Stats Ch. 9 - Estimating Population Proportions & Means

Define these terms : Point Estimate, Confidence Interval, Level of Confidence, & Margin of Error

Point Estimate - The value of a statistic that estimates the value of a parameter.

Confidence Interval - A range of numbers based on a point estimate of an unknown parameter.

Level of Confidence - The expected proportion of intervals that contain the parameter if a large number of different samples is contained.

Margin of Error - Determines the width of the confidence interval.

If you constructed one hundred 93% confidence intervals based on one hundred different random samples, how many of the intervals would you expect to include the unknown​ parameter?

D. 93. - This is because one hundred samples are all using one hundred intervals of the same confidence which makes the confidence % the same as the answer. Using this equation (1−α​)•​100%.

Put these intervals in order from narrowest to widest. 93%,99%,82%,90%

Narrowest is the smallest % and widest is the largest %. So the order is 82%, 90%, 93%, 99%.

(a.) As the number of samples​ increases, the proportion of​ 95% confidence intervals that include the population proportion approaches​ ______.

If a​ 95% confidence interval results in a sample proportion that does not include the population​ proportion, the sample proportion is more than​ ______ standard errors from the proportion

(a.) 0.95 - because the # of samples are becoming larger

(b.) 1.96 (just know it is 1.96 or 1.01 greater than the pop. proportion.

52% of registered voters plan on voting for (name) with a margin of error of ±​3%."

The margin of error was based on a​ 95% confidence level. Fill in the blanks.

This means that people are 95% confident (because of the confidence level) that the Percentage (52 is the percent) of registered voters In The Nation (it's a nationwide sample) who plan to vote for (name) is between 49% and 55% ( this is with the 3% margin of error).

If a sample proportion is 0.64​,

which of the following could be a​ 90% confidence interval for the population​ proportion?

Answers

Lower​ bound: 0.58​;

Upper​ bound: 0.70

OR

Lower​ bound: 0.60​;

Upper​ bound: 0.68

(This results from random appropriate estimation of what the margin of error is)

Compute the critical value z Subscript α/2 that corresponds to a 88​% level of confidence.

Answer 1.55 - because (1-% then / by 2; area for calc. Answer needs to be converted to +)

Find the point estimate, margin of error, and number of individuals in the sample with specified characteristics if you are given: Lower bound=0.116​,

Upper bound=0.704​,

n=1000

Point estimate = 0.41 (because p = sum of upper and lower bounds divided by 2)

Margin of error = 0.294 ( because E = the difference of the upper bound from the lower bound divided by2)

Number of Individuals = 410 (because X = n times p, with p being point estimate and n being the sample size)

Construct a 99​% confidence interval of the population proportion using the given information.

x = 40 , n = 200

Lower bound = 0.127

Upper bound = 0.273

(Because the answers come from the equation

p +/- Z_a/2 x

(p(1-p)/n) in which p is x/n, and the critical value Z_a/2 is given in a table then fill in and solve)

a survey of 2035 adults in a certain country conducted during a period of economic​ uncertainty, 69​%

thought that wages paid to workers in industry were too low. The margin of error was 4 percentage points with 90​% confidence.

Interpretations are flawed IF - The interpretation provides no interval about the population proportion, OR The interpretation indicates that the level of confidence is varying, OR if The interpretation suggests that this interval sets the standard for all the other​ intervals, which is not true.

Interpretations are REASONABLE IF they avoid these issues.

Of 1018 adults​ surveyed, 527

indicated that televisions are a luxury they could do without.

(a.) Point estimate = 0.518 - because p= x/n

(b.)The sample

is stated to be (obvious) a simple random sample, the value of np(1-p) (this is the equation you are going to use) is 254.170​ ( this is the answer) which is greater than or equal to (typically is in order to be right) 10, and the sample size ( the number of people in the survey) can be assumed to be less than or equal to​ 5% of the population size ( the entire population).

(c.) constructing a confidence interval like in Q9.

(d.) Yes it is possible that more than 60% of the population could think of tv as not necessary, but not likely because the 95% confidence interval in (c.) did not contain 0.6.

(e.) 0.452, and 0.513 (because you take the previous bounds from the other interval and subtract from 1)

Explain what

​"90​% confidence" means in a

90​% confidence interval.

If 100 different confidence intervals are​ constructed, each based on a different sample of size n from the same​ population, then we expect 90 of the intervals to include the parameter and

10 to not include the parameter. (self explanatory)

What type of variable is required to construct a confidence interval for a population​ proportion?

Qualitative with 2 possible outcomes

What size sample should be obtained if she wishes the estimate to be within 0.06 with 950% confidence if (a) she uses a previous estimate of 0.23​? (b) she does not use any prior​ estimates?

(a.) 553 (because 1-0.90/2 = 0.05 use as area in calculator = 1.645 convert to +, then use

p(1-p)x(z_a/2/E)^2)

(b.) 600 ( because there are no estimates do the same thing but just with 0.25 infront of (z_a/2/E)^2)

she obtains a simple random sample of 100 adults and constructs a​ 95% confidence interval. WHILE he obtains a simple random sample of 400 adults and constructs a​ 99% confidence interval. Whose estimate will have the smaller margin of​ error?

his estimate will have the smaller margin of error because the (larger sample) size more than (compensates for the higher level of confidence).

Which of the following are properties of the​ Student's t-distribution?

Properties of Student t-distribution include: (1.)The area in the tails of the​ t-distribution is slightly greater than the area in the tails of the standard normal distribution.

(2.)At the sample size n​ increases, the density curve of t gets closer to the standard normal density curve.

(3.)The area under the​ t-distribution curve is 1.

Put the following critical values in order from least to greatest.

t 0.05 with 6 degrees of freedom

t0.05 with 18 degrees of freedom

z 0.05

Answer: Z_0.05 < t_0.05 with 18 degrees < t_0.05 with 6 degrees

(This is because the area in the tails of the​ t-distribution is slightly greater than the area in the tails of the standard normal​ distribution, and as the sample size n​ increases, the density curve of t gets closer to the standard normal density curve.)

(a.) When constructing​ 95% confidence intervals for the mean when the parent population is right skewed and the sample size is​ small, the proportion of intervals that include the population mean is​ (above, below, equal​ to) 0.95.

(b.) When constructing​ 95% confidence intervals for the mean when the parent population is right skewed and the sample size is​ small, the proportion of intervals that include the population mean approaches​ _____ as the sample​ size, n, increases.

(a.) below (because the sample size is small)

(b.) 0.95 (because the sample size, n, is increasing.)

Determine the​ t-value in each of the cases.

(a.) the area in the right tail is 0.25

with 9 degrees of freedom.

(b.) the area in the right tail is 0.01

with 24 degrees of freedom.

(c.) the area left of the​ t-value is

0.20 with 17 degrees of freedom.

(d.) corresponds to 96​% confidence. Assume 30 degrees of freedom.

(a.) 0.703 (be sure to read the chart and find the number that lies where the two given rows meet)

(b.) 2.492 (repeat of a)

(c.) -0.863 (Find the corresponding number and then make it negative)

(d.) 2.147 (take the percent subtract from one then divide by two to find the area, and use the degrees given to find the corresponding number)

The data from a simple random sample with 25 observations was used to construct the plots given below. The normal probability plot that was constructed has a correlation coefficient of 0.937.

Judge whether a​ t-interval could be constructed.

The normal probability plot

does not suggest (because the correlation coefficient for 25 obs. is 0.959 and the correlation from the plot is less.) the data could come from a normal population because 0.937<0.959

and the boxplot

shows ( because there is a dot that does not fall within the interquartile range) outliers, so a​ t-interval could not (because of the flaws in this sample of observations)

be constructed. IF these criteria are AVOIDED then simply flip the wording because the plot now suggests the possibility of making a t-interval.

Determine the point estimate of the population mean and margin of error for the confidence interval. Lower bound is 17​, upper bound is 29.

Point Estimate = 23

(because x = sum of upper and lower/2)

Margin of Error = 6

(because E = the difference of the upper from the lower bounds)

The sample​ mean, x​, is found to be

113​, and the sample standard​ deviation, s, is found to be

88.

(a) Construct a 96​% confidence interval about μ if the sample​ size, n, is 25.

​(b) Construct a 96​% confidence interval about μ if the sample​ size, n, is 15.

​(c) Construct a 90​% confidence interval about μ if the sample​ size, n, is 25.

​(d) Should the confidence intervals in parts​ (a)-(c) have been computed if the population had not been normally​ distributed?

(a) Lower = 109.5 Upper = 116.5

(b) Lower = 108.3

Upper = 117.7

(notice that as the sample size​ decreases, the margin of error increases.)

(c) Lower = 110.3

Upper 115.7

(notice that as the level of confidence​ decreases, the size of the interval decreases.)

(You get these answers by first finding the area which is the equation (1−%)/2, then using the sample size - 1 together to find the t-interval(t) on the chart, then subtracting(for the lower) or adding(for the upper) the sample mean from (t(s/sqr. rt. of the sample size))

(d) No, the population needs to be normally distributed because each sample size is less than 30. (sample sizes must be large to not have to be distributed normally)

An 80​% confidence interval that results from examining 557

customers in Taco​Bell's drive-through has a lower bound of

170.5 seconds and an upper bound of 174.7 seconds.

​(a) What is the mean service time from the 557 customers?

(b) ​What is the margin of error for the confidence​ interval?

(c) Interpret the confidence interval.

(a) 172.6 (find the upper and lower bounds same as Q8 and average them together)

(b) 2.1 (take the sample mean, and subtract it from the lower bound)

(c) One can be 80​%

confident that the mean​ drive-through service time of Taco Bell is between 170.5 seconds and

174.7 seconds. (use the word confident, and then add the lower and upper bounds that you already found)

1100 adult Americans were asked how many hours they worked in the previous week. Based on the​ results, a​ 95% confidence interval for the mean number of hours worked had a lower bound of 42.7 and an upper bound of 44.5. How could you decrease the margin of error of the interval.

You could decrease the margin of error by either Increasing the sample size OR Decreasing the confidence level.

A researcher wishes to estimate the average blood alcohol concentration​... He randomly selects records from

75 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.070 g/dL.

(a) Since the distribution of blood alcohol concentrations is highly skewed​ right, a large sample size is necessary to ensure that the distribution of the sample mean is approximately normal. (self explanatory)

(b) The sample size is likely less than​ 5% of the population. (obvious because 25,000 is less than the entire pop.)

(c) The researcher is

90​% confident that the population mean BAC is between 0.147 and

0.173 for drivers involved in fatal accidents who have a positive BAC value. (Find the confidence interval by repeating steps in Q7)

(d) While it is possible that the population mean is not captured in the confidence​ interval, it is not likely. (the lowest the interval goes is 0.147 which means anything below 0.09 is not included)

The accompanying data represent the total travel tax​ (in dollars) for a​ 3-day business trip in 8 randomly selected cities.

67.67, 79.16, 70.07, 83.35,

79.84, 87.34, 100.53, 97.41

(a) determine a point estimate

(b) Construct and interpret a 90​%

confidence interval for the mean tax paid for a​ three-day business trip.

(c) What would you recommend to a researcher who wants to increase the precision of the​ interval, but does not have access to additional​ data?

(a) 83.17 (add up all the numbers and divide by the total amount of numbers)

(b) One can be

90​% confident that the mean travel tax for all cities is between $75.32 and ​$91.02. (plug in the numbers to calculator list and use 1-var. stats to find s, then follow the same equation to find the upper and lower bounds.)

(c) The researcher could decrease the level of confidence. (Because the researcher can't get anymore data increasing precision can come from reducing the confidence levels.