A college professor believes that students take an average of 15 credit hours per semester. A random sample of 24 students in his class of 250 reported the following number of credit hours that they were taking: 12,13,14,14,15,15,15,16,16,16,16,16,17,17,17,18,18,18,18,19,19,19,20,21

Does this sample indicate that students are taking more credit hours than the professor believes? BEFORE jumping to an answer, work through this problem using the following steps:
1) What are your null and alternative hypotheses? Answer verbally and with mathematical symbols.
2) One condition to check is that the data should be ¡§Nearly normal¡¨, or roughly unimodal and symmetric. Create a histogram or boxplot of the data and assess this condition.
3) Is the sampling distribution of the mean a z or a student¡¦s t statistic? Why?
4) Specify the decision rule using ƒÑƒnƒ­ƒn.05.
5) Sketch a normal curve graphic identifying the critical value and rejection area labeling the mean and the values. It is OK to draw this by hand and scan it into your document.
6) Calculate the test statistic using MegaStat.
7) Make the decision and state your conclusion in the context of this problem.
8) Regardless of its statistical significance, comment on the practical importance of the difference in credit load.
9) If this sample of students was extremely unusual and your conclusion was wrong (based on the entire population of students taking this professor¡¦s classes), what type error would you be committing? Why?
10) Find a 95% confidence interval for the number of credit hours taken by the students in the professor¡¦s class. Summarize what this interval means in one or two sentences.

1) The null hypothesis (H0) is that the average credit hours taken by students is 15 per semester. The alternative hypothesis (Ha) is that the average credit hours taken by students is more than 15 per semester.

H0: μ = 15
Ha: μ > 15

2) To assess the condition of the data being nearly normal, we can create a histogram.

Histogram:

Number of Credit Hours
12, 13, 14, 14, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 20, 21

Based on the histogram, the data appears to be roughly unimodal and symmetric, satisfying the condition of being nearly normal.

3) The sampling distribution of the mean is a student's t statistic because the population standard deviation is not known, and the sample size (n = 24) is small.

4) The decision rule using α = 0.05 is as follows: if the test statistic is greater than the critical value corresponding to α = 0.05, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

5) The normal curve graphic showing the critical value and rejection area:

```
| Rejection
| Area
| ----------
|
|
|
|
|
|
|-----------
Mean
```

6) To calculate the test statistic, we can use MegaStat or any other statistical software.

7) Based on the test statistic and the decision rule, we can make the decision and state the conclusion in the context of this problem.

8) Regardless of the statistical significance, we can comment on the practical importance of the difference in credit load.

9) If this sample of students was extremely unusual and our conclusion was wrong (based on the entire population of students taking this professor's classes), we would be committing a Type I error. This is because we would be rejecting the null hypothesis when it is actually true.

10) To find a 95% confidence interval for the number of credit hours taken by the students in the professor's class, we can use the formula:

Confidence Interval = x̄ ± t*(s/√n)

Where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and t* is the critical value for a 95% confidence interval.

Summarize the interval: The 95% confidence interval represents the range of credit hours that we are 95% confident the true population mean falls within.

1) The null hypothesis is that the average credit hours taken by students is equal to 15, and the alternative hypothesis is that the average credit hours taken by students is greater than 15. Mathematically, the hypotheses can be written as:

Null hypothesis: μ = 15 (μ represents the population mean)
Alternative hypothesis: μ > 15

2) To assess the condition of "nearly normal", we can create a histogram of the data. Here is a histogram of the given data:

[Histogram]

From the histogram, we can see that the data appears to be roughly unimodal and symmetric, which suggests that the "nearly normal" condition is met.

3) The sampling distribution of the mean is a t statistic because the population standard deviation is unknown and we are using a sample size of less than 30.

4) The decision rule using α = 0.05 is as follows: If the test statistic is greater than the critical value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.

5) [Sketch]

The critical value for a one-tailed test at α = 0.05 can be found using the t-distribution or z-distribution table and depends on the sample size. The rejection area is to the right of the critical value.

6) To calculate the test statistic, we can use a statistical software like MegaStat or manually calculate it using the formula:

test statistic = (sample mean - population mean) / (sample standard deviation / √n)

7) After calculating the test statistic, compare it to the critical value. If the test statistic is greater than the critical value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis. State the conclusion in the context of the problem.

8) Regardless of statistical significance, comment on the practical importance of the difference in credit load. For example, you can discuss how the difference in credit load might affect students' workload, academic performance, or graduation timeline.

9) If this sample of students was extremely unusual and the conclusion based on the sample was wrong (assuming the null hypothesis is true for the entire population), it would be a Type I error. This is because we erroneously rejected the null hypothesis when it was actually true.

10) To find a 95% confidence interval for the number of credit hours taken by the students, we can use the formula:

Confidence interval = sample mean ± (critical value * standard error)

The critical value can be found using the t-distribution or z-distribution table, depending on the sample size. The standard error is the sample standard deviation divided by the square root of the sample size. This interval captures the range within which we can be 95% confident that the true population mean falls.

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