TUT week 12

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Feb 20, 2024

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1 EC285 A,B Tutorial Questions Week 12 Conϐidence Intervals and Hypothesis Tests for Means Edda Claus EC285 A,B - Fall 2023 5 December 2023 1. The coffee machine in your company “break room” dispenses coffee into paper cups. You are supposed to get 10 ounces of coffee, but the amount varies slightly from cup to cup. Here are the amounts measured in a random sample of 20 cups. Is there evidence that the machine is short-changing users? 9.9 9.7 10.0 10.1 9.9 9.6 9.8 9.8 10.0 9.5 9.7 10.1 9.9 9.6 10.2 9.8 10.0 9.9 9.5 9.9 Answer Hypothesis H 0 : µ = 100, H A : µ < 10 . 0 Model Random sample; 20 < 10% of all cups (no reason to doubt independence). The histogram of sample data to the right looks roughly unimodal and symmetric, so it’s reasonable to believe that the amount of coffee in all possible cups could be described by a Normal model. Okay to do a t -test for the mean. Mechanics : n = 20, df = 19, y = 9 . 845, s = 0 . 1986 t = 9 . 845 0 √ . 1986 −10 . 0 = −3 . 49 20 P − value = P ( y < 10 . 0) = P ( t 19 < −3 . 49) = 0 . 0012

2 Conclusion : Such a small P-value makes it unlikely that the low sample mean resulted from sampling error, so we reject the null hypothesis. There is strong evidence that the mean amount of coffee dispensed by this machine is less than the stated 10 ϐluid ounces. 2. A company has set a goal of developing a battery that lasts over 5 hours (300 minutes) in continuous use. A ϐirst test of 12 of these batteries measured the following lifespans (in minutes): 321, 295, 332, 351, 281, 336, 311, 253, 270, 326, 311, and 288. Is there evidence that the company has met its goal? Answer We want to know if the mean battery lifespan exceeds the 300-minute goal set by the manufacturer. We have 12 battery lifespans in our sample to test the claim. Hypotheses . The null hypothesis is that the batteries have a mean lifespan of at most 300 minutes. The 300-minute goal has not been met. The alternative hypothesis is that the batteries have a mean lifespan greater than 300 minutes. The 300-minute goal has been met. H 0 : µ = 300, H A : µ > 300 Model Randomization Condition : This is not a random sample of batteries, but merely 12 batteries produced for preliminary testing. However, it is reasonable to assume that these batteries are representative of all batteries. Nearly Normal Condition : The distribution of battery lifespans is roughly unimodal and symmetric, so it’s reasonable to assume that the lifespans of all batteries could be described by a Normal model. Because the conditions have been met, we can do a one sample t -test for the mean, with 11 degrees of freedom. Conclusion . Since the P-value is high, we fail to reject the null hypothesis. There is no evidence to suggest that the mean battery lifespan exceeds 300 minutes. It does not appear that the company has met its goal. Find a 90% conϐidence interval for the mean lifespan of this type of battery. Solution : The conditions have been met, so we can create a one-sample t interval, with 90% conϐidence. We are 90% conϐident that the true mean battery lifespan is between 291 . 05 and 321 . 45 minutes.

3 If we wish to conduct another trial, how many batteries must we test to be 95% sure of estimating the mean lifespan to within 15 minutes? To within 5 minutes? Solution : We want to know how many batteries to test to be 95% sure of estimating the mean lifespan to within 15 minutes. First, do a preliminary estimate using z = 1 . 96 as the critical value. Our ϐirst estimate is about 15 batteries, a small sample (see calculations below.) Now, do a better estimate, using 145 as the critical value. We would need to sample about 18 batteries in order to estimate the mean battery lifespan to within 15 minutes, with 95% conϐidence. Finally, to estimate the mean battery lifespan to within 5 minutes, you could do the entire process again, perhaps using a critical value with much higher degrees of freedom. We know that it’s going to take lots more batteries to cut the margin of error to a third of what it was. Alternatively, we know it will take a sample about 9 times as large, 18(9) = 162 batteries, since the margin of error was decreased to a third of its size. (The standard error involves the square root of the sample size in its denominator.) 3. Suppose a sample is selected from a population with mean µ and standard deviation σ . The sample mean is calculated to estimate µ . The sample standard deviation is also computed and found to be 120. a. What is the standard error if the sample size is 36? b. What is the standard error if the sample size is 64? c. What is the standard error if the sample size is 100? d. How does the standard error differ from the standard deviation of the sampling distribution of the sample mean?

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4 Answer a. 20; b. 15; c. 12; d. The standard error is used when σ is unknown and must be estimated using s . 4. Suppose that you are constructing a conϐidence interval for µ . A sample is taken from a population with a nearly normal distribution and the sample mean and sample standard deviation are calculated. Find the appropriate t -value for each of the following. a. A 90% conϐidence interval based on n = 30; b. A 95% conϐidence interval based on n = 30; c. A 99% conϐidence interval based on n = 30; d. As the level of conϐidence increases, how will the resulting conϐidence interval be affected? Answer a. 1.699; b. 2.045; c. 2.756 d. As the level of conϐidence increases, the interval gets wider. 5. Suppose that you are constructing a conϐidence interval for µ . A sample is taken from a population with a nearly normal distribution and the sample mean and sample standard deviation are calculated. Find the appropriate t-value for each of the following. a. A 95% conϐidence interval based on n = 9; b. A 95% conϐidence interval based on n = 25; c. A 95% conϐidence interval based on n = 101; d. As the sample size increases, how will the resulting conϐidence interval be affected Answer

5 a. 2.306; b. 2.064; c. 1.984; d. As the sample size increases, the interval gets shorter. 6. Suppose that you are constructing a conϐidence interval for µ . A sample is taken from a population with a nearly normal distribution a. If the sample size is 36, and the sample standard deviation is 24, at 99% conϐidence what is the margin of error? b. If the sample size is 36, and the sample standard deviation is 24, at 95% conϐidence what is the margin of error? c. If the sample mean is 250, construct the 95% and 99% conϐidence intervals for µ ? d. Interpret these conϐidence intervals Answer a. 10.9 b. 8.12 c. 95% conϐidence interval is 241 . 88 to 258 . 12; 99% conϐidence interval is 239 . 1 to 260 . 9. d. We are 95% conϐident that the true mean is contained in the interval 241 . 88 to 258 . 12 and 99% conϐident that the true mean is contained in the interval 239 . 1 to 260 . 9. 7. Suppose that you are testing the hypotheses H 0 : µ = 1800 vs. H A : µ > 1800 and a sample of size 25 yields the sample mean of 1895 and the sample standard deviation of 250. a. Compute the value of the t -statistic; b. Find the P -value associated with this t -statistic; c. At α = 0 . 05, what is the critical value for t ? d. State your conclusion at α = 0 . 05. Answer

6 a. 1 . 90; b. between 0 . 025 and 0 . 05 (by the table); 0 . 0348 (by technology); c. 1 . 711; d. Since P -value = 0 . 0348 < 0 . 05 (alternatively, t 24 = 1 . 90 > 1 . 711), we reject H 0 . There is strong evidence that the true mean is higher than 1800. 8. Suppose that you are testing the hypotheses H 0 : µ = 85 vs. H A : µ < 85 and a sample of size 76 yields the sample mean of 84 and the sample standard deviation of 6 . 5. a. Compute the value of the t -statistic; b. Find the P -value associated with this t -statistic; c. At α = 0 . 05, what is the critical value for t ? d. State your conclusion at α = 0 . 05. Answer a. −1 . 34; b. between 0 . 05 and 0 . 10 (by the table); 0 . 0920 (by technology); c. −1 . 665; d. Since P -value = 0 . 0920 > 0 . 05 (alternatively, t 75 = −1 . 34 > −1 . 665), we fail to reject H 0 . There is no evidence to suggest that the true mean is less than 85. 9. Suppose that you are testing the hypotheses H 0 : µ = 15 vs. H A : µ ̸= 15 and a sample size of 16 yields the sample mean of 17 . 5 and the sample standard deviation of 6. a. Compute the value of the t -statistic; b. Find the P -value associated with this t -statistic; c. State your conclusion at α = 0 . 05; d. Would you expect the 95% conϐidence interval for µ to contain 15? Explain. Answer a. 1 . 67;

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7 b. between 0 . 10 and 0 . 20 (by the table); 0 . 1163 (by technology); c. Since P -value = 0 . 1163 > 0 . 05 (alternatively, −2 . 131 < t 15 = 1 . 67 < 2 . 131), we fail to reject H 0 . There is no evidence to suggest that the true mean differs from 15; d. Yes, because we are not able to reject the H 0 at α = 0 . 05.

TUT week 12

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