Chapter 7 Section 2 Hw
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
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.
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. The confidence level is 90%, σ is not known, and the normal quantile plot of the 17 salaries (in thousands of dollars) of basketball players on a team is as shown.
Neither the normal distribution nor the t distribution applies.
The pulse rates of 177 randomly selected adult males vary from a low of 42bpm to a high of 122bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want 95% confidence that the sample mean is within 3bpm of the population mean. Find the sample size using the range rule of thumb to estimate σ. Assume that σ=10.7 bpm, based on the value s=10.7 bpm from the sample of 177 male pulse rates. Compare the results from parts (a) and (b). Which result is likely to be better?
a. 171 Range Rule of Thumb high- low= x/4= standard deviation b. 49 c. The result from part (a) is larger than the result from part (b). The result from part (b) is likely to be better because it uses a better estimate of σ.
Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 9090% confidence level. What does the confidence interval tell about the population of all college students in the state? 3.6, 3.2, 4.0, 4.9, 3.2, 4.1, 3.2, 4.7, 4.8, 4.0, 4.1, 4.0, 3.2, 4.0, 4.0 What is the confidence interval for the population mean μ? What does the confidence interval tell about the population of all college students in the state? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
a. 3.67 <μ< 4.19 b. The results tell nothing about the population of all college students in the state, since the sample is from only one university.
Here are summary statistics for randomly selected weights of newborn girls: n=188, x =32.4hg, s=7.1hg. Construct a confidence interval estimate of the mean. Use a 90% confidence level. Are these results very different from the confidence interval 31.1hg<μ<33.3 hg with only 17 sample values, x=32.2 hg, and s=2.7 hg? a. What is the confidence interval for the population mean μ? b. Are the results between the two confidence intervals very different?
a. 31.4hg <μ< 33.3hg μ = M ± t(sM) M = sample mean t = t statistic determined by confidence level sM = standard error = √(s^2/n) Calculation M = 32.4 t = 1.65 sM = √(7.1^2/187) = 0.52 b. No, because the confidence interval limits are similar. If t distribution table is not provided: https://www.statisticshowto.com/tables/t-distribution-table/
The ages of a group of 142 randomly selected adult females have a standard deviation of 16.1 years. Assume that the ages of female statistics students have less variation than ages of females in the general population, so let σ=16.1 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. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population? The required sample size is Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population?
a. 3983 b. Yes, because statistics students are typically younger than people in the general population.
In a study of speed dating, male subjects were asked to rate the attractiveness of their female dates, and a sample of the results is listed below (1=not attractive; 10=extremely attractive). Construct a confidence interval using a 99% confidence level. What do the results tell about the mean attractiveness ratings of the population of all adult females? 5, 8, 3, 8, 6, 4, 8, 7, 7, 9, 3, 8 What is the confidence interval for the population mean μ? What does the confidence interval tell about the population of all adult females? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
a. 4.4 <μ< 8.2 Plug values into calculator to get mean and standard deviation. μ = M ± t(sM) *the DF is n-1, using the two tails table* b. The results tell nothing about the population of all adult females, because participants in speed dating are not a representative sample of the population of all adult females.
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 99% confidence that the sample mean is within 4IQ 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. The required sample size is Would it be reasonable to sample this number of students?
a. 60 (z alpha/2 x s/x)^2 b. Yes. This number of IQ test scores is a fairly small number.
In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (beforeminus−after)in their levels of LDL cholesterol (in mg/dL) have a mean of 4.7 and a standard deviation of 18.6. Construct a 99% 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? What is the confidence interval estimate of the population mean μ? What does the confidence interval suggest about the effectiveness of the treatment?
a. −3.06mg/dL <μ< 12.46mg/dL b. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.
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. The confidence level is 90%, σ=4165 thousand dollars, and the histogram of 56 player salaries (in thousands of dollars) of football players on a team is as shown.
zα/2=1.64 σ is known. The answer is the confidence level for 90%
Salaries of 49 college graduates who took a statistics course in college have a mean of $63,800. Assuming a standard deviation, σ, of $11,936, construct a 90% confidence interval for estimating the population mean μ.
$61004 <μ<$66596 *use z alpha/2
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. The confidence level is 90%, σ is not known, and the histogram of 64 player salaries (in thousands of dollars) of football players on a team is as shown.
tα/2 = 1.67
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 99% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi? 0.50, 0.80, 0.10, 0.95, 1.34, 0.55, 0.93 What is the confidence interval estimate of the population mean μ? Does it appear that there is too much mercury in tuna sushi?
a. .183ppm <μ< 1.295ppm 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.
Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. 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. b. Identify the best point estimate of muμ and the margin of error. c. In constructing the confidence interval estimate of muμ, why is it not necessary to confirm that the sample data appear to be from a population with a normal distribution?
a. 13.05 Mbps<μ< 22.15Mbps Values in parentheses rounded to two decimal places. b. The point estimate of μ is 17.6 Mbps. The margin of error is E=4.55 Mbps. First value is a. values added together and multiplied by two. Second value is b. minus a. c. Because the sample size of 50 is greater than 30, the distribution of sample means can be treated as a normal distribution.
A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 16 subjects had a mean wake time of 101.0min. After treatment, the 16 subjects had a mean wake time of 76.4min and a standard deviation of 22.5min. Assume that the 16 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 101.0 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. What does the result suggest about the mean wake time of 101.0 min before the treatment? Does the drug appear to be effective?
a. 66.5min <μ< 86.3min b. The confidence interval (does not include) the mean wake time of 101.0min before the treatment, so the means before and after the treatment (are different.) This result suggests that the drug treatment (has) a significant effect.
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3°F and a standard deviation of 0.73°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? What is the confidence interval estimate of the population mean μ? What does this suggest about the use of 98.6°F as the mean body temperature?
a. 98.111°F<μ<98.489°F μ = M ± t(sM) b. This suggests that the mean body temperature could be lower than be lower than 98.6°F.
Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. 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 t Subscript alpha divided by 2tα/2? b. Find the critical value t Subscript alpha divided by 2tα/2 corresponding to a 95% confidence level. c. Give a brief general description of the number of degrees of freedom.
a. df=49 n-1 b. tα/2 = 2.01 Use the table with df 49 and Area in tails being .05 c. The number of degrees of freedom 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.