Review Quiz for Exam 2

Which of the following is NOT an equivalent expression for the confidence interval given by 161.7<μ<​189.5?

C. 161.7±27.8

Assume that females have pulse rates that are normally distributed with a mean of μ=75.0 beats per minute and a standard deviation of σ=12.5 beats per minute. Complete parts​ (a) through​ (c) below.

If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 82 beats per minute. The probability is . 7123. ​(Round to four decimal places as​ needed.) Part 2 b. If 4 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 82 beats per minute. The probability is . 8686 ​(Round to four decimal places as​ needed.) Part 3 c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30? Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. Treating 60 heroin addicts with methadone and asking them how they feel after treatment.

No, because some trials have outcomes classified into more than two categories.

Assume that guesses are made for 2 questions on a medical admissions test such that the probability of success​ (correct) is given by p=0.60​, where there are n=2 trials. Find the probability that the number x of correct answers is exactly 1.

P(1) = 0.480 To use a binomial probabilities​ table, first isolate the row with the desired value of n and the corresponding value of x using the labels on the right and left sides.​ Next, find the column with the desired value of p using the labels across the top and bottom. The number at the intersection of the row and column is the probability of the desired outcome.

Assume that we want to construct a confidence interval. Do one of the​ following, as​ appropriate: (a) find the critical value tα/2​, ​(b) find the critical value zα/2​, or​ (c) state that neither the normal distribution nor the t distribution applies. Here are summary statistics for randomly selected weights of newborn​ girls: n=298​, x=31.2 ​hg, s=7.8 hg. The confidence level is 95​%.

Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. A. tα/2=1.97

What requirements are necessary for a normal probability distribution to be a standard normal probability​ distribution?

The mean and standard deviation have the values of μ=0 and standdev = 1

Fill in the blank. ​_____________ is the distribution of all values of the statistic when all possible samples of the same size n are taken from the same population.

The sampling distribution of a statistic

Which of the following is not a requirement of the binomial probability​ distribution?

The trials must be dependent.

Assume a significance level of α=0.05 and use the given information to complete parts​ (a) and​ (b) below. Original​ claim: The mean pulse rate​ (in beats per​ minute) of a certain group of adult males is 73 bpm. The hypothesis test results in a​ P-value of 0.0074.

. State a conclusion about the null hypothesis.​ (Reject H0 or fail to reject H0​.) Choose the correct answer below. Reject H0 because the​P-value is less than or equal to α. b. Without using technical​ terms, state a final conclusion that addresses the original claim. Which of the following is the correct​ conclusion? There is sufficient evidence to warrant rejection of the claim that the mean pulse rate​ (in beats per​ minute) of the group of adult males is 73 bpm.

Find the indicated critical value. z sub 0.06

= 1.55 The expression zα denotes the z score with an area of α to its right. Use technology or a standard normal distribution table to find the z score.

Find the indicated area under the curve of the standard normal​ distribution; then convert it to a percentage and fill in the blank. About​ ______% of the area is between z=−1 and z=1 ​(or within 1 standard deviation of the​ mean).

About 68.2768.27​% of the area is between z=−1 and z=1 ​(or within 1 standard deviation of the​ mean).

Which of the following would be a correct interpretation of a​ 99% confidence interval such as 4.1<μ<​5.6?

Choose the correct answer below. A. We are​ 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of μ.

Refer to the accompanying data set and construct a 90​% confidence interval estimate of the mean pulse rate of adult​ females; then do the same for adult males. Compare the results. LOADING... Click the icon to view the pulse rates for adult females and adult males. Males 84 73 47 58 54 60 53 77 53 61 75 60 64 75 81 67 64 93 42 84 70 61 70 69 57 64 57 81 71 63 61 94 57 63 55 60 69 71 87 56 Females 82 95 56 69 56 85 81 84 90 53 36 65 86 76 78 64 68 80 62 65 83 84 72 77 85 91 90 90 92 92 67 91 81 81 75 57 102 72 74 78

Construct a 90​% confidence interval of the mean pulse rate for adult females. 7373 bpm<μ<80.380.3 bpm ​(Round to one decimal place as​ needed.) Part 2 Construct a 90​% confidence interval of the mean pulse rate for adult males. 63.363.3 bpm<μ<69.769.7 bpm ​(Round to one decimal place as​ needed.) Part 3 Compare the results. The confidence intervals do not​ overlap, so it appears that adult females have a higher mean pulse rate than adult males.

A ___ random variable has infinitely many values associated with measurements.

Continuous

Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder. Determine whether a probability distribution is given. If a probability distribution is​ given, find its mean and standard deviation. If a probability distribution is not​ given, identify the requirements that are not satisfied. x= 0 P(x) = 0.035 x= 1 P(x) = 0.149 x= 2 P(x) = 0.316 x= 3 P(x) = 0.316 x= 4 P(x) = 0.149 x= 5 P(x) = 0.035

Does the table show a probability​ distribution? Select all that apply. A. ​Yes, the table shows a probability distribution. Find the mean of the random variable x. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. A. μ=2.5 ​child(ren) ​ Find the standard deviation of the random variable x. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. A. σ=1.11.1 ​child(ren)

Determine whether the given procedure results in a binomial distribution​ (or a distribution that can be treated as​ binomial). If it is not​ binomial, explain why. A certain method of gender selection was designed to increase the likelihood that a baby will be a boy. When 323 couples use the method and give birth to 323 ​babies, whether or not the babies are more than 22 inches long is recorded.

Does this procedure result in a binomial distribution​ (or a distribution that can be treated as​ binomial)? Select all that apply. Yes, because the procedure satisfies all of the requirements for a binomial distribution.

Assume that when an adult is randomly​ selected, the probability that they do not require vision correction is 21​%. If 10 adults are randomly​ selected, find the probability that exactly 3 of them do not require a vision correction.

If 10 adults are randomly​ selected, the probability that exactly 3 of them do not require a vision correction is . 213.213.

Weights of adult human brains are normally distributed. Samples of weights of adult human​ brains, each of size n=​15, are randomly collected and the sample means are found. Is it correct to conclude that the sample means cannot be treated as being from a normal distribution because the sample size is too​ small? Explain.

It is not correct. The sample means can be treated as being from a normal distribution because the sample weights come from a population that is normally distributed.

a. Exact weights of the next 500 puppies born in a region b. Number of pets in households c. Shoe sizes such as 8 or 812 of human males d. Responses to the survey question "Are you married?" e. Exact arm span of humans

Since the outcomes are not countable, this is a continuous random variable. Part 2 b. Since the outcomes are countable, this is a discrete random variable. Part 3 c. Since the outcomes are countable, this is a discrete random variable. Part 4 d. Since the outcomes are not numerical, this is not a random variable. Part 5 e. Since the outcomes are not countable, this is a continuous random variable.

Find the area of the shaded region. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. 85 shaded to the left

The area of the shaded region is . 1587 ​(Round to four decimal places as​ needed.)

Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1. z=1.11 shaded to the right

The area of the shaded region is . 8665

Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1. z= -0.88 z= 1.27

The area of the shaded region is 0.7085 The area corresponding to the region between two specific z scores can be found by finding the difference between the area to the left of one z score and the area to the left of the other z score. Use a standard normal table or technology to find these areas.

For bone density scores that are normally distributed with a mean of 0 and a standard deviation of​ 1, find the percentage of scores that are a. significantly high​ (or at least 2 standard deviations above the​ mean). b. significantly low​ (or at least 2 standard deviations below the​ mean). c. not significant​ (or less than 2 standard deviations away from the​ mean).

The percentage of bone density scores that are significantly high is 2.282.28​%. ​(Round to two decimal places as​ needed.) Part 2 b. The percentage of bone density scores that are significantly low is 2.282.28​%. ​(Round to two decimal places as​ needed.) Part 3 c. The percentage of bone density scores that are not significant is 95.4495.44​%. ​(Round to two decimal places as​ needed.)

Assume that 68​% of offspring peas have green pods. Suppose we want to find the probability that when seven offspring peas are randomly​ selected, exactly two of them are green. What is wrong with using the multiplication rule to find the probability of getting two peas with green pods followed by five peas with yellow​ pods: (0.68)(0.68)(0.32)(0.32)(0.32)(0.32)(0.32)=0.00155​?

The probability obtained in this way is too​ low, since it only accounts for the permutation of getting two green followed by five yellow. There are many other permutations through which totals of two green and five yellow can be obtained.

The table to the right lists probabilities for the corresponding numbers of girls in three births. What is the random​ variable, what are its possible​ values, and are its values​ numerical? x= 0 P(x)= 0.125 x= 1 P(x)=0.375 x= 2 P(x)= 0.375 x= 3 P(x)= 0.125

The random variable is​ x, which is the number of girls in three births. The possible values of x are​ 0, 1,​ 2, and 3. The values of the random value x are numerical.

is the distribution of sample​ proportions, with all samples having the same sample size n taken from the same population.

The sampling distribution of the proportion

Which of the following is NOT true about​ P-values in hypothesis​ testing? Question content area bottom Part 1 Choose the correct answer below.

The​ P-value separates the critical region from the values that do not lead to rejection of the null hypothesis.

Refer to the technology output given to the right that results from measured hemoglobin levels​ (g/dL) in 100 randomly selected adult females. The results to the right are based on a 95​% confidence level. Write a statement that correctly interprets the confidence level. TInterval (12.913,13.379) x=13.146 Sx=1.175 n=100

We have 95​% confidence that the limits of 12.913 g/dL and 13.379 ​g/dL contain the true value of the mean hemoglobin level of the population of all adult females.

Twelve different video games showing violence were observed. The duration times of violence were​ recorded, with the times​ (seconds) listed below. What requirements must be satisfied to test the claim that the sample is from a population with a mean greater than 75 ​sec? Are the requirements all​ satisfied? 78 14 586 48 0 57 199 41 186 0 2 57

What requirements must be​ satisfied? Select all that apply. B. Either the population is normally​ distributed, or n>​30, or both. Your answer is correct. C. The sample observations must be a simple random sample. Are the requirements all​ satisfied?No. The sample size is not greater than​ 30, the sample does not appear to be from a normally distributed​ population, and there is not enough information given to determine whether the sample is a simple random sample.

Diastolic blood pressure is a measure of the pressure when arteries rest between heartbeats. Suppose diastolic blood pressure levels in women are normally distributed with a mean of 69.7 mm Hg and a standard deviation of 11 mm Hg. Complete parts​ (a) and​ (b) below.

a. A diastolic blood pressure level above 90 mm Hg is considered to be hypertension. What percentage of women have​ hypertension? 3.253.25​% ​(Round to two decimal places as​ needed.) Notice that an individual value from a normally distributed population has been chosen.​ Therefore, use the population distribution with mean μ and standard deviation σ to determine the probability of values greater than x with technology or a standard normal distribution table. Note that some technologies can find the probability directly without determining the​ z-score. z= x-standev/mean Part 2 b. If we randomly collect samples of women with 25 in each​ sample, what percentage of those samples have a mean above 90 mm​ Hg? 0.000.00​% ​(Round to two decimal places as​ needed.) In this​ case, the desired probability is for the mean of a sample.​ Therefore, use the central limit theorem. The central limit theorem applies when a population has a normal distribution or the sample size n is greater than 30. If all possible simple random samples of size n are selected from a population with mean μ and standard deviation σ​, the mean of all sample means is denoted by μx and the standard deviation of all sample means is denoted by σx. Then the probability can be found with technology or a standard normal distribution table. Note that some technologies can find the probability directly without determining the​ z-score. μx=μ​, σx=σ/squarer root n z=x−μx/σx

Refer to the technology output given to the right that results from measured hemoglobin levels​ (g/dL) in 100 randomly selected adult females. The confidence level of 99​% was used.

a. Express the confidence interval in the format that uses the​ "less than" symbol. Assume that the original listed data use two decimal​ places, and round the confidence interval limits accordingly. 12.70612.706 ​g/dL<muμ<13.46613.466 ​g/dL ​(Round to three decimal places as​ needed.) Part 2 b. Identify the best point estimate of μ and the margin of error. The best point estimate of μ is 13.08613.086. ​(Round to three decimal places as​ needed.) Part 3 The margin of error is . 380.380. ​(Round to three decimal places as​ needed.) Part 4 c. In constructing the confidence interval estimate of μ​, why is it not necessary to confirm that the sample data appear to be from a population with a normal​ distribution? Choose the correct answer below. A. Since the sample size is greater than​ 30, the sample data do not need to be from a population with a normal distribution.

Claim: The mean pulse rate​ (in beats per​ minute) of adult males is equal to 68.7 bpm. For a random sample of 153 adult​ males, the mean pulse rate is 67.9 bpm and the standard deviation is 11.2 bpm. Complete parts​ (a) and​ (b) below.

a. Express the original claim in symbolic form. muμ equals= 68.768.7 bpm ​(Type an integer or a decimal. Do not​ round.) Part 2 b. Identify the null and alternative hypotheses. H0​: muμ equals= 68.768.7 bpm H1​: muμ not equals≠ 68.768.7 bpm ​(Type integers or decimals. Do not​ round.)

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 307 days or longer. b. If the length of pregnancy is in the lowest 3​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

a. The probability that a pregnancy will last 307 days or longer is . 0032.0032. ​(Round to four decimal places as​ needed.) Part 2 b. Babies who are born on or before 238238 days are considered premature. ​(Round to the nearest integer as​ needed.)

Assume that a simple random sample has been selected and test the given claim. Listed below are the lead concentrations in μg/g measured in different traditional medicines. Use a 0.01 significance level to test the claim that the mean lead concentration for all such medicines is less than 15 μg/g. 2.5 6.0 6.0 6.0 20.0 7.0 11.5 20.5 11.0 17.0

identify the null and alternative hypotheses for this test. H0​:μ=15 μg/g H1​: μ<15 μg/g identify the value of the test statistic. = -2.10 The test statistic for conducting a hypothesis test of a claim about a population mean when the population standard deviation is not known can be found using technology or the formula​ below, where x is the sample​ mean, s is the sample standard​ deviation, n is the sample​ size, and μx is the population mean​ (taken from the claim and used in H0​). t= x-μsubx/s/squarroot n Identify the​ P-value. = 0.033 In a hypothesis​ test, the​ P-value is the probability of getting a value of the test statistic that is at least as extreme as the test statistic obtained from the sample​ data, assuming that the null hypothesis is true. Use technology to determine the​ P-value with degrees of freedom given by df=n−​1, where n is the sample size. State the conclusion about the null​ hypothesis, as well as the final conclusion that addresses the original claim. D. Fail to reject H0. There is not sufficient evidence to support the claim that the mean lead concentration for all such medicines is less than 15 μg/g.

The​ _________ hypothesis is a statement that the value of a population parameter is equal to some claimed value.

null

In the binomial probability​ formula, the variable x represents the​ _______.

number of successes

The​ _______ is the best point estimate of the population mean.

sample mean

Use technology to find the​ P-value for the hypothesis test described below. The claim is that for the population of adult​ males, the mean platelet count is μ>212. The sample size is n=57 and the test statistic is t=1.365.

​P-value=0.089 ​(Round to three decimal places as​needed.) First determine the type of test. For a​ left-tailed test, the​ P-value is the area to the left of the test statistic. For a​ right-tailed test, the​ P-value is the area to the right of the test statistic. The​ P-value for a​ two-tailed test is twice the area in the tail beyond the test statistic. Use technology or a t Distribution table to determine the​ P-value for the test statistic t with degrees of freedom given by df=n−​1, where n is the sample size.

Use technology to find the​ P-value for the hypothesis test described below. The claim is that for 12 AM body​ temperatures, the mean is μ>98.6°F. The sample size is n=4 and the test statistic is t=1.029.

​P-value=0.190 ​(Round to three decimal places as​needed.) First determine the type of test. For a​ left-tailed test, the​ P-value is the area to the left of the test statistic. For a​ right-tailed test, the​ P-value is the area to the right of the test statistic. The​ P-value for a​ two-tailed test is twice the area in the tail beyond the test statistic. Use technology to determine the​ P-value for the test statistic t with degrees of freedom given by df=n−​1, where n is the sample size.