Textbook #1: Lane *et al*. **Introduction to Statistics, David M. Lane et al., 2013. (**

**http://onlinestatbook.com/Online_Statistics_Education.pdf**

**)**

Textbook #2:Illowsky

*et al*.

**Introductory Statistics, Barbara Illowsky et al., 2013.**

**(****http://openstaxcollege.org/files/textbook_version/hi_res_pdf/15/col11562-op.pdf****)**

*See the file named “UMUC_STAT200_EXCEL_Tips” at the “Course Materials” menu link to find functions for calculating the Normal Distribution and Student’s T-distribution values needed for this assignment.*

**Lane – Chapter 11: (18)**

- You choose an alpha level of .01 and then analyze your data.

- What is the probability that you will make a Type I error given that the nullhypothesis is true?
- What is the probability that you will make a Type I error given that the nullhypothesis is false?

**Lane – Chapter 12: (7,13)**

7.Below are data showing the results of six subjects on a memory test. The three scores per subject are their scores on three trials (a, b, and c) of a memory task. Are the subjects getting better each trial? Test the linear effect of trial for thedata in the table below.

- Compute L (linear effect of trial) for each subject using the contrast weights -1, 0, and 1. That is,compute (-1)(a) + (0)(b) + (1)(c) for each subject. Make a new column in your table with this result.
- Compute a one-sample t-test on this column (with the L values for eachsubject) you created.

HINT: See the example in the “Specific Comparisons” section of Chapter-12. Find the “t-value” and the “two-tailed probability” using the EXCEL “TDIST” function.Assume the statistic for this problem is “L” and use the following formula for the t-value. Also assume the hypothesized value is 0, since the contrast weighting (-1,0,+1) for a “perfect” set of data would make “L” be 0 in all cases.

t = (statistic – hypothesized value) / (standard error of the statistic)

t = (Mean of L – 0) / (Standard Error of L)

t-value =

X = Sample Mean

S = Sample Standard Deviation

N = Number of L-Samples

13.You are conducting a study to see if students do better when they study all at once or in intervals. One group of 12 participants took a test after studying forone hour continuously. The other group of 12 participants took a test afterstudying for three twenty minute sessions. The first group had a mean score of75 and a variance of 120. The second group had a mean score of 86 and avariance of 100.

HINT: See Chapter-12 section on “Differences between two means (independent groups)”.

- What is the calculated t value? Are the mean test scores of these two groupssignificantly different at the .05 level?
- What would the t value be if there were only 6 participants in each group?Would the scores be significant at the .05 level?

**Lane – Chapter 13: (4)**

4.Rank order the following in terms of power.

Population-1 Mean | n | Population-2 Mean | Standard Deviation | |

A | 29 | 20 | 43 | 12 |

B | 34 | 15 | 40 | 6 |

C | 105 | 24 | 50 | 27 |

D | 170 | 2 | 120 | 10 |

** **

**Illowsky – Chapter 9 (65,71,77) **

** **

Background for problems 65 and 71: Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.

- The null and alternative hypotheses are:
- Ho: x ¯ = 4.5, Ha : x ¯ > 4.5
- Ho: μ ≥ 4.5, Ha: μ < 4.5
- Ho: μ = 4.75, Ha: μ > 4.75
- Ho: μ = 4.5, Ha: μ > 4.5

- The Type-1error is:
- to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
- to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
- to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
- to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher

- An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, onaverage, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the1% level? Solve using the following steps, similar to Appendix-E (Hypothesios Testing with One Sample Mean):

- State the Null Hypothesis (H
_{o}) and Alternate Hypothesis (H_{a}) - Find the random variable X
- State the distribution you will use and why ?
- What is the test statistic (t-value)?
- What is the P-value (probability) ?
- Will you reject or not reject the Null Hypothesis and why ?

**Illowsky – Chapter 10 (79,91,120) **

- A student at a four-year college claims that mean enrollment at four–year colleges is higher than at two–year colleges in the United States. Two surveys are conducted. Of the 35 two–year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191. Test the hypothesis, assuming a 5% significance level. Solve using the following steps, similar to Appendix-E (Hypothesios Testing with Two Sample Means). Use the equations for standard error, test statistic and degrees-of-freedom from Section 10.1 of Illowsky (Two Population Means with unknown Standard Deviations).

- State the Null Hypothesis (H
_{o}) and Alternate Hypothesis (H_{a}) - Find the random variable X (remember that X is the difference between the two sample means)
- State the distribution you will use and why ?
- What is the test statistic (t-value) ?
- What is the P-value (probability) ? HINT: The formula for the “degrees of freedom” is far more complex. See the formula in Section 10.1.
- Will you reject or not reject the Null Hypothesis and why ?

- A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquiddiet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 poundswith a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviationof 14 pounds. Test the hypothesis, assuming a 5% significance level. Solve using the following steps, similar to Appendix-E (Hypothesios Testing with Two Sample Means). Use the equations for standard error, test statistic and degrees-of-freedom from Section 10.1 of Illowsky (Two Population Means with unknown Standard Deviations).

- State the Null Hypothesis (H
_{o}) and Alternate Hypothesis (H_{a}) - Find the random variable X (remember that X is the difference between the two sample means)
- State the distribution you will use and why ?
- What is the test statistic (t-value) ?
- What is the P-value (probability) ? HINT: The formula for the “degrees of freedom” is far more complex. See the formula in Section 10.1.
- Will you reject or not reject the Null Hypothesis and why ?

- A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. Shetakes four new students. She records their 18-hole scores before learning the technique and then after having taken her class.She conducts a hypothesis test. The data are as follows. Is the correct decision to Reject the Null Hypothesis or Not Reject the Null Hypothesis? Test the hypothesis, assuming a 5% significance level. Solve using the following steps:

- State the Null Hypothesis (H
_{o}) and Alternate Hypothesis (H_{a}) - Find the random variable X. Consider we are testing “paired samples”.
- State the distribution you will use and why ?
- What is the test statistic (t-value) ? Assume, the population mean of the differences = 0.
- What is the P-value (probability) ?
- Will you reject or not reject the Null Hypothesis and why ?

Player 1 | Player 2 | Player 3 | Player 4 | |

Mean score before class | 83 | 78 | 93 | 87 |

Mean score after class | 80 | 80 | 86 | 86 |