Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \(B\). Show all details of the test.

Short Answer

Expert verified
The decision on whether there is a significant difference in the means of groups A and B depends on the computed t-value from step 2 and the critical t-value from step 3. If \( |t_{computed}| > |t_{critical}| \), then the conclusion would be that there is a significant difference in means between the two groups.

Step by step solution

01

Compute t-statistic

First, compute the t-statistic for the difference in means between group A and group B. This can be computed using the formula \( t = \frac{{M_A - M_B}}{{\sqrt{{MSE(1/n_A + 1/n_B)}}}} \) where MSE (Mean Squares Within) is the within-group variance, n is the group size, M is the group mean. So, \( M_A = 86.833 \), \( M_B = 76.167 \), \( MSE = 48.3 \), \( n_A = n_B = 6 \).
02

Calculate t-statistic

Now, substitute the known values into the formula: \( t = \frac{{86.833 - 76.167}}{{\sqrt{{48.3(1/6 + 1/6)}}}} \). Solve this equation to get the computed t-value.
03

Determine the critical value

It's necessary to first find the degrees of freedom for the t-distribution. In this case, the degrees of freedom (df) would be: \( df = n_A + n_B - 2 = 6+6-2 = 10 \). Using a significance level of 0.05 (a standard value if none is given), the critical value of t for a two-tailed test from the t-distribution table is approximately ±2.228.
04

Compare t-statistic to critical value

Finally, compare the computed t-value from step 2 with the critical t-value from step 3. If the absolute value of the computed t-value is greater than the absolute value of the critical t-value, then there is a significant difference in means between the two groups.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

We have seen that light at night increases weight gain in mice and increases the percent of calories consumed when mice are normally sleeping. What effect does light at night have on glucose tolerance? After four weeks in the experimental light conditions, mice were given a glucose tolerance test (GTT). Glucose levels were measured 15 minutes and 120 minutes after an injection of glucose. In healthy mice, glucose levels are high at the 15 -minute mark and then return to normal by the 120 -minute mark. If a mouse is glucose intolerant, levels tend to stay high much longer. Computer output is shown giving the summary statistics for both measurements under each of the three light conditions. (a) Why is it more appropriate to use a randomization test to compare means for the GTT-120 data? (b) Describe how we might use the 27 data values in GTT-120 to create one randomization sample. (c) Using a randomization test in both cases, we obtain a p-value of 0.402 for the GTT-15 data and a p-value of 0.015 for the GTT-120 data. Clearly state the results of the tests, using a \(5 \%\) significance level. Does light at night appear to affect glucose intolerance?

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. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 4 & 1200.0 & 300.0 & 5.71 \\ \text { Error } & 35 & 1837.5 & 52.5 & \\ \text { Total } & 39 & 3037.5 & & \end{array}\)

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) What is the pooled standard deviation? What degrees of freedom are used in doing inferences for these means and differences in means?

Drug Resistance and Dosing Exercise 8.39 on page 561 explores the topic of drug dosing and drug resistance by randomizing mice to four different drug treatment levels: untreated (no drug), light ( \(4 \mathrm{mg} / \mathrm{kg}\) for 1 day), moderate \((8 \mathrm{mg} / \mathrm{kg}\) for 1 day), or aggressive ( \(8 \mathrm{mg} / \mathrm{kg}\) for 5 or 7 days). Exercise 8.39 found that, contrary to conventional wisdom, higher doses can actually promote drug resistance, rather than prevent it. Here, we further tease apart two different aspects of drug dosing: duration (how many days the drug is given for) and amount per day. Recall that four different response variables were measured; two measuring drug resistance (density of resistant parasites and number of days infectious with resistant parasites) and two measuring health (body mass and red blood cell density). In Exercise 8.39 we don't find any significant differences in the health responses (Weight and \(R B C)\) so we concentrate on the drug resistance measures (ResistanceDensity and DaysInfectious) in this exercise. The data are available in DrugResistance and we are not including the untreated group. (a) Investigate duration by comparing the moderate treatment with the aggressive treatment (both of which gave the same amount of drug per day, but for differing number of days). Which of the two resistance response variables (ResistanceDensity and DaysInfectious) have means significantly different between these two treatment groups? For significant differences, indicate which group has the higher mean. (b) Investigate amount per day by comparing the light treatment with the moderate treatment (both of which lasted only 1 day, but at differing amounts). Which of the two resistance response variables have means significantly different between these two treatment groups? For significant differences, indicate which group has the higher mean. (c) Does duration or amount seem to be more influential (at least within the context of this study)? Why?

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. \(\begin{array}{lrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } \\ \text { Groups } & 3 & 360.0 & 120.0 & 1.60 \\ \text { Error } & 16 & 1200.0 & 75.0 & \\ \text { Total } & 19 & 1560.0 & & \end{array}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free