For all of the following use the five step process to solve the problem in the question, that is, state the null hypothesis, the alternative hypothesis, the level of significance and the type of test (z-test, t-test, p-value, confidence limit and whether it is two tailed, one-tailed upper or one tailed lower – note that you may need several terms to describe the test) you will use and why, and state your decision rule. As part of your answer sketch out the solution on a normal curve showing the critical value, the calculated value and the p-value.

2. According to a recent survey, Americans get an average of seven (7) hours of sleep each night. A random sample of 150 SDSU students revealed that the average amount of sleep for the students was 6 hours and 48 minutes (6.8 hours) with a standard deviation of .9 hours. From the data is it reasonable to conclude that the students at SDSU sleep less than the average American? Use the .05 level of significance

This questions was submitted to be answered, don't hand me BS about not doing it you bum!

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To solve this problem using the five-step process, we will go through the following steps:

Step 1: State the null hypothesis (H0) and the alternative hypothesis (Ha).
Null Hypothesis (H0): The average amount of sleep for SDSU students is equal to or greater than the average amount of sleep for Americans (μ ≥ 7).
Alternative Hypothesis (Ha): The average amount of sleep for SDSU students is less than the average amount of sleep for Americans (μ < 7).

Step 2: Determine the level of significance.
The level of significance is given in the question as α = 0.05. This means we are willing to accept a 5% chance of making a Type I error (rejecting the null hypothesis when it is actually true).

Step 3: Identify the appropriate test to use.
Based on the information provided in the question and the sample size being greater than 30, we can use a z-test for comparing means.

Step 4: Calculate the test statistic and p-value.
The formula for the z-test for comparing means is:
z = (x̄ - μ) / (σ / √n)
where x̄ is the sample mean, μ is the population mean (7), σ is the standard deviation, and n is the sample size.

Using the given information, we have:
x̄ = 6.8
μ = 7 (from the null hypothesis)
σ = 0.9
n = 150

Calculating the z-test statistic:
z = (6.8 - 7) / (0.9 / √150) ≈ -2.63

To find the p-value, we can use a z-table or a calculator. The p-value corresponds to the probability of observing a test statistic as extreme as -2.63 (in the left tail of the distribution) under the null hypothesis.

Step 5: Make a decision and interpret the results.
Comparing the calculated test statistic (-2.63) with the critical value(s) from the z-table or calculator for a one-tailed test at α = 0.05, we find that the critical value is approximately -1.645.

Decision Rule:
If the calculated test statistic is less than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Interpreting the Results:
Since the calculated test statistic (-2.63) is less than the critical value (-1.645), we reject the null hypothesis. This means there is enough evidence to suggest that the students at SDSU sleep less than the average American.

To sketch out the solution on a normal curve, we would plot the standard normal distribution with the critical value (-1.645) on the left tail, the calculated value (-2.63) to the left of it, and shade the region corresponding to the p-value. The p-value would be calculated as the area under the curve to the left of the calculated value (-2.63).