/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q83E Drug content assessment. Refer t... [FREE SOLUTION] | 91影视

91影视

Drug content assessment. Refer to Exercise 8.16 (p. 467)and the Analytical Chemistry (Dec. 15, 2009) study in which scientists used high-performance liquid chromatography to determine the amount of drug in a tablet. Recall that 25 tablets were produced at each of two different, independent sites. The researchers want to determine if the two sites produced drug concentrations with different variances. A Minitab printout of the analysis follows. Locate the test statistic and p-value on the printout. Use these values =.05and to conduct the appropriate test for the researchers.

Test and CI for two Variances: Content vs Site

Method

Null hypothesis 12=1

Alternative hypothesis 121

F method was used. This method is accurate for normal data only.

Statistics

Site N St Dev Variance 95% CI for St Devs

1 25 3.067 9.406 (2.195,4.267)

2 25 3.339 11.147 (2.607,4.645)

Ratio of standard deviation =0.191

Ratio of variances=0.844

95% Confidence Intervals

Method CI for St Dev Ratio CI Variance Ratio

F (0.610, 1.384) (0.372, 1.915)

Tests

Method DF1 DF2 Test statistic p-value

F 24 24 0.84 0.681

Short Answer

Expert verified

The answer can be reduced from the following steps.

Step by step solution

01

Given information

n=25=.05

Finding the test statistic and p-value on the printout is required for this question. Additionally, you must conduct the test using the alpha value.

02

Explaining the various types of tests

Individuals must choose the proper test because conclusions, interpretations, and suggestions will be based on our analyses.

1.T-tests

Comparing two variables within the same population is done using paired t-tests. For instance, obtained the same group's pre- and post-scores or scores from the same group under various circumstances.

The test aims to determine whether statistical support exists for the claim that the mean difference between observations for a given outcome differs significantly from zero. These t-tests are also referred to as dependent t-tests or repeated measures t-tests.

The variable of interest has to be continuous, the respondents in each situation should be the same, must usually distribute distinction between the pair measures, and your data must not contain any outliers to use this test.

2.Anova-test

The t-tests mentioned above have a few drawbacks. Only two means can be compared, and they can only be applied to one independent variable. ANOVA, which compares various means, is a statistical method that can be used when more than one independent variable has been changed.

3.F-test

An F-test is any statistical test that produces a test statistic with an F-distribution under the null hypothesis. It is frequently used when contrasting models fitted to data sets to determine which statistical model best represents the population from which sampled the data. After the models have been least squares fitted to the data, exact "F-tests" are typically needed.

03

Locating the test statistic, p-value and conducting the test using  α=.05

Test and CI for two Variances: Content vs Site

Method

Null hypothesis 12=1

Alternative hypothesis 121

F method was used. This method is accurate for normal data only.

Statistics

Site N St Dev Variance 95% CI for St Devs

1 25 3.067 9.406 (2.195,4.267)

2 25 3.339 11.147 (2.607,4.645)

Ratio of standard deviation =0.191

Ratio of variances=0.844

95% Confidence Intervals

Method CI for St Dev Ratio CI Variance Ratio

F (0.610, 1.384) (0.372, 1.915)

Tests

Method DF1 DF2 Test statistic p-value

F 24 24 0.84 0.681

The printout displays this F-test. The printout highlights the test statistic (F = 0.84) and p-value (p-value = 0.681). No need to reject the null hypothesis that the population variances of the success indices are not equal because a = 0.05 is less than the p-value.

Let

Null hypothesis 12=1

Alternative hypothesis 121

Here need to ensure that the upper tail is used when forming the rejection region for the two-tailed F-test and include the more considerable sample variance in the F-test statistic's numerator to achieve this.

For example, the numerator localid="1664862541925" S21and denominator S22have df=n-1df=25-1df=24, respectively. The test statistic will be a result

F=LargersamplevarianceSmallersamplevarianceF=S21S22F=3.0673.339F=0.918

When the calculated value of F exceeds the tabulated value, reject12=1

for=0.05

F2=0.025F=1.19

F = 0.918 is in the region of rejection. Therefore the test indicates that the information is sufficient to show that the population variances are different.

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 91影视!

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

The 鈥渨inner鈥檚 curse鈥 in transaction bidding. In transaction bidding, the 鈥渨inner鈥檚 curse鈥 is the miracle of the winning (or loftiest) shot price being above the anticipated value of the item being auctioned. The Review of Economics and Statistics (Aug. 2001) published a study on whether shot experience impacts the liability of the winner鈥檚 curse being. Two groups of a stab in a sealed-shot transaction were compared (1)super-experienced stab and (2) less educated stab. In the super-experienced group, 29 of 189 winning flings were above the item鈥檚 anticipated value; 32 of 149 winning flings were above the item鈥檚 anticipated value in the less-educated group.

  1. Find an estimate of p1, the true proportion of super educated stab who fell prey to the winner鈥檚 curse
  2. Find an estimate of p2, the true proportion of less-educated stab who fell prey to the winner鈥檚 curse.
  3. Construct a 90 confidence interval for p1-p2.
  4. d. Give a practical interpretation of the confidence interval, part c. Make a statement about whether shot experience impacts the liability of the winner鈥檚 curse being.

A paired difference experiment produced the following results:

nd=38,x1=92,x2=95.5,d=-3.5,sd2=21

a. Determine the values zfor which the null hypothesis 12=0would be rejected in favor of the alternative hypothesis 12<0 Use .role="math" localid="1652704322912" 伪=.10

b. Conduct the paired difference test described in part a. Draw the appropriate conclusions.

c. What assumptions are necessary so that the paired difference test will be valid?

d. Find a90% confidence interval for the mean difference d.

e. Which of the two inferential procedures, the confidence interval of part d or the test of the hypothesis of part b, provides more information about the differences between the population means?

Enough money has been budgeted to collect independent random samples of size n1=n2=100from populations 1 and 2 to estimate localid="1664867109106" 1-2. Prior information indicates that 1=2=10. Have sufficient funds been allocated to construct a 90% confidence interval for1-2of width 5 or less? Justify your answer.

Given that xis a binomial random variable, compute P(x)for each of the following cases:

a. n= 7, x= 3, p= .5

b. n= 4, x= 3, p= .8

c. n= 15, x= 1, p= .1

Gouges on a spindle. A tool-and-die machine shop produces extremely high-tolerance spindles. The spindles are 18-inch slender rods used in a variety of military equipment. A piece of equipment used in the manufacture of the spindles malfunctions on occasion and places a single gouge somewhere on the spindle. However, if the spindle can be cut so that it has 14 consecutive inches without a gouge, then the spindle can be salvaged for other purposes. Assuming that the location of the gouge along the spindle is random, what is the probability that a defective spindle can be salvaged?

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.