Stat 200SG chapter 6
Find the margin of error for the given values of c, sigma, and n. cequals0.90, sigmaequals3.4, nequals36
.932 stared calculator
Find the critical value z Subscript c necessary to form a confidence interval at the level of confidence shown below. cequals0.85
1.44 start with (1/2)(1-c) idk what to doafter that sooo
If all other quantities remain the same, how does the indicated change affect the width of a confidence interval? (a) Increase in the level of confidence (b) Increase in the sample size (c) Increase in the population standard deviation
An increase in the level of confidence ___will widen___ the confidence interval. An increase in the sample size ___will narrow___ the confidence interval. An increase in the population standard deviation ___will widen___ the confidence interval.
If all other quantities remain the same, how does the indicated change affect the minimum sample size requirement? (a) Increase in the level of confidence (b) Increase in the error tolerance (c) Increase in the population standard deviation
An increase in the level of confidence will increase the minimum sample size required. An increase in the error tolerance will decrease the minimum sample size required. An increase in the population standard deviation will increase the minimum sample size required.
A cheese processing company wants to estimate the mean cholesterol content of all one-ounce servings of a type of cheese. The estimate must be within 0.65 milligram of the population mean. (a) Determine the minimum sample size required to construct a 95% confidence interval for the population mean. Assume the population standard deviation is 3.03 milligrams. (b) The sample mean is 28 milligrams. Using the minimum sample size with a 95% level of confidence, does it seem likely that the population mean could be within 3% of the sample mean? within 0.3% of the sample mean? Explain.
E = estimate must be within 0.65 so E=.65 95%= alpha = .05 n =84 b.) in calculator starred for easy The 95% confidence interval is ( 27.728.335 27.35, 65 28.65). It does not seem likely that the population mean could be within 3% of the sample mean because 3% off from the sample mean would fall outside the confidence interval. It does seem likely that the population mean could be within 0.3% of the sample mean because 0.3% off from the sample mean would fall inside the confidence interval.
In a random sample of 24 people, the mean commute time to work was 33.3 minutes and the standard deviation was 7.3 minutes. Assume the population is normally distributed and use a t-distribution to construct a 95% confidence interval for the population mean mu. What is the margin of error of mu? Interpret the results.
E = tc(s/(sqrt(n)) n= 24 x = 33.3 s = 7.3 tc=2.069 E = margine of error = part B Part A: (X-E, X+E) With 95% confidence, it can be said that the population mean commute time is between the bounds of the confidence interval.
Find the margin of error for the given values of c, s, and n. cequals0.90, sequals6, nequals27
E = tc(s/sqrt(n)) calc find tc = then do equ
Find the critical values chi Subscript Upper R Superscript 2 and chi Subscript Upper L Superscript 2 for the given confidence level c and sample size n. cequals0.98, nequals16
Free Critical Chi-Square Value Calculator - Free Statistics Calculators for right tail .98=.01 Critical Chi-Square Values - MathCracker.com .01 19 ONLY WORKS IF YOU DO EACH SIDE INDIVIDUALLY ^^^^^^ ONLY WORKS FOR 98%^^
Why is it necessary to check that n ModifyingAbove p with caret greater than or equals 5 and n ModifyingAbove q with caret greater than or equals 5?
It is necessary to check that n ModifyingAbove p with caret greater than or equals 5 and n ModifyingAbove q with caret greater than or equals 5 because, if either of the values are less than 5, the distribution may not be normally distributed, thus z Subscript c cannot be used to calculate the confidence interval
You construct a 95% confidence interval for a population mean using a random sample. The confidence interval is 24.9less thanmuless than31.5. Is the probability that mu is in this interval 0.95? Explain.
No. With 95% confidence, the mean is in the interval (24.9,31.5).
A magazine includes a report on the energy costs per year for 32-inch liquid crystal display (LCD) televisions. The article states that 14 randomly selected 32-inch LCD televisions have a sample standard deviation of $3.58. Assume the sample is taken from a normally distributed population. Construct 90% confidence intervals for (a) the population variance sigmasquared and (b) the population standard deviation sigma. Interpret the results.
SV=3.58^2 = 12.8164 n=14 c=90% use calc 90% between lower on calc under SD 90% between
What happens to the shape of the chi-square distribution as the degrees of freedom increase?
The distribution approaches the shape of a normal curve.
Use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a random sample of 17 mortgage institutions, the mean interest rate was 3.48% and the standard deviation was 0.36%. Assume the interest rates are normally distributed.
Use a t-distribution because it is a random sample, sigma is unknown, and the interest rates are normally distributed. n=17 x=3.48 s=.36 With 95% confidence, it can be said that the population mean interest rate is between the bounds of the confidence interval.
Which statistic is the best unbiased estimator for mu?
X with a line over it
Does a population have to be normally distributed in order to use the chi-square distribution?
Yes, in order to use the chi-square distribution, the population must be normally distributed.
From a random sample of 81 dates, the mean record high daily temperature in a certain city has a mean of 83.87degreesF. Assume the population standard deviation is 14.24degreesF. This creates a 95% confidence interval for the population mean of left parenthesis 80.77 comma 86.97 right parenthesis. Does it seem possible that the population mean could be less than 86degreesF? Explain.
Yes, it seems possible because 86degreesF is between the endpoints of the confidence interval.
In a survey of 3005 adults, 1465 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
calc With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.
Construct the confidence interval for the population mean mu. cequals0.98, x overbar equals 7.8, sigmaequals0.4, and nequals40
calculator stared Fre con,,, (7.63, 7.97)
Find the critical value tc for the confidence level cequals0.90 and sample size nequals20.
degrees of freedom (DOF) = n-1 = 19 use find critical t calculator starred USE THE TWO TAILED OR IT WILL NOT WORK
A publisher wants to estimate the mean length of time (in minutes) all adults spend reading newspapers. To determine this estimate, the publisher takes a random sample of 15 people and obtains the results below. From past studies, the publisher assumes sigma is 1.9 minutes and that the population of times is normally distributed.
find mean of all numbers use calculator starred to find intervals 99% is wider
The gas mileages (in miles per gallon) of 25 randomly selected sports cars are listed in the accompanying table. Assume the mileages are not normally distributed. Use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results.
n= 25 95%=.05 Let sigma be the population standard deviation and let n be the sample size. Which distribution should be used to construct the confidence interval? Neither distribution can be used to construct the confidence interval, since the population is not normally distributed and n less than 30. Neither the standard normal distribution nor the t-distribution can be used to construct the interval. Neither the standard normal distribution nor the t-distribution can be used to construct the interval.
A researcher wishes to estimate, with 99% confidence, the population proportion of adults who think Congress is doing a good or excellent job. Her estimate must be accurate within 2% of the true proportion. (a) No preliminary estimate is available. Find the minimum sample size needed. (b) Find the minimum sample size needed, using a prior study that found that 40% of the respondents said they think Congress is doing a good or excellent job. (c) Compare the results from parts (a) and (b).
n= use calc use calc "sample" change pop por to 40% Having an estimate of the population proportion reduces the minimum sample size needed.
Find the minimum sample size n needed to estimate mu for the given values of c, sigma, and E. cequals0.95, sigmaequals8.7, and Eequals1
n=(zcSD/E)^2 use marked calculator alpha/sig. level = .05 ALWAYS
In a random sample of eleven people, the mean driving distance to work was 25.1 miles and the standard deviation was 5.5 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 90% confidence interval for the population mean mu. Interpret the results.
n=11 s= 5.5 x=25.1 90% = .1 e= 3 miles ME= (X-E, X+E) With 90 90% confidence, it can be said that the population mean driving distance to work (in miles) is between the interval's endpoints.
The state test scores for 12 randomly selected high school seniors are shown on the right. Complete parts (a) through (c) below. Assume the population is normally distributed.
put calculator on SAMPLE Use tc calc n=12 x=911.1 s=303 tc= calc E=157.093544145 x-E, x+E
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 32 business days, the mean closing price of a certain stock was $111.11. Assume the population standard deviation is $10.98.
same as question above Which interval is wider? Choose the correct answer below. 95% You can be 90% confident that the population mean price of the stock is between the bounds of the 90% confidence interval, and 95% confident for the 95% interval.
For each level of confidence c below, determine the corresponding normal confidence interval. Assume each confidence interval is constructed for the same sample statistics.
size order smallest goes with smallest c
In a survey of 2293 adults in a recent year, 1303 say they have made a New Year's resolution. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.
starred calc starred calc With the given confidence, it can be said that the population proportion of adults who say they have made a New Year's resolution is between the endpoints of the given confidence interval. The 95% confidence interval is wider.
Determine whether the statement is true or false. If it is false, rewrite it as a true statement. The point estimate for the population proportion of failures is 1minusModifyingAbove p with caret.
true
Use technology to construct the confidence intervals for the population variance sigmasquared and the population standard deviation sigma. Assume the sample is taken from a normally distributed population. cequals0.95, ssquaredequals18.49, nequals27
use Confidence Interval for Variance Calculator - MathCracker.com lower on same calc is SD
Construct the indicated confidence interval for the population mean mu using the t-distribution. Assume the population is normally distributed. cequals0.99, x overbarequals13.6, sequals3.0, nequals6
use cal to find E (X-E, X+E)
The number of hours of reserve capacity of 10 randomly selected automotive batteries is shown to the right. 1.74 1.83 1.52 1.66 1.71 1.99 1.32 1.53 1.46 2.02 Assume the sample is taken from a normally distributed population. Construct 98% confidence intervals for (a) the population variance sigmasquared and (b) the population standard deviation sigma.
use calc With 98% confidence, it can be said that the population variance is between .022 . 022 and .223 . 223. ------------------------ lower on calc is SD interval With 98% confidence, you can say that the population standard deviation is between .147 . 147 and .472 . 472 hours of reserve capacity.
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 60 home theater systems has a mean price of $149.00. Assume the population standard deviation is $18.70.
use calculator starred Easy Confidence Interval Calculator (145.03, 152.97) (144.27,153.73) With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.
The table to the right shows the results of a survey in which 2590 adults from Country A, 1101 adults from Country B, and 1074 adults from Country C were asked if human activity contributes to global warming. Complete parts (a), (b), and (c).
use conf calc use conf calc use calc for all take total pop*%
Determine the minimum sample size required when you want to be 90% confident that the sample mean is within one unit of the population mean and sigmaequals17.3. Assume the population is normally distributed.
use miminum starred calc signif level = .1 (90%=.1, 95=.05 99%=.01) 472
A researcher wishes to estimate, with 95% confidence, the population proportion of adults who support labeling legislation for genetically modified organisms (GMOs). Her estimate must be accurate within 3% of the true proportion. (a) No preliminary estimate is available. Find the minimum sample size needed. (b) Find the minimum sample size needed, using a prior study that found that 72% of the respondents said they support labeling legislation for GMOs. (c) Compare the results from parts (a) and (b).
use sample calc use sample calc PP=72% not 50 Having an estimate of the population proportion reduces the minimum sample size needed.
Use the given confidence interval to find the margin of error and the sample mean. (14.6,24.4)
use starred calc
Let p be the population proportion for the following condition. Find the point estimates for p and q. In a survey of 1148 adults from country A, 489 said that they were not confident that the food they eat in country A is safe.
uzse calc .095 q=1-p
Use the standard normal distribution or the t-distribution to construct a 90% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a recent season, the population standard deviation of the yards per carry for all running backs was 1.35. The yards per carry of 25 randomly selected running backs are shown below. Assume the yards per carry are normally distributed.
x= add all/total number of numbers = 4.992 n= 25 s= 1.35 90%=.1 Use a normal distribution because sigma is known and the data are normally distributed. use the starred "easy blah blah calculator" With 90% confidence, it can be said that the population mean yards per carry is between the bounds of the confidence interval.
For the same sample statistics, which level of confidence would produce the widest confidence interval? Explain your reasoning.
99%, because as the level of confidence increases, z Subscript c increases
Determine whether the statement is true or false. To estimate the value of p, the population proportion of successes, use the point estimate x.
False, to estimate the value of p, use the point estimate p=(x/n)
The table to the right shows the results of a survey in which 2580 adults from Country A, 1100 adults from Country B, and 1083 adults from Country C were asked if they believe climate change poses a large threat to the world. Complete parts (a), (b), and (c).
No, because the two confidence intervals do not overlap. No, because the two confidence intervals do not overlap. es, because the two confidence intervals overlap.
