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

Is there any systematic tendency for part-time college faculty to hold their students to different standards than do full-time faculty? The article “Are There Instructional Differences Between Full-Time and Part-Time Faculty?” (College Teaching, \(2009: 23 - 26)\) reported that for a sample of \(125 \) courses taught by fulltime faculty, the mean course \(GPA\) was \(2.7186\) and the standard deviation was \(.63342\), whereas for a sample of \(88\) courses taught by part-timers, the mean and standard deviation were \(2.8639\) and \(.49241,\) respectively. Does it appear that true average course \(GPA\) for part-time faculty differs from that for faculty teaching full-time? Test the appropriate hypotheses at significance level \( .01\).

Short Answer

Expert verified

the solution is

There is little evidence to support the notion that part-time faculty's genuine average course GPA differs from that of full-time professors.

Step by step solution

01

determine the value of the test statistic

\(\begin{array}{l}{{\bar x}_1} = 2.7186\\{{\bar x}_2} = 2.8639\\{s_1} = 0.63342\\{s_2} = 0.49241\end{array}\)

\(\begin{array}{l}{n_1} = 125\\{n_2} = 88\\\alpha = 0.01\end{array}\)

We may use the \(z\)-test because the samples are large \(\left( {n > 30} \right)\). (instead of a t-test).

Assumption:

Either the null hypothesis or the alternative hypothesis is asserted. The null hypothesis and the alternative hypothesis are diametrically opposed. An equality must be included in the null hypothesis.

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _2}\\{H_a}:{\mu _1} \ne {\mu _2}\end{array}\)

Determine the value of the test statistic:

\(z = \frac{{{{\bar x}_1} - {{\bar x}_2}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{2.7186 - 2.8639}}{{\sqrt {\frac{{0.6334{2^2}}}{{125}} + \frac{{0.4924{1^2}}}{{88}}} }} \approx - 1.88\)

If the null hypothesis is true, the \(P\)-value is the probability of getting a result more extreme or equal to the standardized test statistic \(z\). Using the normal probability table, determine the probability.

\(P = P(Z < - 1.88 or Z > 1.88) = 2P(Z < - 1.88) = 2(0.0301) = 0.0602\)

The null hypothesis is rejected if the P-value is less than the alpha significance level.

\(P > 0.01 \Rightarrow \)fail to reject \({H_0}\)

There is insufficient data to support the notion that part-time faculty have a different average course GPA than full-time faculty.

02

conclusion

There is little evidence to support the notion that part-time faculty's genuine average course GPA differs from that of full-time professors.

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

Antipsychotic drugs are widely prescribed for conditions such as schizophrenia and bipolar disease. The article "Cardiometabolic Risk of SecondGeneration Antipsychotic Medications During First-Time Use in Children and Adolescents"\((J\). of the Amer. Med. Assoc., 2009) reported on body composition and metabolic changes for individuals who had taken various antipsychotic drugs for short periods of time.

a. The sample of 41 individuals who had taken aripiprazole had a mean change in total cholesterol (mg/dL) of\(3.75\), and the estimated standard error\({s_D}/\sqrt n \)was\(3.878\). Calculate a confidence interval with confidence level approximately\(95\% \)for the true average increase in total cholesterol under these circumstances (the cited article included this CI).

b. The article also reported that for a sample of 36 individuals who had taken quetiapine, the sample mean cholesterol level change and estimated standard error were\(9.05\)and\(4.256\), respectively. Making any necessary assumptions about the distribution of change in cholesterol level, does the choice of significance level impact your conclusion as to whether true average cholesterol level increases? Explain. (Note: The article included a\(P\)-value.)

c. For the sample of 45 individuals who had taken olanzapine, the article reported\(99\% CI\)as a\(95\% \)CI for true average weight gain\((kg)\). What is a $\(99\% CI\)?

Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor air pollution. The article “Domestic Fuels, Indoor Air Pollution, and Children’s Health” (Annals of the N.Y. Academy of Sciences, \(2008:209 - 217\)) presented information on various pulmonary characteristics in samples of children whose households in India used either biomass fuel or liquefied petroleum gas (\(LPG\)). For the \(755\) children in biomass households, the sample mean peak expiratory flow (a person’s maximum speed of expiration) was \(3.30L/s\), and the sample standard deviation was \(1.20\). For the \(750\) children whose households used liquefied petroleum gas, the sample mean \(PEF\) was \(4.25\) and the sample standard deviation was \(1.75\).

a. Calculate a confidence interval at the \(95\% \) confidence level for the population mean \(PEF\) for children in biomass households and then do likewise for children in \(LPG\) households. What is the simultaneous confidence level for the two intervals?

b. Carry out a test of hypotheses at significance level \(.01\) to decide whether true average \(PEF\) is lower for children in biomass households than it is for children in \(LPG\) households (the cited article included a P-value for this test)

c. \(FE{V_1}\), the forced expiratory volume in \(1\) second, is another measure of pulmonary function. The cited article reported that for the biomass households the sample mean FEV1 was \(2.3L/s\) and the sample standard deviation was \(.5L/s\). If this information is used to compute a \(95\% \) \(CI\) for population mean \(FE{V_1}\), would the simultaneous confidence level for this interval and the first interval calculated in (a) be the same as the simultaneous confidence level determined there? Explain

The invasive diatom species Didymosphenia geminata has the potential to inflict substantial ecological and economic damage in rivers. The article "Substrate Characteristics Affect Colonization by the BloomForming Diatom Didymosphenia geminata" (Acquatic Ecology, 2010:\(33 - 40\)) described an investigation of colonization behavior. One aspect of particular interest was whether the roughness of stones impacted the degree of colonization. The authors of the cited article kindly provided the accompanying data on roughness ratio (dimensionless) for specimens of sandstone and shale.

\(\begin{array}{*{20}{l}}{ Sandstone: }&{5.74}&{2.07}&{3.29}&{0.75}&{1.23}\\{}&{2.95}&{1.58}&{1.83}&{1.61}&{1.12}\\{}&{2.91}&{3.22}&{2.84}&{1.97}&{2.48}\\{}&{3.45}&{2.17}&{0.77}&{1.44}&{3.79}\end{array}\)

\(\begin{array}{*{20}{l}}{ Shale: }&{.56}&{.84}&{.40}&{.55}&{.36}&{.72}\\{}&{.29}&{.47}&{.66}&{.48}&{.28}&{}\\{}&{.72}&{.31}&{.35}&{.32}&{.37}&{.43}\\{}&{.60}&{.54}&{.43}&{.51}&{}&{}\end{array}\)

Normal probability plots of both samples show a reasonably linear pattern. Estimate the difference between true average roughness for sandstone and that for shale in a way that provides information about reliability and precision, and interpret your estimate. Does it appear that true average roughness differs for the two types of rocks (a formal test of this was reported in the article)? (Note: The investigators concluded that more diatoms colonized the rougher surface than the smoother surface.)

Using the traditional formula, a \(95\% \) CI for \({p_1} - {p_2}\)is to be constructed based on equal sample sizes from the two populations. For what value of \(n( = m)\)will the resulting interval have a width at most of .1, irrespective of the results of the sampling?

The article "Pine Needles as Sensors of Atmospheric Pollution" (Environ. Monitoring, 1982: 273-286) reported on the use of neutron-activity analysis to determine pollutant concentration in pine needles. According to the article's authors, "These observations strongly indicated that for those elements which are determined well by the analytical procedures, the distribution of concentration is lognormal. Accordingly, in tests of significance the logarithms of concentrations will be used." The given data refers to bromine concentration in needles taken from a site near an oil-fired steam plant and from a relatively clean site. The summary values are means and standard deviations of the log-transformed observations.

Site Sample Size Mean Log Concentration SD of Log Concentration

Steamplant 8 18.0 4.9

Clean 9 11.0 4.6

Let \(\mu _1^*\) be the true average log concentration at the first site, and define \(\mu _2^*\) analogously for the second site.

a. Use the pooled t test (based on assuming normality and equal standard deviations) to decide at significance level .05 whether the two concentration distribution means are equal.

b. If \(\sigma _1^8\) and \(\sigma _2^*\) (the standard deviations of the two log concentration distributions) are not equal, would \({\mu _1} and {\mu _2}\) (the means of the concentration distributions) be the same if \(\mu _1^* = \mu _2^*\) ? Explain your reasoning.

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