Stony Brook University 3

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Stony Brook University *

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201

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Statistics

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Jan 9, 2024

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pdf

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Uploaded by MegaKomodoDragonPerson678

[\ I\: \’\\\ Stony Brook University “ Name: \PxO For the sole use of students at Stony Brook University. Copyright © 2019 (Copyright Dr. Brenda J Anderson) All Rights Reserved Chapter 9 PRACTICE PROBLEM 1: Fifteen years ago, the average weight of a select subset of the US population between the ages of 30 and 50 was 166 |b. (p=166). A researcher would like to determine whether there has been any change in this figure during the past 15 years. A sample of n = 100 people within the same subset and between 30 and 50 years old was randomly chosen. The average weight for the sample was 173 Ib with a standard deviation of 23. Can the researcher conclude that there is a significant change in weight? Use a two tailed test with alpha =.05 1.1a State the Null Hypothesis in words: M€ b N0 dnong( e (] T p ‘ 1.1b State the Null Hypothesis in mathematical terms: ) My 1.2a What is your alpha level? O5 df? 1o \ 1.2b What is your criterion t value? _= ) \) Calculate your sample statistic: Under these conditions, we use the t-test. The strategy for testing our hypothesis with a z-score or t-score is the same. And that similarity is very evident in this chapter, where we assume we still know the . We can use the same numerator, M -y, but we have to replace g, as the denominator. Just as M estimates |, we'll calculate the sample standard deviation, s, to estimate. 0. We'll use s to calculate Sm_to replace o, as the denominator. 13 | : [ oD £ 1 I \“ \ 1.4a How would you describe this to your aunt? 1.4b How would you report this in a research paper? »

g b | Name: A S . ! For the sole use of students at Stony Brook University. Copyright © 2019 (Copyright Dr. Brenda J Anderson) All Rights Reserved \V Stony Brook University | [t(23)= >, p 05; __ vt the Null Hypothesis]. Is this a “statistically significant difference”? PRACTICE PROBLEM 2: Same idea but starting with the raw data for the sample. “A major corporation in the Northeast noted its employees averaged 5.8 absences during the winter season (December to February) last year (u = 5.8). The distribution is normal. In an attempt to reduce absences, the company offered free flu shots to a small random sample of employees. For a sample of 16 people who took flu shots, the number of absences can be seen in the box below. Use a 1-tailed test and alpha .05, 2.1a State the alternative hypothesis: i Pl shy 2.1b State the Null Hypothesis in words: The A v Y redwu S 2.1c State the Null Hypothesis in mathematical terms: _ |4, 2 M, = Au- 2.2 aWhat is your alpha level? . 05 df?_Lb -1 =15 2.2 b What is your criterion t value? (the direction/sign is determined by the alternative hypothesis) — 1. 157 (note that in the t-table, like the unit normal table, only positive values are provided. However, the t-distribution, like the normal distribution has negative values. When conducting a one-tailed test, you must decide whether the criterion t-score should be positive or negative. Like before, draw the distribution. Place the alpha in the side stated by the alternative hypothesis). If you aren’t sure, see the help item (page 2) in the chapter folder. Calculate your sample statistic: Under these conditions, we use the t-test. The strategy for testing our hypothesis with a z-score or t-score is the same. And that similarity is very evident in this chapter, where we assume we still know the u We can use the same numerator, M -, but we have to replace a,, as the denominator. Just as M estimates u, we'll calculate the sample standard deviation, s, to estimate o. We'll use s to calculate sy to replace @, as the denominator.

"\% ; \\ Stony Brook University Name: For the sole use of students at Stony Brook University. Copyright ® 2019 (Copyright Dr. Brenda J Anderson) All Rights Reserved X X2 To calculate SS, s2, s and finally 0 ) Sy, you first need to find ¥ X and 0 (T X)? i 1 ~ 2 SS= - 2 3 < 3 2 = ——— - 35S §2 = 3 4 4 / \ 4 r \ ] AL 5 6 e 6 S J oAk 6 Sm — E - . \ \ A\ \k: \ 9 S \ =9 2 /’-—,— 5 Y. X=5% XX= 193 - (05 M =105 Calculate the t-score (A7 L c.l . a S@lol 20 - 1 pf e M- pu T =T (,05 Sm To make your decision about the null hypothesis, you need to compare the statistic from your sample to the criterion score. When we report the t-score, we list the degrees of freedom in parentheses after the “t” then place an “=" sign and list the value of the score you calculated. Finally, if your score is more extreme than the criterion score, then state the following, p <.05, because more extreme scores have smaller tails in the distribution. If your sample statistic is closer to O (in the center) than the criterion score, then it is highly likely your sample came from the original t-distribution. In that case, you don’t have good enough evidence that the null is false, and thus you fail to reject the null hypothesis. In that case, you state p >.05.

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