not \(=):\) T-Value \(=-0.45\) P-Value \(=0.653 \quad \mathrm{DF}=60\)

Ron flips a coin \(n_{1}\) times and Freda flips a coin \(n_{2}\) times. We can assume all coin flips are fair: The coin has an equal chance of landing heads or tails. In each of the following cases, state whether inference for a difference in proportions is appropriate using the methods of this section. If so, give the mean and standard error for the distribution of the difference in proportions \(\left(\hat{p}_{1}-\hat{p}_{2}\right)\) and state whether the normal approximation is appropriate. (a) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Freda's flips that land heads; \(n_{1}=100\) and \(n_{2}=50\). (b) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Ron's flips that land tails; \(n_{1}=100\). (c) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Freda's flips that land tails; \(n_{1}=200\) and \(n_{2}=200\). (d) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land tails and \(\hat{p}_{2}\) be the proportion of Freda's flips that land tails; \(n_{1}=5\) and \(n_{2}=10\).
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