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The following information was obtained from two independent samples selected from two normally distributed populations with unequal and unknown population standard deviations. $$ \begin{array}{lll} n_{1}=14 & \bar{x}_{1}=.109 .43 & s_{1}=2.26 \\ n_{2}=15 & \bar{x}_{2}=.113 .88 & s_{2}=5.84 \end{array} $$ Test at a \(1 \%\) significance level if \(\mu_{1}\) is less than \(\mu_{2}\).

Short Answer

Expert verified
Carry out the steps mentioned above to test whether \(\mu_{1}\) is less than \(\mu_{2}\) at a 1% significance level. Due to the nature of statistics, a numeric value cannot be produced without performing the calculations with the data given. After calculating the test statistic and the degrees of freedom, check these values against a t-distribution table to make a conclusion.

Step by step solution

01

Null and Alternative Hypothesis

Formulate the null hypothesis (H0) and alternative hypothesis (H1). The null hypothesis is that the means are equal, i.e. \(\mu_{1} = \mu_{2}\) and the alternative hypothesis is that the mean of the first population is lesser than the mean of the second population, i.e. \(\mu_{1} < \mu_{2}\).
02

Calculate Test Statistic

Calculate the test statistic using the formula for Welch's t-test which is \[t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}}\]. Using the provided data: \(n_{1}=14, \bar{x}_{1}=.109.43, s_{1}=2.26, n_{2}=15, \bar{x}_{2}=.113.88, s_{2}=5.84\), calculate the test statistic.
03

Calculate Degrees of Freedom

Calculate degrees of freedom using the formula for the Welch-Satterthwaite equation which is \[df = \frac{((s_1^2/n_1 + s_2^2/n_2)^2)}{((s_1^2/n_1)^2/(n_1-1) + (s_2^2/n_2)^2/(n_2-1))}\]. Use the provided data to calculate df.
04

Determine the Critical Value

Look up the critical value in the t-distribution table using 1% significance level (0.01) and the degrees of freedom calculated in Step 3. The critical value is the t score such that the area to its right is 0.01.
05

Compare Test Statistic and Critical Value

If the test statistic calculated in Step 2 is less than the negative of the t critical value, then reject the null hypothesis in favor of the alternative hypothesis.

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Key Concepts

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

Welch's t-test
Welch's t-test is a statistical test used to compare the means of two independent groups, especially when the population variances are unequal and unknown. This test is particularly helpful in situations where you are dealing with two normally distributed populations, but have no assumption of equal variance. Instead of pooling variances as in a traditional t-test, Welch's approach adapts the standard error term to account for unequal variances. This results in a more accurate estimation of the p-value.

It's particularly useful when comparing two independent samples. Unlike a regular t-test, Welch's t-test does not assume that the variance of the two groups is equal, which ensures a more robust outcome in varied conditions. It provides a test statistic that is used to determine whether there is a statistically significant difference between the sample means. In practice, you would calculate the Welch's t-test statistic using the formula:
  • \( t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}} \)
Where \( \bar{x}_{1} \) and \( \bar{x}_{2} \) are the sample means, \( s_{1} \) and \( s_{2} \) are the sample standard deviations, and \( n_{1} \) and \( n_{2} \) are the sample sizes of the two groups, respectively.
Degrees of freedom
Degrees of freedom are an essential component in calculating statistics, especially in hypothesis testing. For Welch's t-test, degrees of freedom are not simply calculated as the sum of the sample sizes minus two (as in the case of the equal variance t-test) but through a more complex formula given by the Welch-Satterthwaite equation. This formula allows for an accurate calculation of the degrees of freedom when variances are unequal.

The formula to find degrees of freedom in the context of Welch's t-test is:
  • \( df = \frac{((s_1^2/n_1 + s_2^2/n_2)^2)}{((s_1^2/n_1)^2/(n_1-1) + (s_2^2/n_2)^2/(n_2-1))} \)
This calculation impacts the critical value that you will reference later in the t-distribution table. In scenarios where sample sizes and variances are notably different, this formula ensures that the test is not biased towards either sample.
Significance level
The significance level is a threshold used to determine whether a test result is statistically significant. It is usually denoted as \( \alpha \), and common choices are 0.05 (5%) or 0.01 (1%). In this exercise, a significance level of 0.01 is employed, which indicates a high standard for rejecting the null hypothesis. The choice of the significance level balances the chances of making a Type I error, which is rejecting a true null hypothesis.

A 1% significance level means that there is only a 1% probability of incorrectly rejecting the null hypothesis when it is true. This stringent level minimizes the risk of claiming a difference when there is none. It is important to choose the significance level wisely, taking into account the context and consequences of potential errors in the hypothesis test.
Critical value
The critical value is a point in the distribution of the test statistic beyond which the null hypothesis is rejected. To find this value, you compare it to a t-distribution based on the degrees of freedom calculated earlier and at the chosen significance level. For a one-tailed test such as in this exercise, the critical value is that t-score which leaves 1% of the tail on the distribution side.

You will use a t-distribution table or statistical software to find this critical value. If the calculated test statistic is less than this critical value (flipped for left-tailed tests), it indicates strong evidence against the null hypothesis, prompting its rejection. Understanding where and how to find critical values is crucial for evaluating your hypothesis test results accurately. If the test statistic falls beyond the critical value, you have sufficient evidence to support your alternative hypothesis.

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Most popular questions from this chapter

Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. The fuel consumption, in \(\mathrm{mpg}\), will be calculated for each test run. The average mpg for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg}\). Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.e., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

Maine Mountain Dairy claims that its 8-ounce low-fat yogurt cups contain, on average, fewer calories than the 8 -ounce low-fat yogurt cups produced by a competitor. A consumer agency wanted to check this claim. A sample of 27 such yogurt cups produced by this company showed that they contained an average of 141 calories per cup. A sample of 25 such yogurt cups produced by its competitor showed that they contained an average of 144 calories per cup. Assume that the two populations are approximately normally distributed with population standard deviations of \(5.5\) and \(6.4\) calories, repectively. a. Make a \(98 \%\) confidence interval for the difference between the mean number of calories in the 8 -ounce low-fat yogurt cups produced by the two companies. b. Test at a \(1 \%\) significance level whether Maine Mountain Dairy's claim is true. c. Calculate the \(p\) -value for the test of part b. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.05 ?\) What if \(\alpha=.025 ?\)

The manager of a factory has devised a detailed plan for evacuating the building as quickly as possible in the event of a fire or other emergency. An industrial psychologist believes that workers actually leave the factory faster at closing time without following any system. The company holds fire drills periodically in which a bell sounds and workers leave the building according to the system. The evacuation time for each drill is recorded. For comparison, the psychologist also records the evacuation time when the bell sounds for closing time each day. A random sample of 36 fire drills showed a mean evacuation time of \(5.1\) minutes with a standard deviation of \(1.1\) minutes. A random sample of 37 days at closing time showed a mean evacuation time of \(4.2\) minutes with a standard deviation of \(1.0\) minute. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Test at a \(5 \%\) significance level whether the mean evacuation time is smaller at closing time than during fire drills. Assume that the evacuation times at closing time and during fire drills have equal but unknown population standard deviations.

The manager of a factory has devised a detailed plan for evacuating the building as quickly as possible in the event of a fire or other emergency. An industrial psychologist believes that workers actually leave the factory faster at closing time without following any system. The company holds fire drills periodically in which a bell sounds and workers leave the building according to the system. The evacuation time for each drill is recorded. For comparison, the psychologist also records the evacuation time when the bell sounds for closing time each day. A random sample of 36 fire drills showed a mean evacuation time of \(5.1\) minutes with a standard deviation of \(1.1\) minutes. \(\mathrm{A}\)random sample of 37 days at closing time showed a mean evacuation time of \(4.2\) minutes with a standard deviation of \(1.0\) minute. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Test at a \(5 \%\) significance level whether the mean evacuation time is smaller at closing time than during fire drills. Assume that the evacuation times at closing time and during fire drills have unknown and unequal population standard deviations.

Describe the sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) for two independent samples when \(\sigma_{1}\) and \(\sigma_{2}\) are known and either both sample sizes are large or both populations are normally distributed. What are the mean and standard deviation of this sampling distribution?

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