7.2 Stats HW

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 95​% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi? 0.55 0.75 0.09 0.97 1.37 0.55 0.90 a. What is the confidence interval estimate of the population mean μ​? b. Does it appear that there is too much mercury in tuna​ sushi?

.368 ppm < μ < 1.112 ppm Enter data in L1 then use TInterval TInterval List: L1 Freq: 1 C-Level: .95 calculate --------------------------- TInterval (.3681 , 1.1119) x̄ = .74 Sx = .4021193853 n = 7 b. ​Yes, because it is possible that the mean is greater than 1 ppm.​ Also, at least one of the sample values exceeds 1​ ppm, so at least some of the fish have too much mercury.

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 2.9 and a standard deviation of 15.6. Construct a 95​% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL​ cholesterol? a. What is the confidence interval estimate of the population mean μ​? b. What does the confidence interval suggest about the effectiveness of the​ treatment?

​-1.79 mg/dL < μ < 7.59 mg/dL TInterval x̄ = 2.9 Sx = 15.6 n = 45 C-Level: .95 calculate --------------------------- TInterval (-1.787 , 7.5868) x̄ = 2.9 Sx = 15.6 n = 45 C-Level: .95 b. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.

A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is the​ 90% confidence interval for the mean​ score, μ​, of all students taking the​ test?

28.4 < μ < 30.6 ZInterval σ = 5.2 x̄ = 29.5 n = 59 C-Level: .90 calculate (28.386 , 30.614) x̄ = 29.5 n = 59

A confidence interval for a population mean​ __________.

A confidence interval for a population mean​ gives possible values the true population mean will be with a certain level of confidence.

Which of the following is NOT an observation about critical​ values?

A critical value is the area in the​ right-tail region of the standard normal curve.

A simple random sample of 150 adults is obtained and each​ person's red blood cell count​ (in cells per​ microliter) is measured. The sample mean is 4.63. The population standard deviation for red blood cell counts is 0.54. Which of the following is true regarding the distribution of sample​ means?

Even though the distribution of the population data is not​ given, the distribution of sample means will be approximately normal because the sample size is large enough according to the Central Limit Theorem. (The distribution of sample means will be normal or approximately normal if the sample size is large enough​ (typically more than 30 or​ so) and/or the population data follow a normal distribution. The first condition is met.)

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

Percentage, probability, and proportion

What does it mean to say that the confidence interval methods for the mean are robust against departures from​ normality?

The confidence interval methods of this section are robust against departures from​ normality, meaning they work well with distributions that​ aren't normal, provided that departures from normality are not too extreme. Confidence interval methods are robust against departures from normality if either the sample size is greater than​ 30, the population is normally​ distributed, or the departure from normality is not too​ extreme, which can be checked by using a histogram or a dotplot.

Birth weights in the United States are normally distributed with σx = 500 grams. A random sample of 15 babies was taken. The average birth weight of these 15 babies was 3450 grams. Which of the following is true regarding the distribution of sample​ means?

The distribution of sample means will be normal since birth weights of all babies are normally distributed. (The distribution of sample means will be normal or approximately normal if the sample size is large enough​ (typically more than 30 or​ so) and/or the population data follow a normal distribution. The second condition is met.)

Researchers studied the mean egg length​ (in millimeters) for a bird population. After taking a random sample of​ eggs, they obtained a​ 95% confidence interval of​ (45,60). What is the value of the margin of​ error?

The margin of error is half the width of the confidence interval. (45, 60) 60-45 = 15 15/2 = 7.5 Solving this system of equations, we find that MOE = 7.5

A college student was interested in the average amount college students spend on entertainment each week. He randomly sampled 200 students and found the following​ 95% confidence​ interval: (24,28) in dollars per week. If this student randomly sampled 100 students​ instead, what would happen to the margin of error​ (assuming everything else remained the​ same)?

The margin of error would increase. (Decreasing the sample size will increase the margin of error not only because the denominator would​ decrease, but also the​ t* critical value would increase. If the​ z-methods were used​ instead, the​ z* critical value would not change for different sample sizes.)

An IQ test is designed so that the mean is 100 and the standard deviation is 24 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 4 IQ points of the true mean. Assume that σ = 24 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation. Would it be reasonable to sample this number of​ students? 90% α = 0.10 = z α/2 = 1.645 95% α = 0.05 = z α/2 = 1.96 99% α = 0.01 = z α/2 = 2.575

The required sample size is 139. z α/2 = 1.96 n = [ (z α/2 × σ) ÷ E ] 2 = [ (1.96 × 24) ÷ 4 ] 2 = 138.2976 Yes. This number of IQ test scores is a fairly small number.

Researchers studied the mean egg length​ (in millimeters) for a bird population. After taking a random sample of​ eggs, they obtained a​ 95% confidence interval of​ (45,60). What is the value of the sample​ mean?

The sample mean is the point estimate. A certain amount is added and subtracted from the sample mean to obtain the bounds of the confidence interval.​ Therefore, the sample mean is​ half-way between the lower and upper bounds. 45+60 = 105 105/2 = 52.5 95% confidence interval X shrtaa -1.96 standard divination/ square n=45 X shrtaa+1.96 standard sicinTION/sqaue root n=60 ---------------------- 2x shrtaa=105 sp x=52.5

When analyzing​ polls, which of the following is NOT a​ consideration?

The sample should be a voluntary response or convenience sample.

Which of the following is NOT true of the confidence level of a confidence​ interval?

There is a 1 - α ​chance, where α is the complement of the confidence​ level, that the true value of p will fall in the confidence interval produced from our sample.

Which concept below is NOT a main idea of estimating a population​ proportion?

Using a sample statistic to estimate the population proportion is not utilizing descriptive​ statistics

A graduate student wanted to estimate the average time spent studying among graduate students at her school. She randomly sampled graduate students from her school and obtained a​ 99% confidence interval of​ (17.3,22.5) hours/week. In the context of the​ problem, which of the following interpretations is​ correct?

We are​ 99% sure that the average amount of time spent studying among graduate students at this​ student's school is between 17.3 and 22.5 hours per week. (Since 99 out of every 100 random samples of the same sample size from the same population will generate confidence intervals that capture the population​ mean, one can be​ 99% confident that the interval constructed will be one of those that will capture the population mean.)

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) x̄ = 17.598 Sx = 16.01712719 n = 50

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. Step-by-step explanation: We know that a 95% confidence interval gives a range of data values that we are 95% sure that the true mean of the population lies in it. Given : The results in the screen display are based on 95% confidence level. Confidence interval : (13.046 , 22.15) It means that we are 95% confident that the true value of population mean lies in (13.046 , 22.15) . Thus, the statement that correctly interprets the confidence interval will be: 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.

A data set includes 110 body temperatures of healthy adult humans having a mean of 98.1°F and a standard deviation of 0.64°F. Construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body​ temperature? a. What is the confidence interval estimate of the population mean μ​? b. What does this suggest about the use of 98.6°F as the mean body​ temperature?

a. 97.94 °F < μ < 98.26 °F Using TInterval TInterval x̄ = 98.1 Sx = .64 n = 110 C-Level: .99 calculate --------------------------- TInterval (97.94 , 98.26) x̄ = 98.1 Sx = .64 n = 110 -------------------------- b. This suggests that the mean body temperature could be lower than 98.6°F.

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts​ (a) through​ (c) below. TInterval ​(13.046 , 22.15) x̄ = 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. b. Identify the best point estimate of μ and the margin of error. 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?

a. 13.046 Mbps < μ < 22.15 Mbps b. The point estimate of μ is 17.6 Mbps. 22.15 - 13.05 = 9.1 9.1 / 2 = 4.55 4.55 + 13.05 = 17.6 The Margin of error is E = 4.55 Mbps. c. Because the sample size of 50 is greater than​ 30, the distribution of sample means can be treated as a normal distribution. (In order to construct a confidence​ interval, either the population from which the sample comes must be normally​ distributed, or the sample size must be greater than 30 to allow the distribution of sample means to be treated as a normal​ distribution, or both.)

A critical​ value, zα​, denotes the​

z-score with an area of α to its right.

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

point estimate

Which of the following is NOT needed to determine the minimum sample size required to estimate a population​ proportion?

standard deviation σ

Which of the following is NOT a requirement for constructing a confidence interval for estimating the population​ proportion?

trials are done without replacement