I need help answering: part H and part I I also need a picture sketched of the situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value. (Upload your file below.)In a particular year, 68% of online courses taught at a system of community colleges were taught by full-time faculty. To test if 68% also represents a particular state's percent for full-time faculty teaching the online classes, a particular community college from that state was randomly selected for comparison. In that same year, 35 of the 44 online courses at this particular community college were taught by full-time faculty. Conduct a hypothesis test at the 5% level to determine if 68% represents the state in question.Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) A Hypothesis Tests - 1 sample - Xb Answered: State the distributio XE Rounding Numbers CalculatorFrom Generation To Generation XNew Tabwebassign.net/web/Student/Assignment-Responses/submit?dep=24266164&tags=autosave#question3174743_13Part (h)Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.(i) Alpha:a =(ii) Decision:reject the null hypothesisdo not reject the null hypothesis(iii) Reason for decision:Since a > p-value, we do not reject the null hypothesis.Since a < p-value, we do not reject the null hypothesis.Since a > p-value, we reject the null hypothesis.Since a < p-value, we reject the null hypothesis.(iv) Conclusion:There is sufficient evidence to conclude that the percent of online courses taught at community colleges in the state is not equal to 68%.There is not sufficient evidence to conclude that the percent of online courses taught at community colleges in the state is not equal to 68%.Part (i)Construct a 95% confidence interval for the true proportion. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to four decimal places.)95% C.I.

Part (h) (i) The alpha value is 0.05. (ii) The proportion of online courses at this particular community college were taught by full-time faculty is, p^=xn=3544=0.7955 The test statistic is, z=p^-pp1-pn=0.7955-0.680.681-0.6844=0.11550.0703=1.6430 The test statistic is 1.6430. The hypothesis is two tail. The p-value for two tailed test is, p-value=Pz<1.6430 or z>1.6430=2×Pz<-1.6430 The probability of z less than –1.6430 can be obtained using the excel formula “=NORM.S.DIST(–1.6430,TRUE)”. The probability value is 0.0502. The p-value is, p-value=2×Pz<-1.6430=2×0.0502=0.1004 The p-value is 0.1004. The graph is, Decision rule: If p-value≤α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: The p-value (0.1004) is greater than the level of significance (0.05). Based on the decision rule, do not reject the null hypothesis. (iii) Since α<p-value, we do not reject the null hypothesis. (iv) There is not sufficient evidence to conclude that the percent of online courses taught at community colleges in the state is not equal to 68%.Part (i) Computation of critical value: 1-α=0.95α=0.05α2=0.0251-α2=0.975 The critical value of z-distribution can be obtained using the excel formula “=NORM.S.INV(0.975)”. The critical value is 1.96. The 95% confidence interval for the true proportion is, CI=p^±zα2p^1-p^n=0.7955±1.960.79551-0.795544=0.7955±0.1192=0.7955-0.1192,0.7955+0.1192=0.6763,0.9147 Thus, the 95% confidence interval for the true proportion is (0.6763,0.9147). The graph is,

Answered: In a particular year, 68% of online…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Concept explainers

Question

I need help answering: part H and part I

I also need a picture sketched of the situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value. (Upload your file below.)

In a particular year, 68% of online courses taught at a system of community colleges were taught by full-time faculty. To test if 68% also represents a particular state's percent for full-time faculty teaching the online classes, a particular community college from that state was randomly selected for comparison. In that same year, 35 of the 44 online courses at this particular community college were taught by full-time faculty. Conduct a hypothesis test at the 5% level to determine if 68% represents the state in question.

Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

A Hypothesis Tests - 1 sample - X
b Answered: State the distributio X
E Rounding Numbers Calculator
From Generation To Generation X
New Tab
webassign.net/web/Student/Assignment-Responses/submit?dep=24266164&tags=autosave#question3174743_13
Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
(i) Alpha:
a =
(ii) Decision:
reject the null hypothesis
do not reject the null hypothesis
(iii) Reason for decision:
Since a > p-value, we do not reject the null hypothesis.
Since a < p-value, we do not reject the null hypothesis.
Since a > p-value, we reject the null hypothesis.
Since a < p-value, we reject the null hypothesis.
(iv) Conclusion:
There is sufficient evidence to conclude that the percent of online courses taught at community colleges in the state is not equal to 68%.
There is not sufficient evidence to conclude that the percent of online courses taught at community colleges in the state is not equal to 68%.
Part (i)
Construct a 95% confidence interval for the true proportion. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to four decimal places.)
95% C.I.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Charts and Graphs for Data Visualization$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.