5.1_hw_solutions

5.1_HW_Solutions
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MATH 115, PROBABILITY & STATISTICS N. PSOMAS Ch. 5.1 - HOMEWORK (SOLUTIONS) Section 5.1 5.9 Generating a sampling distribution. Let’s illustrate the idea of a sampling distribution in the case of a very small sample from a very small population. The population is the 10 scholarship players currently on your men’s basketball team. For convenience, the 10 players have been labeled with the integers 0 to 9. For each player, the total amount of time spent (in minutes) on Facebook during the last month is recorded in the table below. The parameter of interest is the average amount of time on Facebook. The sample is an SRS of size n = 3 drawn from this population of players. Because the players are labeled 0 to 9, a single random digit from Table B chooses one player for the sample. (a) Find the mean of the 10 players in the population. This is the population mean μ . (b) Use Table B to draw an SRS of size 3 from this population (Note: you may sample the same player’s time more than once). Write down the three times in your sample and calculate the sample mean . This statistic is an estimate of μ . (c) Repeat this process 10 times using different parts of Table B . Make a histogram of the 10 values of . You are constructing the sampling distribution of . (d) Is the center of your histogram close to μ ? Would it get closer to μ the more times you repeated this sampling process? Explain. Solution (a) Mean time for ten players =338.8
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(b) , (c), (d) Average of 10 sample means =338.0 5.10 Total sleep time of college students. In Example 5.1 , the total sleep time per night among college students was approximately Normally distributed with mean μ = 7.02 hours and standard deviation σ = 1.15 hours. Suppose you plan to take an SRS of size n = 200 and compute the average total sleep time. (a) What is the standard deviation for the average time? (b) Use the 95 part of the 68–95–99.7 rule to describe the variability of this sample mean. (c) What is the probability that your average will be below 6.9 hours? Solution (a) 1.15/sqrt(200) = 0.08132 (b) 95% of sample means fall in the interval μ±2σ/sqrt(200) or 7.02 ± 2*( 0.081332) = (6.8657, 7.1826) (c) P[x-bar < 6.9] = 0.07002 (why?) Sample 1 Sample 2 Sample 3 Sample 4 Sample 5 Sample 6 Sample 7 Sample 8 Sample 9 Sample 10 370 319 319 358 290 319 327 323 309 370 366 358 327 358 370 368 327 309 309 366 309 370 327 358 323 319 358 290 358 366 348.33 349.00 324.33 358.00 327.67 335.33 337.33 307.33 325.33 367.33
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5.11 Determining sample size. Recall the previous exercise. Suppose you want to use a sample
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