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Exercises 8.46 to 8.52 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } & \\ \text { A } & 6 & 86.833 & 5.231 & \\ \text { B } & 6 & 76.167 & 6.555 & \\ \text { C } & 6 & 80.000 & 9.230 & \\ \text { D } & 6 & 69.333 & 6.154 & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array}\) Test for a difference in population means between groups \(A\) and \(D .\) Show all details of the test.

Short Answer

Expert verified
To sum up, the t-test is performed to check for a significant difference in population means between groups A and D. Begin by defining the hypotheses, calculate the mean difference and standard error, then compute the t-value. Finally, find the p-value using the degrees of freedom and make a decision based on this p-value.

Step by step solution

01

Define the Hypothesis

The first step is to develop the null and alternative hypothesis. The null hypothesis (H0) states there's no significant difference between the population means, whereas the alternative hypothesis (H1) says there is a significant difference between the population means. In this case, H0: μA = μD, H1: μA ≠ μD.
02

Calculate the Difference in Means and Standard Error

The difference in means is computed by taking the mean of group A and subtracting the mean of group D. Similarly, the standard error (SE) can be calculated using the formula: SE = sqrt[(s1²/n1) + (s2²/n2)], where s1 and s2 are the standard deviations of the first and second group, and n1 and n2 are the sizes of the groups. This produces the formula: SE = sqrt[(5.231²/6) + (6.154²/6)].
03

Calculate the t-Value

The t-value is calculated using the difference in means (from Step 2), divided by the standard error (SE). The formula is: t = (mean difference) / SE.
04

Determine Degrees of Freedom and Find the p-value

Degrees of Freedom (DF) for an independent samples t-test is generally given by n1 + n2 - 2. In this case: DF = 6 + 6 - 2 = 10. Using the calculated t-value and the DF, the p-value can be obtained from the t-distribution table or a calculator that provides p-values.
05

Decision Making

If the p-value is less than the significance level (0.05), the null hypothesis can be rejected. Otherwise, we fail to reject the null hypothesis. This decision will lead to conclusions about whether there's a significant difference between the population means of groups A and D.

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Key Concepts

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

Null and Alternative Hypothesis
The null and alternative hypotheses are foundational components in the realm of hypothesis testing. In a statistical test, the null hypothesis, often denoted as H0, asserts the absence of a particular effect, difference, or relationship. In the case of the exercise on comparing population means, the null hypothesis posits that there is no significant difference between the means of groups A and D—mathematically represented as H0: μA = μD.

The alternative hypothesis, denoted as H1 or Ha, takes a contrary stance, implying the presence of an effect or difference. In this scenario, the alternative hypothesis claims there is a meaningful discrepancy between the means, stipulated as H1: μA ≠ μD. This hypothesis acts as a research query, aiming to demonstrate a statistical significance that will influence the decision to accept or reject the null hypothesis.
t-Value Calculation
Computing the t-value is a critical step when conducting a t-test, which is used to compare the means from two groups when sample sizes are small and the population variance is unknown. The t-value is a ratio that compares the deviation between the group means against the variation within the groups.

To calculate the t-value, one must first determine the mean difference between the two groups and the standard error of this difference. The formula for the t-value is: \( t = \frac{\text{mean difference}}{\text{SE}} \). In our exercise, the mean difference is obtained by subtracting the mean of group D from group A. The standard error, on the other hand, encapsulates the variability of sample means by combining the variances of both groups, adjusting them for their respective sample sizes. A higher t-value suggests a greater difference between groups, relative to the variability within groups, indicating a potential statistical significance.
Degrees of Freedom
The concept of degrees of freedom (DF) is an integral part of statistical tests, including the t-test. Degrees of freedom refer to the number of independent values in a calculation that can vary while arriving at a statistic. In the context of a t-test for comparing two independent groups, the degrees of freedom is typically calculated as the sum of the sample sizes of both groups minus two (DF = n1 + n2 - 2).

In our exercise, the degrees of freedom for the t-test is 10, derived from six participants in each group minus two. The DF is used to reference the appropriate t-distribution, which is crucial when determining the critical t-value for hypothesis testing. By locating the corresponding critical value in a t-distribution table or using statistical software, researchers assess if the calculated t-value falls within the rejection region, thereby informing the decision on the null hypothesis.
p-Value Significance
The p-value is a pivotal metric in hypothesis testing that quantifies the probability of observing the test results—or more extreme values—assuming the null hypothesis is true. It provides a bridge from statistical calculation to inference, offering a measure of the strength of the evidence against the null hypothesis.

A p-value is interpreted in the context of a pre-specified significance level, often denoted as α, with 0.05 being a common threshold. If the p-value is less than or equal to the significance level (p ≤ α), the result is deemed statistically significant, leading researchers to reject the null hypothesis. Conversely, if the p-value is greater than the significance level (p > α), insufficient evidence exists to discard the null hypothesis. In our exercise, with a p-value of 0.003 against a standard α of 0.05, we find compelling evidence to reject the null hypothesis, indicating a statistically significant difference between the means of groups A and D.

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Most popular questions from this chapter

Some computer output for an analysis of variance test to compare means is given. (a) How many groups are there? (b) State the null and alternative hypotheses. (c) What is the p-value? (d) Give the conclusion of the test, using a \(5 \%\) significance level. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 3 & 360.0 & 120.0 & 1.60 \\ \text { Error } & 16 & 1200.0 & 75.0 & \\ \text { Total } & 19 & 1560.0 & & \end{array}\)

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Find a \(95 \%\) confidence interval for the mean of population \(\mathrm{A}\).

Data 4.1 introduces a study in mice showing that even low-level light at night can interfere with normal eating and sleeping cycles. In the full study, mice were randomly assigned to live in one of three light conditions: LD had a standard light/dark cycle, LL had bright light all the time, and DM had dim light when there normally would have been darkness. Exercises 8.28 to 8.34 in Section 8.1 show that the groups had significantly different weight gain and time of calorie consumption. In Exercises 8.55 and \(8.56,\) we revisit these significant differences. When Calories Are Consumed Researchers hypothesized that the increased weight gain seen in mice with light at night might be caused by when the mice are eating. Computer output for the percentage of food consumed during the day (when mice would normally be sleeping) for each of the three light conditions is shown, along with the relevant ANOVA output. Which light conditions give significantly different mean percentage of calories consumed during the day? \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 55.516 & 10.881 \\ \text { LD } & 9 & 36.485 & 7.978 \\ \text { LL } & 9 & 76.573 & 9.646\end{array}\) One-way ANOVA: Day/night consumption versus Light \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Light } & 2 & 7238.4 & 3619.2 & 39.01 & 0.000 \\ \text { Error } & 25 & 2319.3 & 92.8 & & \\ \text { Total } & 27 & 9557.7 & & & \end{array}\)

Exercises 8.46 to 8.52 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } & \\ \text { A } & 6 & 86.833 & 5.231 & \\ \text { B } & 6 & 76.167 & 6.555 & \\ \text { C } & 6 & 80.000 & 9.230 & \\ \text { D } & 6 & 69.333 & 6.154 & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array}\) Find a \(99 \%\) confidence interval for the mean of population \(\mathrm{A}\). Is 90 a plausible value for the population mean of group \(\mathrm{A}\) ?

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Find a \(90 \%\) confidence interval for the difference in the means of populations \(\mathrm{B}\) and \(\mathrm{C}\).

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