STA 6.3,6.4,7 Review

Express the confidence interval (0.062,0.130) in the form of p−E<p<p+E.

.62 < p < .130

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

161.7±27.8 The interval 161.7±27.8 would actually produce 133.9<μ<189.5.

Find the critical value zα/2 that corresponds to the given confidence level. 98​%

2.33

C levels

80% = 1.28 90% = 1.64 95% = 1.96 98% = 2.33 99% = 2.58 example find za/2 1-.99=.01 .01/2=.005 1-.005=.995 invNorm(.995 za/2 = 2.57

Listed below are the amounts of net worth​ (in millions of​ dollars) of the ten wealthiest celebrities in a country. Construct a 90​% confidence interval.

Clear lists first 1. Enter data in L1 2. TInterval Data 3. Scroll past "Freq" item to the "C-Level" row and enter the level of confidence 4. Calculate and the output will be displayed that has the desired CI 253 215 180 166 162 158 154 154 149 149 154.2 million < u < 193.8 million

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 14 subjects had a mean wake time of 105.0 min. After​ treatment, the 14 subjects had a mean wake time of 81.7 min and a standard deviation of 20.5 min. Assume that the 14 sample values appear to be from a normally distributed population and construct a 90​% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 105.0 min before the​ treatment? Does the drug appear to be​ effective?

Construct the 90​% confidence interval estimate of the mean wake time for a population with the treatment. TInterval Stats xbar: 81.7 Sx: 20.5 n: 14 c: .9 72 min < u < 91.4 min What does the result suggest about the mean wake time of 105.0 min before the​ treatment? Does the drug appear to be​ effective? The confidence interval does not include the mean wake time of 105.0 min before the​ treatment, so the means before and after the treatment are different. This result suggests that the drug treatment has a significant effect.

6.3 Which of the following is a biased​ estimator? That​ is, which of the following does not target the population​ parameter?

Median is biased. ___________________ Mean Proportion Variance Unbiased

Which of the following groups has terms that can be used interchangeably with the​ others?

Percentage, Probability, and Proportion

A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 98​% confidence interval estimate of the mean amount of mercury in the population.

Please copy and paste the link for the instructions of a similar problem with different data values: https://youtu.be/GAK799Ts_jY Mercury (ppm) 0.52 0.81 0.11 0.97 1.31 0.53 0.80 Enter data in L1 TInterval .268 ppm < u < 1.175 ppm

Which of the following statistics are unbiased estimators of population​ parameters?

Sample​ means, sample​ proportions, and sample variances are unbiased estimators of population parameters.

In a test of the effectiveness of garlic for lowering​ cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes ​(before−​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 5.2 and a standard deviation of 17.2. Construct a 90​% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment.

TInterval Stats xbar: 5.2 Sx: 17.2 n: 45 C-level: .9 Calculate .89 mg/dL < u < 9.51 mg/dL

7.2 Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. The results in the screen display are based on a​ 95% confidence level. Write a statement that correctly interprets the confidence interval.

TInterval ​(13.046,22.15) x=17.598 Sx=16.01712719 n=50 Choose the correct answer below. We have​ 95% confidence that the limits of 13.05 Mbps and 22.15 Mbps contain the true value of the mean of the population of all data speeds at the airports.

Instructor questions Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts​ (a) through​ (b) below.

TInterval ​(13.046,22.15) x=17.598 Sx=16.01712719 n=50 a. Express the confidence interval in the format that uses the​ "less than" symbol. Given that the original listed data use one decimal​ place, round the confidence interval limits accordingly. 13.05 Mbps < u < 22.15 Mbps b. Identify the best point estimate of μ and the margin of error. Best point estimate xbar = 17.6 mbps Margin of error is E = 22.15 - 13.046 / 2 =4.55 mbps

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts​ (a) through​ (b) below.

TInterval ​(13.046,22.15) x=17.598 Sx=16.01712719 n=50 a. What is the number of degrees of freedom that should be used for finding the critical value tα/2​? Degrees of freedom = n-1 b. Find the critical value tα/2 corresponding to a​ 95% confidence level. TInterval Stats xbar: 0 Sx: sqrt n = 7.07 n: 50 C-level = .95 Calculate (-2.009 , 2.0093) ta/2 = 2.01

7.1 A newspaper provided a​ "snapshot" illustrating poll results from 1910 professionals who interview job applicants. The illustration showed that​ 26% of them said the biggest interview turnoff is that the applicant did not make an effort to learn about the job or the company. The margin of error was given as ±3 percentage points. What important feature of the poll was​ omitted?

The confidence level

Which of the following is NOT a property of the Student t​ distribution?

The standard deviation of the Student t distribution is s=1.

Here are summary statistics for randomly selected weights of newborn​ girls: n=160​, xbar=28.8 hg, s=7.3 hg. Construct a confidence interval estimate of the mean. Use a 95​% confidence level. Are these results very different from the confidence interval 26.8 hg<μ<29.6 hg with only 19 sample​ values, x=28.2 hg, and s=2.9 hg?

What is the confidence interval for the population mean μ​? TInterval Stats xbar: 28.8 Sx: 7.3 n: 160 C: .95 27.7 hg < u < 29.9 hg

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=169​, x=32.1 ​hg, s=6.2 hg. The confidence level is 99​%.

When the standard​ deviation, σ​, is​ known, use the normal distribution. When it is not​ known, use the t distribution. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. o is not known so must find ta/2 TInterval xbar: 0 Sx: sqrt n = 13 n: 169 C-level: .99 Calculate ta/2 = high interval level of 2.61

The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be 95​% confident that his estimate is within six percentage points of the true population​ percentage? Complete parts​ (a) through​ (c) below.

a) Assume that nothing is known about the percentage of adults who have heard of the brand. Find n E = .06 C = .95 za/2=1.96 n = [za/2]^2(0.25)/E^2 1.96^2(.25) /.06^2 =267 rounded up ​b) Assume that a recent survey suggests that about 79​% of adults have heard of the brand. n = p^(q^)(za/2)^2/E^2 p^=.79 q^=.21 .79(.21) (1.96^2) /.06^2 =178 rounded up

You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you​ survey? Assume that you want to be 90​% confident that the sample percentage is within 2.5 percentage points of the true population percentage. Complete parts​ (a) and​ (b) below.

a. Assume that nothing is known about the percentage of passengers who prefer aisle seats. Find n n = za/2^2(0.25)/E^2 za/2 = 1.64 E = .025 1.64^2(0.25)/.025^2 1.64^2 (.25) /0.25^2 n=1076 rounded up b. Assume that a prior survey suggests that about 36​% of air passengers prefer an aisle seat. n = p^(q^)(za/2)^2/E^2 p^=.36 q^=.64 .36(.64) (1.64^2) /.025^2 =992 rounded up

In a study of government financial aid for college​ students, it becomes necessary to estimate the percentage of​ full-time college students who earn a​ bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.04 margin of error and use a confidence level of 99​%. Complete parts​ (a) through​ (b) below.

a. Assume that nothing is known about the percentage to be estimated. Find n E = .04 C = .99 find za/2 1-.99=.01 .01/2=.005 1-.005=.995 invNorm(.995 za/2 = 2.57 n = [za/2]^2(0.25)/E^2 2.57^2(.025) /.04^2 n=1033 rounded up b. Assume prior studies have shown that about 60​% of​ full-time students earn​ bachelor's degrees in four years or less. n = p^(q^)(za/2)^2/E^2 p^=.6 q^=.4 .6(.4) (2.57^2) /.04^2 =991 rounded up

Fill in the blank. The number of​ _______ for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values.

degrees of freedom

The ages of a group of 146 randomly selected adult females have a standard deviation of 16.4 years. Assume that the ages of female statistics students have less variation than ages of females in the general​ population, so let σ=16.4 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics​ students? Assume that we want 95​% confidence that the sample mean is within​ one-half year of the population mean.

n = [za/2(o)/E]^2 1.96(16.4)/0.5 then ^2 =4133 rounded up

An IQ test is designed so that the mean is 100 and the standard deviation is 12 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 95​% confidence that the sample mean is within 7 IQ points of the true mean. Assume that σ=12 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.

n = [za/2(o)/E]^2 1.96^2(12)/7 then ^2 = 12 rounded up

The ages of a group of 138 randomly selected adult females have a standard deviation of 18.4 years. Assume that the ages of female statistics students have less variation than ages of females in the general​ population, so let σ=18.4 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics​ students? Assume that we want 99​% confidence that the sample mean is within​ one-half year of the population mean.

n = [za/2(o)/E]^2 2.58(18.4)/.5 then ^2 = 9015 rounded up

Salaries of 42 college graduates who took a statistics course in college have a​ mean, xbar​, of $61,800. Assuming a standard​ deviation, σ​, of ​$15,366​, construct a 99​% confidence interval for estimating the population mean μ.

o is known so zinterval ZInterval Stats o: 15366 xbar: 61800 n: 42 c: .99 55693 < u < 67907

A​ _______ is a single value used to approximate a population parameter.

point estimate

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

sample mean

Instructor questions An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and 201 lb. The new population of pilots has normally distributed weights with a mean of 160 lb and a standard deviation of 34.8 lb.

u = 160 o = 34.8 x = 150,201 a. If a pilot is randomly​ selected, find the probability that his weight is between 150 lb and 201 lb. z = x - u / o 150-160 /34.8 = -.29 normalcdf(-.29,999999,0,1 .6141 1-.6141 =.3859 201-160 /34.8 =1.18 normalcdf(-999999,1.18,0,1 =.881 .881-.3859 =.4951 b. If 32 different pilots are randomly​ selected, find the probability that their mean weight is between 150 lb and 201 lb. n: 32 uxbar: 160 oxbar: o / sqrt n = 6.152 xbar: 150,201 z = xbar - uxbar / oxbar 150-160 /6.152 = -1.63 normalcdf(-1.63,999999,0,1 =.9484 1-.9484 =.0516 201-160 /6.152 =6.66 =.9999 .9483

In a survey of 3281 adults aged 57 through 85​ years, it was found that 83.1​% of them used at least one prescription medication. Complete parts​ (a) through​ (b) below.

xbar = 57,85 a. How many of the 3281 subjects used at least one prescription​ medication? 3281 x .831 =2727 rounded up b. Construct a​ 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication. 1-PropZInt x: 2727 n: 3281 C: .9 82% < p < 84.2%

Use the sample data and confidence level given below to complete parts​ (a) through​ (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll, n=1079 and x=569 who said​ "yes." Use a 90% confidence level.

​a) Find the best point estimate of the population proportion p. 1-PropZInt Calculate p^=.527 ​b) Identify the value of the margin of error E. E = za/2(sqrt p^q^/n p^=.527 q^=.473 n=1079 sqrt(.527x.473/1079) x1.64 =.025 ​c) Construct the confidence interval. 1-PropZInt .502 < p < .552 ​d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below. One has 90​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.