Stats 227 Ch 8
Describe the sampling distribution of p hat. Assume the size of the population is 25,000. n=1200, p= 0.526 Describe the shape of the sampling distribution of p hat. Choose the correct answer below. Determine the mean of the sampling distribution of p hat. Determine the standard deviation of the sampling distribution of p hat.
- The shape of the sampling distribution of p hat is approximately normal because n≤0.05N and np(1-p)>10. - mean of the sampling distribution of p hat = 0.526 - To find the str dev of the sampling distribution of p hat do the below. first 1-0.526= 0.474 then use desmos calculator enter sqrt 0.526 (p) - 0.474 (mean p hat) / 1200 (n)
What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample mean?
Since the sample size if large enough, the population distribution does not need to be normal.
A simple random sample of size n=36 is obtained from a population with μ=89 and σ=24. (a) Describe the sampling distribution of x. (b) What is P x>94.6? (c) What is P x≤80? (d) What is P 85.4<x<96.6?
a) The distribution is approximately normal. Find the mean and standard deviation of the sampling distribution of x. μx = 89 σx= 4 (find using desmos, 24/sqrt(36) b) 0.0808 Find using statcrunch, below steps. Stat > Calculators> Normal> Standard> Enter mean (89) Enter Std. dev (4) and enter P(x>94.6) and compute. round 4 decimals c) 0.0122 follow stat crunch steps above just change the P part to get correct answer. round 4 decimals d) 0.7872 Stat crunch steps above but instead of standard, choose between and enter P and compute. round 4 decimals.
The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is unknown. However, records indicate that the mean time is 16.3 minutes, and the standard deviation is 3.8 minutes. Complete parts (a) through (c). (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? (b) What is the probability that a random sample of n=45 oil changes results in a sample mean time less than 15 minutes? (c) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 45 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value? This will be the goal established by the manager.
a) The sample size needs to be greater than or equal to 30. b) 0.0109 Stat > Cal> Normal>standar>enter mean> strdev (3.8/sqrt(45) > P X < 15 > Compute 0.01086909 Round to 4th decimal 0.0109. c) 15.6 mins Stat > Cal> Normal> Standard> Enter mean (same) > Enter str dev (w sqrt) > P X < LEAVE BLANK = 0.10> compute and answer will be in the "leave blank area"
Determine μx and σx from the given parameters of the population and sample size. μ=71, σ=21, n=49
a) μx = 71 b) σx = 3 To find b do the below on the desmos scientific calculator online. - 21/sqrt(49) = 3.
Suppose a simple random sample of size n=1000 is obtained from a population whose size is N=2,000,000 and whose population proportion with a specified characteristic is p=0.78. Complete parts (a) through (c) below. (a) Describe the sampling distribution of p. (b) What is the probability of obtaining x=810 or more individuals with the characteristic? (c) What is the probability of obtaining x=740 or fewer individuals with the characteristic?
a)Approximately normal, μp=0.78 and σp≈ 0.0131 how you get the above. Sqrt 0.78 (1-.78) / 1000 b) = 0.0110 (rounded 4 decimals) To get answer use stat crunch normal calculator. In the P(x> enter 810/1000) = 0.01100845. c) 0.011 Use same stat details as part b, except this time its <.
Suppose a simple random sample of size n=200 is obtained from a population whose size is N=30,000 and whose population proportion with a specified characteristic is p=0.6. Complete parts (a) through (c) below. (a) Describe the sampling distribution of p hat. Choose the phrase that best describes the shape of the sampling distribution below. Determine the mean of the sampling distribution of p hat. Determine the standard deviation of the sampling distribution of p hat. b) What is the probability of obtaining x=128 or more individuals with the characteristic? That is, what is P(p≥0.64)? (c) What is the probability of obtaining x=118 or fewer individuals with the characteristic? That is, what is P(p≤0.59)?
a) Approximately normal because n≤0.05N and np(1-p)>10 - mean of the sampling distribution of p hat is 0.6. (p=0.6) - from the question. - str dev of the sampling distribution of p hat is = 0.03641 (found in desmos enter sqrt 0.6 (1-0.6) / 200 to get the answer. b) P(p≥0.64) = 0.1241 use stat > cal> normal> enter mean above & str dev > then > 0.64 and compute. c)P(p≤0.59) = 0.3864 use stat > cal> normal> enter mean above & str dev > then <0.59 and compute.
Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean μ=225 days and standard deviation σ=11 days. Complete parts (a) through (f) below. (a) What is the probability that a randomly selected pregnancy lasts less than 221 days? The probability that a randomly selected pregnancy lasts less than 221 days is approximately Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Round to the nearest integer as needed.) (b) Suppose a random sample of 21 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. c) What is the probability that a random sample of 21 pregnancies has a mean gestation period of 221 days or less? Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (d) What is the probability that a random sample of 47 pregnancies has a mean gestation period of 221 days or less?
a) 0.3581 Stat > cal> normal > standard> enter mean, str.dev and P x <221 compute. - Interpretation - If 100 pregnant individuals were selected independently from this population, we would expect 36 (this was rounded from 0.3581) pregnancies to last less than 221 days b) The sampling distribution of x is normal with μx=225 and σx=2.4004 c) 0.0478 - use stat crunch (stat,cal,normal..etc), but instead enter mean & str.dev from pt B. Interpretation - If 100 independent random samples of size n=21 pregnancies were obtained from thispopulation, we would expect 55 sample(s) to have a sample mean of 221 days or less. d)
Without doing any computation, decide which has a higher probability, assuming each sample is from a population that is normally distributed with μ=100 and σ=15. Explain your reasoning. (a) P(90≤x≤110) for a random sample of size n=10 (b) P(90≤x≤110) for a random sample of size n=20
a) = 0.9650 b) = 0.9971 Answer: P(90≤x≤110) for a random sample of size n=20 has a higher probability. As n increases, the standard deviation decreases.
Suppose Jack and Diane are each attempting to use a simulation to describe the sampling distribution from a population that is skewed left with mean 60 and standard deviation 15. Jack obtains 1000 random samples of size n=3 from the population, finds the mean of the means, and determines the standard deviation of the means. Diane does the same simulation, but obtains 1000 random samples of size n=30 from the population. Complete parts (a) through (c) below. (a) Describe the shape you expect for Jack's distribution of sample means. Describe the shape you expect for Diane's distribution of sample means. Choose the correct answer below. b) What do you expect the mean of Jack's distribution to be? What do you expect the mean of Diane's distribution to be? (c) What do you expect the standard deviation of Jack's distribution to be? What do you expect the standard deviation of Diane's distribution to be?
a) Jack's distribution is expected to be skewed left, but not as much as the original distribution. Diane's distribution is expected to be approximately normal. b) Jack's distribution is expected to have a mean of 60. Diane's distribution is expected to have a mean of 60. c) Jack's distribution is expected to have a standard deviation of 2.74. Diane's distribution is expected to have a standard deviation of 8.66. found using desmos - str dev / sqrt sample size to find each new str dev.
A simple random sample of size n=46 is obtained from a population with μ=68 and σ=17. (a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample mean? Assuming that this condition is true, describe the sampling distribution of x. (b) Assuming the normal model can be used, determine P(x<71.6). (c) Assuming the normal model can be used, determine P(x≥69.1).
a) Since the sample size is large enough, the population distribution does not need to be normal Approximately normal, with μx=68 and σx= 17/sqrt(46) b) P(x<71.6) = 0.945 Steps: STAT Calculator Normal Enter mean Enter str.dev as square root (17/sqrt(46)) Stat will take it as sqrt written out like this. Compute. c)P(x≥69.1)= 0.3304 Same steps as above, just change the less/greater than symbol.
According to a study conducted by a statistical organization, the proportion of people who are satisfied with the way things are going in their lives is 0.74. Suppose that a random sample of 100 people is obtained. Complete parts (a) through (e) below. (a) Suppose the random sample of 100 people is asked, "Are you satisfied with the way things are going in your life?" Is the response to this question qualitative or quantitative? Explain. (b) Explain why the sample proportion, p hat, is a random variable. What is the source of the variability? (c) Describe the sampling distribution of p hat, the proportion of people who are satisfied with the way things are going in their life. Be sure to verify the model requirements. (d) In the sample obtained in part (a), what is the probability that the proportion who are satisfied with the way things are going in their life exceeds 0.77? e) Using the distribution from part (c), would it be unusual for a survey of 100 people to reveal that 66 or fewer people in the sample are satisfied with their lives?
a) The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not. b)The sample proportion p hat is a random variable because the value of p hat varies from sample to sample. The variability is due to the fact that different people feel differently regarding their satisfaction. c) Since the sample size is no more than 5% of the population size and np(1−p)=19.24≥10, the distribution of p hat is approximately normal with μp=0.74 and σp=0.044 Use desmos to find - 19.24 you multiply the np(1-p) = 100x0.74(1-0.74) 0.74 is given in the sentence as well as the sample size. to find the str dev you sqrt 0.74(1-0.74 / 100. d) The probability that the proportion who are satisfied with the way things are going in their life exceeds 0.77 is 0.2477 find using stat>cal>normal>enter mean>enter str dev>enter p>0.77> compute e) The probability that 66 or fewer people in the sample are satisfied is 0.0345, which is unusual because this probability is less than 5% 0.0345 - use statcrunch to find.
The shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility is unknown. However, records indicate that the mean time is 21.7 minutes, and the standard deviation is 4.9 minutes. Complete parts (a) through (c). a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? b) What is the probability that a random sample of n=40 oil changes results in a sample mean time less than 20 minutes? c) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value? This will be the goal established by the manager.
a) The sample size needs to be greater than or equal to 30. b) The probability is approximately: 0.0141 Use following statcrunch steps to get above: STAT Calculator Normal Enter mean Enter str.dev as square root (4.9/sqrt(40)) Stat will take it as sqrt written out like this. Compute. c) There is a 10% chance of being at or below a mean oil-change time of 20.7 minutes. To find 20.7 use statcrunch, steps below: Enter mean in this case was (21.7) Enter str. dev in this case was 4.9/sqrt(40) Leave P(X <= blank) =.10 (the 10% oil-change). Compute Use generated answer in the P(X<= 20.7)
The acceptable level for insect filth in a certain food item is 3 insect fragments (larvae, eggs, body parts, and so on) per 10 grams. A simple random sample of 60 ten-gram portions of the food item is obtained and results in a sample mean of x=3.6 insect fragments per ten-gram portion. Complete parts (a) through (c) below. (a) Why is the sampling distribution of x approximately normal? (b) What is the mean and standard deviation of the sampling distribution of x assuming μ=3 and σ=sqrt3 c) What is the probability a simple random sample of 60 ten-gram portions of the food item results in a mean of at least 3.6 insect fragments? P(x≥3.6)= Is this result unusual? What might we conclude?
a) The sampling distribution of x is approximately normal because the sample size is large enough. b) mean: 3 str dev: 0.224 Found the str dev using the desmos scientific calculator sqrt3/sqrt60. c) P(x≥3.6)= 0.0037. - The result is unusual because its probability is small. - Since this result if unusual, it is reasonable to conclude that the population mean is higher than 3.
A simple random sample of size n=38 is obtained from a population with μ=61 and σ=14. (a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample mean? Assuming that this condition is true, describe the sampling distribution of x. (b) Assuming the normal model can be used, determine P(x<65.1). (c) Assuming the normal model can be used, determine P(x≥63.3).
a) since the sample size is large enough, the population distribution does not need to be normal Assuming the normal model can be used, describe the sampling distribution x. Choose the correct answer below. Approximately normal, with μx=61 and σx= 14/sqrt (38) b) 0.9645 use statcrunch - Stat>Cal>Normal>Standard>enter mean > enter str. dev you can enter as 14/sqrt(38) enter P and compute. round 4 decimals c) 0.1556 - use statcrunch steps above. round 4 decimals
The reading speed of second grade students in a large city is approximately normal, with a mean of 88 words per minute (wpm) and a standard deviation of 10 wpm. Complete parts (a) through (f). (a) What is the probability a randomly selected student in the city will read more than 92 words per minute? Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (b) What is the probability that a random sample of 11 second grade students from the city results in a mean reading rate of more than 92 words per minute? The probability is _____ Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (c) What is the probability that a random sample of 22 second grade students from the city results in a mean reading rate of more than 92 words per minute? The probability is _______ (d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. (e) A teacher instituted a new reading program at school. After 10 weeks in the program, it was found that the mean reading speed of a random sample of 20 second grade students was 90.8 wpm. What might you conclude based on this result? Select the correct choice below and fill in the answer boxes within your choice. (Type integers or decimals rounded to four decimal places as needed.)
a) the probability is 0.3346 Use the stat normal calculator. Interpretation: If 100 different students were chosen from this population, we would expect 3434 to read more than 92 words per minute. b) The probability is 0.0923. Use the stat normal calculator, but instead you std dev will be 10/sqrt(11). Interpretation: If 100 independent samples of n=11 students were chosen from this population, we would expect 99 sample(s) to have a sample mean reading rate of more than 92 words per minute. c) 0.0303 Use the stat normal calculator, but instead you std dev will be 10/sqrt(22). Interpretation: If 100 independent samples of n=22 students were chosen from this population, we would expect 33 sample(s) to have a sample mean reading rate of more than 92 words per minute. d) Increasing the sample size decreases the probability because σx decreases as n increases. e)