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Conduct the following tests of hypotheses, assuming that the populations of paired differences are normally distributed. a. \(H_{0} \cdot \mu_{d f}=0, \quad H_{1}: \mu_{d} \neq 0, \quad n=26, \quad \bar{d}=9.6, \quad s_{d}=3.9, \quad \alpha=.05\) b. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}>0, \quad n=15, \quad \bar{d}=8.8, \quad s_{d}=4.7, \quad \alpha=.01\) c. \(H_{0}=\mu_{d}=0, \quad H_{1}: \mu_{d}<0, \quad n=20, \quad \bar{d}=-7.4, \quad s_{d}=2.3, \quad \alpha=.10\)

Short Answer

Expert verified
Hypothesis testing led to calculated T values for each part. Confirm by comparing these T values with the respective critical values from the T-distribution table with df = \(n-1\) under given significance levels. Based on these comparisons, a decision can be made on whether to accept or reject the null hypothesis for each part.

Step by step solution

01

Hypothesis Testing for Part a

Start by setting up the null hypothesis \(H_0: \mu_d = 0\) and the alternative hypothesis \(H_1: \mu_d \neq 0\). The test statistic for the given hypotheses can be calculated using the formula \[T = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}}\]. Therefore, substitute the given values: \(T = (9.6 - 0) / (3.9 / \sqrt{26})\). If the calculated test statistic, T, falls outside the critical values from the T-distribution table with degree of freedom (df) = \(n-1\) and \(\alpha/2\), we reject the null hypothesis.
02

Hypothesis Testing for Part b

Set up the null hypothesis \(H_0: \mu_d = 0\) and the alternative hypothesis \(H_1: \mu_d > 0\). Compute the test statistic using the above formula: \(T = (8.8 - 0) / (4.7 / \sqrt{15})\). We compare calculated T with critical value from the T-distribution table with degree of freedom (df) = \(n-1\) and \(\alpha\). If T > critical value, reject null hypothesis.
03

Hypothesis Testing for Part c

Set up the null hypothesis \(H_0: \mu_d = 0\) and the alternative hypothesis \(H_1: \mu_d < 0\). Compute the test statistic: \(T = (-7.4 - 0) / (2.3 / \sqrt{20})\). Compare T with critical value from the T-distribution table with degree of freedom (df) = \(n-1\) and \(\alpha\). If T < critical value, reject null hypothesis.

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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 crucial in hypothesis testing, especially when dealing with small sample sizes. It is a probability distribution that is symmetric and bell-shaped, much like the normal distribution, but with thicker tails. These thicker tails help accommodate the uncertainty in estimates of the population standard deviation. This feature makes the T-distribution particularly useful when sample sizes are small (generally n < 30). When conducting hypothesis tests, you use the T-distribution to determine critical values that decide whether you reject the null hypothesis. These critical values depend on two key parameters:
  • Degrees of freedom (df): For a T-distribution, this is usually one less than the sample size (n-1).
  • Significance level (\( \alpha \)): The probability of rejecting the null hypothesis when it is actually true. Commonly used significance levels are 0.05, 0.01, and 0.10.
Once you calculate your test statistic, you compare it to the critical value from the T-distribution table to decide if you should reject the null hypothesis.
Null Hypothesis
In hypothesis testing, the null hypothesis is a statement that there is no effect or no difference. It's a default assumption that stands unless evidence suggests otherwise. In the context of the paired differences exercises above:
  • The null hypothesis (\( H_0 \)) posits that the mean difference (\( \mu_d \)) is equal to zero, suggesting no real change or effect in the data being tested.
When performing hypothesis tests, your goal is to either reject or fail to reject the null hypothesis. You need a calculated test statistic and a critical value to assess the validity of this assumption. The decision to reject the null hypothesis depends on the critical region defined by the T-distribution.For example, in part a of the exercise, the null hypothesis is that the mean difference is zero (\( H_0: \mu_d = 0 \)). If the test shows significant enough evidence against this, the result will "reject the null hypothesis." Otherwise, we would "fail to reject the null hypothesis," implying the evidence is not strong enough.
Alternative Hypothesis
The alternative hypothesis is a statement that contradicts the null hypothesis. It reflects what you aim to prove through statistical evidence. In the exercises:
  • Part a's alternative hypothesis (\( H_1 \)) suggests that the mean difference is not equal to zero (\( \mu_d eq 0 \)).
  • Part b's alternative hypothesis is that the mean difference is greater than zero (\( \mu_d > 0 \)).
  • Part c's alternative hypothesis is that the mean difference is less than zero (\( \mu_d < 0 \)).
The direction of the alternative hypothesis determines whether the hypothesis test is two-tailed, left-tailed, or right-tailed:
  • Two-tailed: Tests whether the parameter is significantly different in either direction, used in part a.
  • Right-tailed: Tests if the parameter is greater, used in part b.
  • Left-tailed: Tests if the parameter is less, used in part c.
Upon computing the test statistic, if the statistic falls in the region defined by the alternative hypothesis, the null hypothesis can be rejected in favor of the alternative hypothesis.
Test Statistic
The test statistic is a crucial part of hypothesis testing. It helps to determine if the data provides enough evidence to reject the null hypothesis. You calculate it based on the sample data and use it to make statistical inferences about the population.To calculate the test statistic (T), use the formula:\[T = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}}\]Where:
  • \( \bar{d} \) = sample mean difference
  • \( \mu_d \) = mean difference under the null hypothesis (often 0)
  • \( s_d \) = standard deviation of the differences
  • \( n \) = sample size
After computing the test statistic, you compare it against a critical value from the T-distribution (based on the chosen significance level and degrees of freedom). If the test statistic is in the rejection region, you reject the null hypothesis; otherwise, you fail to reject it. This comparison helps you intimately connect the calculated output to your hypothesis choices and ultimately draw conclusions about your data's significance.

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Most popular questions from this chapter

One type of experiment that might be performed by an exercise physiologist is as follows: Each person in a random sample is tested in a weight room to determine the heaviest weight with which he or she can perform an incline press five times with his or her dominant arm (defined as the hand that a person uses for writing). After a significant rest period, the same weight is determined for each individual's nondominant arm. The physiologist is interested in the differences in the weights pressed by each arm. The following data represent the maximum weights (in pounds) pressed by cach arm for a random sample of 18 fifteen-year old girls. Assume that the differences in weights pressed by each arm for all fifteenyear old girls are approximately normally distributed. \begin{tabular}{cccccc} \hline Subject & Dominant Arm & Nondominant Arm & Subject & Dominant Arm & Nondominant Arm \\ \hline 1 & 59 & 53 & 10 & 47 & 38 \\ 2 & 32 & 30 & 11 & 40 & 35 \\ 3 & 27 & 24 & 12 & 36 & 36 \\ 4 & 18 & 20 & 13 & 21 & 25 \\ 5 & 42 & 40 & 14 & 51 & 48 \\ 6 & 12 & 12 & 15 & 30 & 30 \\ 7 & 29 & 24 & 16 & 32 & 31 \\ 8 & 33 & 34 & 17 & 14 & 14 \\ 9 & 22 & 22 & 18 & 26 & 27 \\ \hline \end{tabular} a. Make a \(99 \%\) confidence interval for the mean of the paired differences for the two populations, where a paired difference is equal to the maximum weight for the dominant arm minus the maximum weight for the nondominant arm. b. Using a \(1 \%\) significance level, can you conclude that the average paired difference as defined in part a is positive?

A company that has many department stores in the southern states wanted to find at two such stores the percentage of sales for which at least one of the items was returned. A sample of 800 sales randomly selected from Store A showed that for 280 of them at least one item was returned. Another sample of 900 sales randomly selected from Store B showed that for 279 of them at least one item was returned. A. Construct a \(98 \%\) confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned. b. Using a \(1 \%\) significance level, can you conclude that the proportions of all sales for which at least one item is returned is higher for Store A than for Store B?

Briefly explain the meaning of independent and dependent samples. Give one example of each.

A car magazine is comparing the total repair costs incurred during the first three years on two sports cars, the T-999 and the XPY. Random samples of 45 T-999s and 51 XPY's are taken. All 96 cars are 3 years old and have similar mileages. The mean of repair costs for the 45 T-999 cars is \(\$ 3300\) for the first 3years. For the 51 XPY cars, this mean is \(\$ 3850\). Assume that the standard deviations for the two popuLations are \(\$ 800\) and \(\$ 1000\), respectively. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Using a \(1 \%\) significance level, can you conclude that such mean repair costs are different for these two types of cars? c. What would your decision be in part \(b\) if the probability of making a Type I error were zero? Explain.

Perform the following tests of hypotheses, assuming that the populations of paired differences are normally distributed. a. \(H_{0} \cdot \mu_{d f}=0, \quad H_{1}: \mu_{d} \neq 0, \quad n=9, \quad \bar{d}=6.7, \quad s_{d}=2.5, \quad \alpha=.10\) b. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}>0, \quad n=22, \quad \bar{d}=14.8, \quad s_{d}=6.4, \quad \alpha=.05\) c. \(H_{0}=\mu_{d}=0, \quad H_{1}: \mu_{d}<0, \quad n=17, \quad \bar{d}=-9.3, \quad s_{d}=4.8, \quad \alpha=.01\)

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