Transcription
Stats ReviewChapter 9Mary Stangler Center for Academic SuccessRevised 8/16
Note:This review is meant to highlight basic concepts fromthe course. It does not cover all concepts presentedby your instructor. Refer back to your notes, unitobjectives, handouts, etc. to further prepare for yourexam.The questions are displayed on one slide followed bythe answers are displayed in red on the next.This review is available in alternate formats uponrequest.Mary Stangler Center for Academic Success
Confidence Interval for Population ProportionIn a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the90% confidence interval for the population proportion, p?Mary Stangler Center for Academic Success9.1
Confidence Interval for Population ProportionIn a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the90% confidence interval for the population proportion, p? Condition 1: n(.05) N The sample size (10) is less than 5% of the population (millions of musicians), sothe condition is met. Condition 2: np(1-p) 10 π 210 .2 ππ 1 π 10 .2 1 .2 1.6 . This is less than 10 so this condition is notmet.It would not be practical to construct the confidence interval.Mary Stangler Center for Academic Success9.1
Confidence Interval for Population ProportionA poll conducted found that 944 of 1748 adults do not believe that people with tattoos aremore rebellious. If appropriate construct a 90% confidence interval.Mary Stangler Center for Academic Success9.1
Confidence Interval for Population ProportionA poll conducted found that 944 of 1748 adults do not believe that people with tattoos are morerebellious. If appropriate construct a 90% confidence interval.Is it appropriate? Yes, it satisfies both conditions.1) point estimate944π 1748 .542) Find π§πΌ/2 . First take1 .902 .05The corresponding z is -1.645. We will ignore the negative and just use 1.645.3) Find the Margin of ErrorπΈ π§πΌ/2π(1 π)π 1.645.54(1 .54)1748 .01964) Take the point estimate and add/subtract the margin of error.54-.0196 .5204, .54 .0196 .5596The confidence interval is (.5204,.5596)Mary Stangler Center for Academic Success9.1
Sample Size Needed of Population ProportionA researcher wants to estimate the proportion of Americans that have sleep deprivation. Howlarge a sample is needed in order to be 95% confident and within 3% ifa) the researcher used a previous estimate of 60%?b) the researcher doesnβt use a previous estimate?Mary Stangler Center for Academic Success
Sample Size Needed of Population ProportionA researcher wants to estimate the proportion of Americans that have sleep deprivation. Howlarge a sample is needed in order to be 95% confident and within 3% ifa) the researcher used a previous estimate of 60%?π§πΌ21.962π π 1 π .60 1 .60πΈ.03b) the researcher doesnβt use a previous estimate?π§πΌπ .252πΈ21.96 .25.032 1024.4 ππππ2 1067.1 ππππRemember to always round up when finding sample sizes.Mary Stangler Center for Academic Success
Find the t-value (using the table)a) Right tail .1, n 6b) Left tail .05, n 16c) 90% confidence, n 21d) 95% Confidence , n 83e) 99% Confidence, n 1200Mary Stangler Center for Academic Success
Find the t-value (using the table)a)Right tail .1, n 6Degrees of Freedom (df) n-1 5, t 1.476b)Left tail .05 n 16df 15, t -1.753 (t is negative for left tail)c)90% confidence, n 21df 20,d) .05, π π. πππ95% Confidence , n 83df 82 (this is not in the table so choose the closest one: 80),1 .952e)1 .902 .025, π π. πππ99% Confidence, n 1200df 1199, since this is more than 1000 we use the z-value1 .992 .005, π ππ π π. πππMary Stangler Center for Academic Success
Estimating a population meanIn a sample of 81 SARS patients, the mean incubation period was 4.6 days with a standarddeviation of 15.9 days. Construct a 95% confident interval and interpret.Mary Stangler Center for Academic Success9.2
Estimating a population meanIn a sample of 81 SARS patients, the mean incubation period was 4.6 days with a standarddeviation of 15.9 days. Construct a 95% confident interval and interpret. We cannot use z because we do not have the population standard deviation. Find π‘πΌ/2 :1 .952 .025, degrees of freedom n-1 81-80. The corresponding t-value is 1.99 Find the Margin of ErrorπΈ π‘πΌ/2π 15.9 1.99 3.516π81 Add/subtract the margin of error from the sample mean4.6-3.516 1.084, 4.6 3.516 8.116We are 95% confident that the mean incubation period for SARS patients is between1.084 and 8.116.Mary Stangler Center for Academic Success9.2
Sample Size for Estimating Population MeanHow large must a sample be in order to be 95% confident within 2 points given a samplestandard deviation of 13.67?Mary Stangler Center for Academic Success
Sample Size for Estimating Population MeanHow large must a sample be in order to be 95% confident within 2 points given a samplestandard deviation of 13.67?π§πΌ π π 2πΈ21.96 13.67 22 πππ. π πππMary Stangler Center for Academic Success
Width of the Confidence IntervalWhat Happens to the width of the Confidence Intervala) As the sample size increasesb) As the level of confidence increaseMary Stangler Center for Academic Success
Width of the Confidence IntervalWhat Happens to the width of the Confidence Intervala) As the sample size increasesThe width decreasesb) As the level of confidence increaseThe width IncreasesTo find out which margin of error is smaller (smaller width) with different sample sizes or level ofconfidence when everything else is the same calculateπ§πΌ/2π(for proportion) orπ‘πΌ/2π(for mean) for both cases to see which is smaller.Mary Stangler Center for Academic Success
Reasonable Interpretation of Confidence IntervalsA 90% confidence interval for the hours that college students sleep during the weekday is (6.8, 10.8).Which interpretation(s) are correct?a)90% of college students sleep between 6.8 and 10.8 hoursb)Weβre 90% confident that the mean number of hours of sleep that college students get any dayof the week is between 6.8 and 10.8 hoursc)There is a 90% probability that the mean hours of sleep that college students get during aweekday is between 6.8 and 10.8 hours.d)Weβre 90% confident that the mean number of hours of sleep that college students during aweekday is between 6.8 and 10.8 hours.Mary Stangler Center for Academic Success
Reasonable Interpretation of Confidence IntervalsA 90% confidence interval for the hours that college students sleep during the weekday is (6.8, 10.8).Which interpretation(s) are correct?a)90% of college students sleep between 6.8 and 10.8 hoursFlawed: makes an implication about individuals rather than the meanb)Weβre 90% confident that the mean number of hours of sleep that college students get any dayof the week is between 6.8 and 10.8 hoursFlawed: should be about the weekday, not any day of the weekc)There is a 90% probability that the mean hours of sleep that college students get during aweekday is between 6.8 and 10.8 hours.Flawed: implies the population mean varies rather than the intervald)Weβre 90% confident that the mean number of hours of sleep that college students during aweekday is between 6.8 and 10.8 hours.Correct!Mary Stangler Center for Academic Success
Chi-Squared ValuesFind πππ πΆ/π ππ§ππππΆ/π90% confidence, n 2095% confidence, n 25Mary Stangler Center for Academic Success
Chi-Squared ValuesFind πππ πΆ/π ππ§ππππΆ/π90% confidence, n 20Ξ± 1-.90 .1; df n-1 19222 π1 πΌ/2 π1 .1/2 π.95 ππ. πππ222 ππΌ/2 π.1/2 π.05 ππ. πππ95% confidence, n 25Ξ± 1-.95 .05; df n-1 24222 π1 πΌ/2 π1 .05/2 π.975 ππ. πππ222 ππΌ/2 π.05/2 π.025 ππ. πππ If the degrees of freedom is not on the table, use the closest degrees of freedom If the degrees of freedom is directly between 2 values, find the mean of the values. Ex: For 65,take the Ο2 values for both 60 and 70 and average their Ο2 values.Mary Stangler Center for Academic Success
Estimating a Population Standard DeviationA student randomly selects 10 paperback books at a store. The mean price is 8.75 with astandard deviation of 1.50. Construct a 95% confidence interval for the population standarddeviation, Ο. Interpret the confidence interval. Assume the data are normally distributedMary Stangler Center for Academic Success
Estimating a Population Standard DeviationA student randomly selects 10 paperback books at a store. The mean price is 8.75 with astandard deviation of 1.50. Construct a 95% confidence interval for the population standarddeviation, Ο. Interpret the confidence interval. Assume the data are normally distributed221. Find π1 πΌ/2andππΌ/2for 95% confidence level and df 92π1 .05/2 2.72.2π.05/2 19.023Find the lower and upper bounds of Ο2Lower:(π 1)π 22ππΌ/2Upper:(π 1)π 2 (10 1)1.52 2.7 7.52π1 πΌ/2 (10 1)1.5219.023 1.0653.Since we want the confidence interval for Ο and we have the lower/upper bounds ofΟ2, we need to take the square roots of the lower and upper bound.Lower: 1.065 1.03Upper: 7.5 2.74The confidence interval is (1.03,2.74)Weβre 95% confident that the population standard deviation of price of paperback books atthis store is between 1.03 and 2.74 hours.Mary Stangler Center for Academic Success
A student randomly selects 10 paperback books at a store. The mean price is 8.75 with a standard deviation of 1.50. Construct a 95% confidence interval for the population standard deviation, Ο. Interpret the confidence interval. Assume the data are normally distributed 1. Fin
