Deliverable 4 - Hypothesis Test Assignment Content
Instructions - Read First
Instructions: The following questions on the next six tabs are shown to you by a student who is asking for help. Your job is to help the student walk through the problems by showing the student how to solve each problem in detail. You are expected to explain all of the steps in your own words. Key: <i> - This problem is an incorrect. Your job is to find the errors, correct the errors, and explain what they did wrong. <p> - This problem is partially finished. You must complete the problem by showing all steps and explain your corrections. <b> - This problem is blank. You must start from scratch and explain how you will approach the problem, how you solve it, and explain why you took each step.
Question 1
| lower | 0.1515050028 | ||||||||||||||
| Upper | 0.9777844056 | ||||||||||||||
| Probability of bone density | 0.8262794028 | 0.8263 |
<b> Assume that a randomly selected subject is given a bone density test. Those tests follow a standard normal distribution. Find the probability that the bone density score for this subject is between -1.03 and 2.01. Show your calculations using Excel functions in the area to the right of this text box and answer the problem with explanations below. Assuming the scores given above are standard scores then the following steps follow step 1: use the standard distribution formula to get the probability of the lower and upper standard score step 2: since the two probabilities involve cummulative entries with areas from the left side of normal curve, we subtract the lower standard score probability from the upper to get the probability of the bone density
Question 2
| lower height difference | -1.8 | |||||||||||||
| upper height difference | 12.2 | |||||||||||||
| lower standard score | -0.6206896552 | |||||||||||||
| upper standard score | 4.2068965517 | |||||||||||||
| lower | 0.2674019186 | |||||||||||||
| upper probability | 0.9999870549 | |||||||||||||
| 73.26% |
<b> The U.S. Airforce requires that pilots have a height between 62 in. and 76 in. If women’s heights are normally distributed with a mean of 63.8 in. and a standard deviation of 2.9 in, find the percentage of women that meet the height requirement. Show your calculations using Excel functions in the area to the right of this text box and answer the problem with explanations below. step 1: we are given the population mean and and SD, therefore, the normal distribution will be used Step 2: subtract the population mean from the sample means Step 3: find standard scores by dividing the diference with the population standard deviation step 4: use the standard distribution formula to get the probability of the lower and upper standard score step 2: since the two probabilities involve cummulative entries with areas from the left side of normal curve, we subtract the lower standard score probability from the upper and multiply by 100% to get the %
Question 3
| z-score for woman | -0.6880733945 |
Question 4
| P(Z<-0.18) | 0.4285762841 | |||||||||||||
| P(Z>-0.18) | 0.5714237159 |
<p> What is the cumulative area from the left under the curve for a z-score of -0.18? What is the area on the right of that z-score? Hint: You will have two answers (one for the left and one for the right). Student’s partially finished answer: The first part is to find the cumulative area from the left under the curve for a z-score of -0.18. This means that I need to find the probability that a z-score is less than -0.18. Show your calculations using Excel functions in the area to the right of this text box and finish answering the problem with explanations below. Step 1: probability that the z-score is less than -0.18 gives the cumulative area to the left step 2: probability that the Z-score is greater than -0.18 gives the cumulative area to the right Step 3: to get area to the subtract the above probability from 1
Question 5
| area to the right | 0.337 | |||||||||||||
| correct standard score | -0.4206646196 |
<i> If the area under the standard normal distribution curve is 0.6630 from the right, what is the corresponding z-score? Student’s incorrect answer: We plug in “=NORM.S.INV(0.6630)” into Excel and get a z-score of 0.42. Identify where the student went wrong when solving the problem above. Show how to correctly calculate the answer using Excel in the area to the right of this text box. Finish the problem by stating how to correctly set up the problem and solve it below. Step 1: the student used the right formula but the formula provided the standard score to the left side. Step 2: To get correct answer, subtract 0.663 from 1 Step 3: use the formula with the above result to get the correct standard score
Question 6
| difference | 0.808 | |||||||||||||
| division by SD | 1.3138211382 | |||||||||||||
| P(x<2) | 0.9055467988 | |||||||||||||
| 90.55% |
<p> A research shows that the Richter scale magnitudes of earthquakes are normally distributed with a mean of 1.192 and a standard deviation of 0.615. Earthquakes with magnitudes less than 2.00 are considered "microearthquakes" that humans do not feel. What percentage of earthquakes fall into this category? Student’s partially finished answer: We need to find the probability that earthquakes fall into the microearthquake category. The first step is to find the probability that the Richter scale is less than 2.00. Show your calculations using Excel functions in the area to the right of this text box and finish answering the problem with explanations below. step 1: to find the probability that the Richter scale is less than 2.00, we first subtract the population mean (1.192) from 2 and divide the result by the population standard deviation (0.615) Step 2: find the cummulative probability to the left step 3: convert to percentage using excel (%) button