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.* Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value. (Upload your file below.)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 Sixty-eight percent of online cc XToday's Christmas Radio41 x b My Questions | bartlebyConfidence Interval Calculator x +webassign.net/web/Student/Assignment-Responses/submit?dep=24266164&tags=autosave#Q8Indicate 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.Additional Materials A Hypothesis Tests - 1 sample - Xb Sixty-eight percent of online cc XToday's Christmas Radio41 x b My Questions | bartlebyConfidence Interval Calculator x +webassign.net/web/Student/Assignment-Responses/submit?dep=24266164&tags=autosave#Q8State the distribution to use for the test. (Round your standard deviation to four decimal places.)P'-Part (e)What is the test statistic? (Round your answer to two decimal places.)---Select--- vPart (f)What is the p-value? (Round your answer to four decimal places.)Explain what the p-value means for this problem.O If Ho is true, then there is a chance equal to the p-value that the proportion of online courses taught by full-time faculty is at least as different as the sample proportion is from 68%.O If Ho is true, then there is a chance equal to the p-value that the proportion of online courses taught by full-time faculty is not at least as different as the sample proportion is from 68%.O If Ho is false, then there is a chance equal to the p-value that the proportion of online courses taught by full-time faculty is at least as different as the sample proportion is from 68%.O If Ho is false, then there is a chance equal to the p-value that the proportion of online courses taught by full-time faculty is not at least as different as the sample proportion is from 68%.Part (g)O 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 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.

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Description: 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 communi

Given Information: p=68%=0.68 Sample size (n) = 44 X = 35 were taught by full-time faculty.…

Answered: State the distribution to use for the…

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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.

* Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value. (Upload your file below.)

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 Sixty-eight percent of online cc X
Today's Christmas Radio
41 x b My Questions | bartleby
Confidence Interval Calculator x +
webassign.net/web/Student/Assignment-Responses/submit?dep=24266164&tags=autosave#Q8
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.
Additional Materials

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.

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