STT homework/quizzes/test questions

In an exit​ poll, for which the sample size is 3037​, the sampling distribution of the sample proportion voting for the incumbent has mean 0.497 and standard deviation 0.009. Find an interval of values within which the sample proportion will fall with​ 99.7% certainty.

(0.470, 0.524)

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion. When 293 college students are randomly selected and​ surveyed, it is found that 114 own a car. Construct a​ 99% confidence interval for the percentage of all college students who own a car.

(31.6%, 46.2%)

A​ speed-dating agency claims that 60​% of the participants in its​ speed-dating evenings leave with at least one second date arranged. Suppose that 180 people participate in a​ speed-dating evening. Let X be the number who leave with at least one second date arranged. Assume that this scenario satisfies the conditions for the binomial distribution and use the normal distribution approximation to give an interval in which you would expect X almost certainly to​ fall, if it is true that p=0.6

(88, 128)

Find the margin of error for a​ 95% confidence interval for estimating the population mean when the sample standard deviation equals 90​, with a sample size of​ (i) 441 and​ (ii) 1600. What is the effect of the sample​ size?

(i) Find the margin of error for a​ 95% confidence interval for estimating the population mean when the sample standard deviation equals 90 with a sample size of 441. 8.40 (ii) Find the margin of error for a​ 95% confidence interval for estimating the population mean when the sample standard deviation equals 90 with a sample size of 1600. 4.41 What is the effect of the sample size on the margin of​ error? Choose the correct answer below. The larger sample size decreases the margin of error.

Suppose a coin is tossed four times. Let X denote the total number of tails obtained in the four tosses. What are the possible values of the random variable​ X?

0, 1,​ 2, 3, 4

Out of 400​ trials, 60 turned out positive. Find the standard error for the sample proportion.

0.0179

For a normal​ distribution, find the probability that an observation is at least 2 standard deviations below the mean. Sketch a curve and show the tail probability.

0.0228

For a normal​ distribution, find the probability that an observation falls within 0.92 standard deviations of the mean. Sketch a graph and shade the appropriate region.

0.6424

Use a table of areas to find the specified area under the standard normal curve. The shaded area shown (curve with -1.88< >1.88 shaded)

0.9398

The body temperatures of adults have a mean of 98.6° F and a standard deviation of 0.60° F. If 36 adults are randomly​ selected, find the probability that their mean body temperature is greater than 98.4° F.

0.9772

On a​ 10-question multiple choice​ test, each question has four possible​ answers, one of which is correct. For students who guess at all​ answers, find the mean for the random variable​ X, the number of correct answers.

2.5

According to a college​ survey, 22% of all students work full time. Find the mean for the random variable​ X, the number of students who work full time in samples of size 16.

3.52

Use the empirical rule to solve the problem. The lifetimes of light bulbs of a particular type are normally distributed with a mean of 360 hours and a standard deviation of 8 hours. What percentage of the bulbs have lifetimes that lie within 2 standard deviations of the​ mean?

95%

Use the empirical rule to solve the problem. The systolic blood pressure of​ 18-year-old women is normally distributed with a mean of 120 mmHg and a standard deviation of 12 mmHg. What percentage of​ 18-year-old women have a systolic blood pressure that lies within 3 standard deviations of the​ mean?

99.7%

Does a​ P-value of 0.002 give strong evidence or not especially strong evidence against the null​ hypothesis?

The​ P-value gives strong evidence against the null hypothesis because it is small.

A researcher would like to estimate the population proportion of adults living in a certain town who have at least a high school education. No information is available about its value. How large a sample size is needed to estimate it to within 0.11 with 95​% confidence?

n=80

How large a sample size is needed to estimate the mean annual income of people in a certain​ county, correct to within $1100 with probability​ 0.99? No information is available about the standard deviation of their income. It is estimated that nearly all of the incomes fall between​ $0 and $240,000 and that this distribution is approximately​ bell-shaped.

n=8774

You must pick one of two​ wagers, for an outcome based on flipping a fair coin. 1. You win ​$320 if it comes up heads and lose ​$80 if it comes up tails. 2. You win ​$480 if it comes up heads and lose ​$160 if it comes up tails. Find the expected outcome for each wager. Which wager is better in this​ sense?

The expected outcome for the first wager is ​$120 The expected outcome for the second wager is ​$160 Which wager is better in this​ sense? The second wager

For one​ test's distribution ​(μ=374, σ=94) and another​ test's distribution ​(μ=23, σ=4) which score is relatively​ higher, a score of 541 on the first test or a score of 22 on the second​ test? Explain.

The first score is relatively higher than the second score because the​z-score of the score on the first test is greater than the​ z-score of the score on the second test.

When two balanced dice are​ rolled, 36 equally likely outcomes are possible as shown below. ​ (1, 1)​ (1, 2)​ (1, 3)​ (1, 4)​ (1, 5)​ (1, 6) ​ (2, 1)​ (2, 2)​ (2, 3)​ (2, 4)​ (2, 5)​ (2, 6) ​ (3, 1)​ (3, 2)​ (3, 3)​ (3, 4)​ (3, 5)​ (3, 6) ​ (4, 1)​ (4, 2)​ (4, 3)​ (4, 4)​ (4, 5)​ (4, 6) ​ (5, 1)​ (5, 2)​ (5, 3)​ (5, 4)​ (5, 5)​ (5, 6) ​ (6, 1)​ (6, 2)​ (6, 3)​ (6, 4)​ (6, 5)​ (6, 6) Let X denote the absolute value of the difference of the two numbers. Find the probability distribution of X. Give the probabilities as decimals rounded to three decimal places.

x | P(X=x) ----------- 0 | 0.167 1 | 0.278 2 | 0.222 3 | 0.167 4 | 0.111 5 | 0.056

A multiple choice test consists of four questions. Each question has five possible​ answers, of which only one is correct. A student guesses on every question. Find the probability distribution of​ X, the number of questions she answers correctly.

x | P(X=x) ----------- 0 | 0.4096 1 | 0.4096 2 | 0.1536 3 | 0.0256 4 | 0.0016

For the normal distribution shown​ below, find the probability that an observation falls in the shaded region. (curve with >1.57)

0.0582

The probability of winning a certain lottery is 1/51,949. For people who play 560​ times, find the standard deviation for the random variable​ X, the number of wins. Round your answer to four decimal places.

0.1038

Police estimate that​ 25% of drivers drive without their seat belts. If they stop 6 drivers at​ random, find the probability that all of them are wearing their seat belts. Round your answer to four decimal places.

0.1780

The gestation time for humans has a mean of 266 days and a standard deviation of 25 days. If 100 women are randomly​ selected, find the probability that they have a mean pregnancy between 266 days and 268 days.

0.2881

Find the standard deviation of the binomial random variable. According to a college​ survey, 22% of all students work full time. Find the standard deviation for the random variable​ X, the number of students who work full time in samples of size 16. Round your answer to two decimal places.

1.66

Use the empirical rule to solve the problem. At one​ college, GPAs are normally distributed with a mean of 2.6 and a standard deviation of 0.4. What percentage of students at the college have a GPA between 2.2 and​ 3?

68%

The percent of the​ voting-age population that is registered to vote for the 50 states and the District of Columbia has a mean of 65 with a standard deviation of 7.1. Assuming that the distribution is​ normal, what percentage of states had between 50 and 70 percent of​ it's voting-age population who are registered to​ vote? Round your answer to two decimal places.

74.20%

The average financial aid package for students admitted to a particular college five years ago was $46,750. A student organization plans to investigate whether this average has decreased for students admitted this year. Define the population parameter of interest and state the null and alternative hypotheses for this investigation.

Define the population parameter of interest and state the null and alternative hypotheses for this investigation. H0​:μ=46750 Ha​:μ<46750

A person who claims to be psychic says that the​ probability, p, that he can correctly predict the outcome of the birthday of a person in another room is greater than 1/365​, the value that applies with random guessing. If we want to test this​ claim, we could use the data from an experiment in which he predicts the outcomes for n trials. State hypotheses for a significance​ test, letting the alternative hypothesis reflect the​ psychic's claim.

Which of the following is the hypothesis test to be​ conducted? H0​:p=1/365 Ha​:p>1/365

In an exit​ poll, suppose that the mean of the sampling distribution of the proportion of the 3140 people in the sample who voted for recall was 0.55 and the standard deviation was 0.0089. Answer the following questions.

a. Based on the approximate normality of the sampling​ distribution, give an interval of values within which the sample proportion will almost certainly fall. Choose the correct interval below. 0.523 to 0.577 b. Based on the result in ​(a​), if you take an exit poll and observe a sample proportion of 0.53​, would this be a rather unusual​ result? ​ Why? No​, because the observed sample proportion lies within the interval.

Consider the test of H0​:the new drug is safe against Ha​:the new drug is not safe. Complete parts a through d below.

a. Explain in context the conclusion of the test if H0 is rejected. Choose the correct answer below. It is likely that the drug is not safe b.Describe, in​ context, a Type I error. A Type I error results in finding that the drug is not safe when it actually is safe. c.Explain in context the conclusion of the test if you fail to reject H0. It is plausible to believe that the drug may be safe d.Describe, in​ context, a Type II error. A Type II error results in finding that the drug is safe when it is actually not safe.

A social scientist uses a survey to study how much time per day people spend talking on the phone. The variable given on the survey website measures this using the values​ 0, 1,​ 2, ..., 24. Complete parts a through c below

a. Explain​ how, in​ theory, talking is a continuous random variable. Choose the correct answer below. Someone could talk for exactly 1 hour or for 1,8643 hours. b.An article about the study shows two​ histograms, both skewed to the​ right, to summarize talking for females and males. Since talking is in theory​ continuous, why were histograms used instead of​ curves? Each person reported their talking by rounding to the nearest whole number c.If the article instead showed two​ curves, explain what they would represent. Choose the correct answer below. The curves would approximate the two histograms to represent the approximate distribution if the study measured talking in a continuous manner.​ Then, the area below the curve over any interval represents the proportion of people whose talking fell in that interval.

The figure illustrates two sampling distributions for sample proportions when the population proportion p=0.45. Complete parts a through c. (n=2000 and n=200)

a. Find the standard deviation for the sampling distribution of the sample proportion with (i) n=200, (ii) n=2000. ​(i) standard deviation=0.0352 ​(ii) standard deviation=0.0111 b. Explain why the sample proportion would be very likely to fall​ (i) between 0.34 and 0.56 when n=200​, and (ii) between 0.42 and 0.48 when n=2000 The sample proportion is very likely to fall within three standard deviations of the mean. c.Explain how the results in (b​) indicate that the sample proportion is closer to the population proportion when the sample size is larger. When n is​ larger, the standard deviation is​ smaller, so the interval is smaller.

A study has a random sample of 48 subjects. The test statistic for testing H0: μ=150 is t=1.86. Find the approximate​ P-value for the alternative a.Ha: μ≠150​, b.Ha: μ>150​, and c.Ha: μ<150.

a. Find the​ P-value for the alternative Ha: μ≠150. ​P-value=0.069 b. Find the​ P-value for the alternative Ha: μ>150. ​P-value=0.035 c. Find the​ P-value for the alternative Ha: μ<150. ​P-value=0.965

Jan's All You Can Eat Restaurant charges ​$8.70 per customer to eat at the restaurant. Restaurant management finds that its expense per​ customer, based on how much the customer eats and the expense of​ labor, has a distribution that is skewed to the right with a mean of ​$8.10 and a standard deviation of ​$4. Complete parts​ (a) and​ (b).

a. If the 100 customers on a particular day have the characteristics of a random sample from their customer​ base, find the mean and standard deviation of the sampling distribution of the​ restaurant's sample mean expense per customer. The mean is 8.10 The standard deviation is 0.4 b. Find the probability that the restaurant makes a profit that​ day, with the sample mean expense being less than ​$8.70. ​(Hint: Apply the central limit theorem to the sampling distribution in​ (a).) 0.933 ​

Let p=0.3 be the proportion of smart phone owners who have a given app. For a particular smart phone​ owner, let x=1 if they have the app and x=0 otherwise. For a random sample of 80 owners, complete parts a through e below.

a. State the population distribution​ (that is, the probability distribution of X for each​ observation). ​P(X=​1)=0.3 and ​P(X=​0)=0.7 b. state the data distribution if 56 of the 80 owners sampled have the app.​ (That is, give the sample proportions of observed 0s and 1s in the​ sample.) The proportion for x=1 is 0.7 and the proportion for x=0 is 0.3 c.Find the mean of the sampling distribution of the sample proportion who have the app among the 80 people. mean=0.3 d.Find the standard deviation of the sampling distribution of the sample proportion who have the app among the 80 people. standard deviation = 0.051 e. Explain what the standard deviation in part d describes. Which sentence below best describes the standard​ deviation? The variability in the sample proportion across all possible samples of size 80.

Consider a sampling distribution with p=0.13 and samples of size n each. Using the appropriate​ formulas, find the mean and the standard deviation of the sampling distribution of the sample proportion. a. For a random sample of size n=5000. b. For a random sample of size n=1000. c. For a random sample of size n=500.

a. The mean is 0.13 The standard deviation is 0.0048 b. The mean is 0.13 The standard deviation is 0.0106 c. The mean is 0.13 The standard deviation is 0.0150 Summarize the effect of the sample size on the size of the standard deviation. As the sample size gets​ larger, the standard deviation gets smaller.

According to a recent​ survey, the population distribution of number of years of education for​ self-employed individuals in a certain region has a mean of 13.4 and a standard deviation of 3.0. a. Identify the random variable X whose distribution is described here. b. Find the mean and the standard deviation of the sampling distribution of x for a random sample of size 100. Interpret them. c. Repeat​ (b) for n=400. Describe the effect of increasing n.

a. The number of years of education b. The mean of the sampling distribution of size 100 is 13.4 The expected value for the mean of a sample of size 100 The standard deviation is 0.3 The variablility of the mean for samples of size 100 c.The mean of the sampling distribution of size 400 is 13.4 The standard deviation is 0.15 The mean of the sampling distribution stays the same as n increases. The standard deviation of the sampling distribution decreases as n increases.

Use the graph to the right to answers parts a and b. (curve with >0.2 shaded)

a. The​ z-score is 0.84 b. The​ z-score that corresponds to a​ right-tail probability of 0.05 is 1.64 ​The​ z-score that corresponds to a​ right-tail probability of 0.011 is 2.29

The table shows the probability distribution of the number of bases for a randomly selected time at bat for a random player on a certain baseball team. Complete parts a through c below. 0=0.7447, 1=0.1703, 2=0.0525, 3=0.0026, 4=0.0299

a. Verify that the probabilities give a legitimate probability distribution. Choose the correct answer below. The sum of the probabilities is 1 and each probability falls between 0 and 1. b. Find the mean of this probability distribution 0.4027 c. Interpret the​ mean, explaining why it does not have to be a whole​ number, even though each possible value for the number of bases is a whole number. Choose the correct answer below. Over the course of many​ at-bats, the average number of bases reached by the batter will approach the mean. Since the mean is not a measure of an individual​ at-bat, it does not have to be a whole number.

Assume each newborn baby has a probability of approximately 0.48 of being female and 0.52 of being male. For a family with three ​children, let X=number of children who are girls. a. Identify the three conditions that must be satisfied for X to have the binomial distribution. b. Identify n and p for the binomial distribution. c. Find the probability that the family has two girls and one boy.

a. Which of the below are the three conditions for a binomial​ distribution? I. The n trials are independent. II. Each trial has at least two possible outcomes. III. The n trials are dependent. IV. Each trial has the same probability of a success. V. There are two trials. VI. Each trial has two possible outcomes. I, IV, and VI b. n=3 p=0.48 c. The probability that the family has two girls and one boy is 0.3594

In a certain​ country, the mean birth weight for boys is 3.34 kg, with a standard deviation of 0.54 kg. Assuming that the distribution of birth weight is approximately​ normal, complete parts a through e below.

a.A baby is considered of low birth weight if it weighs less than 2.5 kg. What proportion of baby boys in this country are born with low birth​ weight? The proportion of baby boys that are born with a low birth weight is 0.060 b.What is the​ z-score for a baby boy that weighs 1.5 kg​ (defined as extremely low birth​ weight)? The z-score is -3.41 c. Typically, birth weight is between 2.5 kg and 4.0 kg. Find the probability a baby is born with typical birth weight. The probability that a baby is born with typical birth weight is 0.829 d. Matteo weighs 3.7 kg at birth. He falls at what​ percentile? The percentile that Matteo falls at is 74.8 e.Max's parents are told that their newborn son falls at the 96th percentile. How much does Max​ weigh? Max weighs 4.29 kg

Let X represent the number of homes a real estate agent sells during a given month. Based on previous sales​ records, she estimates that P(0)=0.58​, P(1)=0.25​, P(2)=0.12​, P(3)=0.03​, P(4)=0.02​, with negligible probability for higher values of x. a. Explain why it does not make sense to compute the mean of this probability distribution as (0+1+2+3+4)/5=2.0. b. Find the correct mean and interpret.

a.Choose the reason why the mean is not the simple average of the possible values. The probabilities of each possible value are not all the same. b.Select the correct choice below and fill in the answer box to complete your choice. The​ long-term average number of homes the real estate agent expects to sell each month is 0.66

Therapeutic touch​ (TT) practitioners claim to perceive an energy field above a​ patient's skin. To test this claim a TT practitioner was​ blind-folded and the researcher placed her hand over either the right or left hand of the practitioner. Let p denote the probability of a correct prediction by a TT practitioner. An experiment used 12 TT practitioners who had to predict the correct hand in each of 11 trials. Complete parts a through e.

a.Define the parameter of interest and set up hypotheses to test that the probability of a correct guess is 0.50 against the TT​ practitioners' claim that it exceeds 0.50. Define a parameter of interest p=probability of a correct guess Set up the hypotheses for the hypothesis test. H0​:p=0.5 Ha​:p>0.5 b.The 132 trials had 74 correct guesses. Find and interpret the test statistic. The test statistic is z=1.39, which represents the number of standard errors that the estimate falls from the null hypothesis value. c. P-value=0.082 Indicate your​ decision, in the context of this​ experiment, using a 0.05 significance level. Since the​ P-value is greater than the significance level of​ 0.05, do not reject the null hypothesis. There is not evidence TT is more effective than just guessing. d.​Yes, the sample size was large enough to make the inference in part c. Indicate what assumptions you would need to make for your inferences to apply to all TT practitioners. We would need to assume randomly selected subjects. e.Find the​ P-value for the​ two-sided alternative that the probability of a correct guess is different from 0.50. P-value =0.164

In a national basketball​ association, the top​ free-throw shooters usually have probability of about 0.90 of making any given free throw. Complete parts a through c.

a.During a​ game, one such player shot 9 free throws. Let X=number of free throws made. What must you assume in order for X to have a binomial​ distribution? It is assumed that the data are​ binary, that there is the same probability of success for each trial​ (free throw), and that the trials are independent. b. n=9 p=0.9 c. The probability that the player made all 9 free throws is 0.387 The probability that he made 8 free throws is 0.387 The probability that he made more than 6 free throws is 0.947

In response to a survey question about the number of hours daily spent watching​ TV, the responses by the eight subjects who identified themselves as Hindu were 3​, 2​, 1​, 3​, 2​, 0​, 4​, 1 . Complete parts​ (a) and​ (b).

a.Find a point estimate of the population mean for Hindus. 2 The margin of error at the​ 95% confidence level for this point estimate is 0.91. Explain what this represents. The margin of error indicates we can be 95% confident that the sample mean falls within 0.91 of the population mean

In​ basketball, the top free throw shooters usually have a probability of about 0.85 of making any given free throw. Over the course of a​ season, one such player shoots 400 free throws. a. Find the mean and standard deviation of the probability distribution of the number of free throws he makes. b. By the normal distribution​ approximation, within what range would the number of free throws made almost certainly​ fall? Why? c. Within what range would the proportion made be expected to​ fall?

a.Find the mean and standard deviation of the probability distribution of the number of free throws he makes. μ=340 σ=7.141 b.By the normal distribution​ approximation, within what range would the number of free throws made almost certainly fall​, if p=0.85​? The number of free throws made would almost certainly fall between 319 and 361. c.Within what range would the proportion of free throws made be expected to​ fall? The proportion of free throws made would almost certainly fall between 0.80 and 0.90

For a normal​ distribution, verify that the probability​ (rounded to two decimal​ places) within a. 1.88 standard deviations of the mean equals 0.94. b. 0.59 standard deviations of the mean equals 0.44. c. Find the probability that falls within 0.35 standard deviations of the mean. d. Sketch these three cases on a single graph.

a.Looking up 1.88 in a standard normal distribution​ table, the cumulative probability is 0.9699. Likewise, the cumulative probability is 0.0301 for −1.88. 0.9699−0.0301=0.9398, which rounds to 0.94. b.By a standard normal distribution​ table, the cumulative probability to the left of 0.59 is 0.7224. The cumulative probability to the left of −0.59 is 0.2776. 0.7224−0.2776=0.4448, which rounds to 0.44. c.The probability that falls within 0.35 standard deviations of the mean is 0.27 d. graph with pink in middle, thin dark blue on both sides, and then light blue on both side

When a survey asked subjects whether they would be willing to accept cuts in their standard of living to protect the​ environment, 385 of 1170 subjects said yes. a. Find the point estimate of the proportion of the population who would answer yes. b.Find the margin of error for a​ 95% confidence interval. c. Construct the​ 95% confidence interval for the population proportion. What do the numbers in this interval​ represent? d. State and check the assumptions needed for the interval in ​(c) to be valid.

a.Find the point estimate of the proportion of the population who would answer yes. p=0.32906 b.Find the margin of error for a​ 95% confidence interval. 0.02692 ​ c.Construct the​ 95% confidence interval for the population proportion. ​(0.30214​, 0.35598) What do the numbers in this interval​ represent? The numbers represent the most believable values for the population proportion. d.State and check the assumptions needed for the interval in ​(c) to be valid. The data must be obtained​ randomly, and the expected numbers of successes and failures must both be at least 15.

In kidney​ transplantations, compatibility between donor and receiver depends on such factors as blood type and antigens. Suppose that for a randomly selected donor from a large national kidney​ registry, there is a 21​% chance that he or she is compatible with a specific receiver. Three donors are randomly selected from this registry. Complete parts a and b below.

a.Find the probability that 0​, 1​, 2​, or all 3 selected donors are compatible by constructing a sample space and finding the probability for each possible outcome of choosing three donors. Let S denote a successful match and F denote a failure. What is the sample space for this​ data? The sample space for this data is SFF, FSF, FFS, SSF, FSS, SFS, SSS, FFF. Next, find the probability of each outcome. P(FFF)=0.493 ​P(FFS)=0.131 ​P(FSF)=0.131 ​P(SFF)=0.131 ​P(FSS)=0.035 ​P(SFS)=0.035 ​P(SSF)=0.035 ​P(SSS)=0.009 Use these probabilities to construct the probability distribution. Let x be the number of successful matches. x P(x) 0 0.493 1 0.393 2 0.105 3 0.009 b.Find the probability that 0​, 1​, 2​, or all 3 selected donors are compatible by using the formula for the binomial distribution. ​P(0)= 3!/0!3! (0.21)^0(0.79)^(3−0)=0.493 ​P(1)= 3!/1!2! (0.21)^1(0.79)^(3−1)=0.393 ​P(2)= 3!/2!1! (0.21)^2(0.79)^(3−2)=0.105 ​P(3)= 3!/3!0! (0.21)^3(0.79)^(3−3)=0.009

For a normal​ distribution, answer the questions below. Answer parts a and b.

a.How would you show that a total probability of 0.28 falls more than z=1.08 standard deviations from the​ mean? Divide the probability 0.28 by two to find the amount in each​ tail, 0.14. Then subtract this from 1.0 to determine the cumulative probability associated with this​ z-score, 0.86. Look up this probability on a standard normal probability table to find the​ z-score of 1.08. b.Find the​ z-score for which the​ two-tail probability that falls more than z standard deviations from the mean in either direction equals​ (a) 0.55​, ​(b) 0.73. Sketch the two cases on a single graph. The​ z-score with a​ two-tail probability of 0.55 is 0.59 The​ z-score with a​ two-tail probability of 0.73 is 0.34 graph with white middle, pink of side of that, and light blue on side of that

A study considered whether daily consumption of garlic could reduce tick bites. The study used a crossover design where half of the subjects used placebo first and garlic second and half the reverse. The authors described garlic being more effective with 31 subjects and placebo being more effective with 29 subjects. Does this suggest a real difference between garlic and​ placebo, or are the results consistent with random​ variation? Complete parts a through d below.

a.Identify the relevant variable and parameter. The relevant variable is whether garlic or placebo is more​ effective, and the parameter is the population​ proportion, p, those for whom garlic is more effective than placebo. b.State hypotheses for a​ large-sample two sided test. H0​:p=0.5 Ha​:p (/=) 0.5 Yes, the sample size was large enough to make the inference. c.Find the test statistic value. z=0.26 d. Find the​ P-value. P-value=0.79 Interpret the​ P-value and state the conclusion in context. Use a significance level of 0.05. The​ P-value is greater than the significance​ level; do not reject the null hypothesis. There is not sufficient evidence that the proportion who think garlic more effective than a placebo is greater than 0.5.

Many students brag that they have more than 100 friends on a social media website. For a class​ project, a group of students asked a random sample of 13 students at their college who used the social media website about their number of friends and got the data available below. Is there strong evidence that the mean number of friends for the student population at the college who use the social media website is larger than 100​? Complete parts a through d below. 110, 20, 185, 40, 140, 205, 20, 185, 240, 195, 30, 265, 205

a.Identify the relevant variable and parameter. What is the relevant​ variable? The number of friends students have on the social media website. What is the​ parameter? The mean number friends for students at the college who use the social media website. b. State the null and alternative hypotheses H0​:μ=100 Ha​:μ>100 c.Find and interpret the test statistic value. t=1.70 What does the test statistic value​ represent? The test statistic value is the number of standard errors from the null hypothesis value to the sample mean. d.Report and interpret the​ P-value and state the conclusion in context. Use a significance level of 0.05. The P-value is 0.057 What does the​ P-value represent? The​ P-value is the probability of observing a sample mean this high or higher if the null hypothesis is true. Since the​ P-value is greater than the level of​ significance, the sample does not provide evidence to reject the null hypothesis. There is insufficient evidence to conclude that the mean number of friends on the social media website is larger than 100.

For​ (a) and​ (b), is the statement a null hypothesis or an alternative​ hypothesis? a. The proportion of adults who favor legalized gambling equals 0.10. b. The proportion of all college students who are regular smokers is less than 0.15​, the value it was ten years ago. c. Introducing notation for a​ parameter, state the hypotheses in​ (a) and​ (b) in terms of the parameter values.

a.Is the statement in part​ (a) the null or alternative​ hypothesis? Null hypothesis b.Is the statement in part​ (b) the null or alternative​ hypothesis? alternative hypothesis c. Choose the correct null and alternative hypotheses for part​ (a). H0​:p=0.10 Ha​:p (/=) 0.10 H0​:p=0.15 Ha​:p<0.15

Let X denote the response of a randomly selected person to the question​ "What is the ideal number of children for a family to​ have?" Suppose the table shows the probability distribution of X for a certain​ region, according to the gender of the person asked the question. Complete parts a and b below. x=0 females=0.00 males=0.02, x=1 females=0.12 males=0.12, x=2 females=0.64 males=0.57, x=3 females=0.11 males=0.11, x=4 females=0.13 males=0.18

a.Show that the means are​ similar, 2.25 for females and 2.31 for males. Begin by showing the mean for females is 2.25. Choose the correct answer below. 0*0.00+1*0.12+2*0.64+3*0.11+4*0.13=2.25 Now show the mean for males is 2.31. Choose the correct answer below. 0*0.02+1*0.12+2*0.57+3*0.11+4*0.18=2.31 b.The standard deviation for females is 0.829 and 0.966 for the males. Explain why a practical implication of the values for the standard deviations is that males hold slightly less consistent views than females about the ideal family size. Choose the correct answer below. The responses for females tend to be closer to the mean than the responses for the males

A study dealing with health care issues plans to take a sample survey of 1500 Americans to estimate the proportion who have health insurance and the mean dollar amount that Americans spent on health care this past year. a. Identify the population parameters that this study will estimate. b. Identify the statistics that can be used to estimate these parameters.

a.The population mean dollar amount spent on health care this past year The population proportion who have health insurance b.The sample proportion The sample mean

For the normal distributions shown to the​ right, find the probability that an observation falls in the shaded region. (curve with >0.91 and curve with -1.39< >1.39)

a.The probability that an observation falls in the shaded area is 0.1814 b.The probability that an observation falls in the shaded area is 0.8355

A new roller coaster at an amusement park requires individuals to be at least​ 4' 8" ​(56 ​inches) tall to ride. It is estimated that the heights of​ 10-year-old boys are normally distributed with μ=53.5 inches and σ=4 inches. a. What proportion of​ 10-year-old boys is tall enough to ride the​ coaster? b. A smaller coaster has a height requirement of 50 inches to ride. What proportion of​ 10-year-old boys is tall enough to ride this​ coaster? c. What proportion of​ 10-year-old boys is tall enough to ride the coaster in part b but not tall enough to ride the coaster in part​ a?

a.The proportion of​ 10-year-old boys tall enough to ride the coaster is 0.2660 b.The proportion of 10​ year-old-boys tall enough to ride the smaller coaster is 0.8092 c.The proportion of​ 10-year-old boys tall enough to ride the coaster in part b but not tall enough to ride the coaster in part a is 0.5432

Researchers are interested in the effect of a certain nutrient on the growth rate of plant seedlings. Using a hydroponics grow procedure that utilized water containing the​ nutrient, they planted six tomato plants and recorded the heights of each plant 14 days after germination. Those​ heights, measured in​ millimeters, were 55.2​, 59.2​, 61.4​, 62.2​, 66.3​, and 68.5 . Complete parts a through d below.

a.Using​ technology, find the​ 95% confidence interval for the population mean μ. ​(57.1, 67.2​) b.Name two things you could do to get a narrower interval than the one in part a Decrease the confidence level or increase the sample size. c.Using​ technology, construct a​ 99% confidence interval. ​(54.2, 70.0) Why is the​ 99% confidence interval wider than the​ 95% interval? The​ t-distribution critical value is larger with a higher confidence level. d.On what assumptions is the interval in part a​ based? The data are obtained by randomization and the population distribution is approximately normal.

A survey asked the question​ "What do you think is the ideal number of children for a family to​ have?" The 581 females who responded had a median of​ 2, mean of 3.45​, and standard deviation of 1.65. Answer parts ​a-d.

a.What is the point estimate of the population​ mean? 3.45 b. Find the standard error of the sample mean. standard error=0.068 c. The​ 95% confidence interval is ​(3.32​, 3.58​). Interpret. We can be​ 95% confident that the mean number of children that females would like to have is between 3.32 and 3.58. d. Is it plausible that the population mean μ=​2? No, because 2 falls outside the confidence interval.

A health study reported​ that, in one​ country, systolic blood pressure readings have a mean of 119 and a standard deviation of 14. A reading above 140 is considered to be high blood pressure. Complete parts a through d below.

a.What is the​ z-score for a blood pressure reading of​ 140? z=1.5 b.If systolic blood pressure in that country has a normal​ distribution, what proportion of the population suffers from high blood​ pressure? The proportion with high blood pressure is 0.0667 c.What proportion of the population has systolic blood pressure in the range from 104 to​ 140? The proportion with systolic blood pressure between 104 and 140 is 0.7913 d. Find the 90th percentile of blood pressure readings. The 90th percentile of blood pressure readings is 137

An experiment tests whether or not dogs can detect cancer by sniffing 7 urine​ vials, one of which was from a cancer patient. When the hypothesis is tested that the dogs randomly​ guess, H0​: p=1/7​, against the hypothesis that the dogs do not randomly​ guess, Ha​: p≠1/7​, a​ P-value of 0.04 is obtained. a. What would the decision be for a significance level of 0.06​? b. If the decision in part a is in​ error, what type of error is​ it? Explain using the context of the experiment.

a.What would the decision be for a significance level of 0.06​? The null hypothesis would be rejected and the conclusion would be that the dogs must be able to detect which vial is from the cancer patient. b.If the decision in part a is in​ error, what type of error is​ it? Explain using the context of the experiment. A type I error. The​ P-value would cause the rejection of the hypothesis that the dogs randomly guess when in reality they do randomly guess.

The body temperatures of adults have a mean of 98.6°F and a standard deviation of 0.60° F. Describe the center and variability of the sampling distribution of the sample mean for a random sample of 75 adults. Round the variability to two decimal places.

center=​98.6, variability=0.07