Stats chapter 7 online HW

You want to estimate the percentage of adults who believe that passwords should be replaced with biometric security​ (such as​ fingerprints). How many randomly selected adults must you​ survey? Assume that you want to be 99​% confident that the sample percentage is within 4.3 percentage points of the true population percentage. Complete parts​ (a) through​ (c) below.

a. Assume that nothing is known about the percentage of adults who believe that passwords should be replaced with biometric security. assume p hat and q hat are 0.5 because no info is known. Use equation: n= ((Zα/2)/E)^2 p̂q̂ n=(2.576/0.043)^2 (0.5)(0.5)= 897 sample size needed b. Assume that a prior survey suggests that about 47​% of adults believe that biometric security should replace passwords. n=(2.576/0.043)^2 (0.47)(0.53)= 894 sample size c.Does the additional survey information from part​ (b) have much of an effect on the sample size that is​ required? The additional survey information from part​ (b) causes the required sample size to change by less than 10%. Based on​ this, the additional survey information causes no significant change in the sample size that is required.

A physician wants to develop criteria for determining whether a​ patient's pulse rate is​ atypical, and she wants to determine whether there are significant differences between males and females. Use the sample pulse rates below. Using spss: Male/Female 96 72 64 84 88 104 64 68 56 68 64 104 56 68 76 96 60 96 72 124

a. Construct a 95​% confidence interval estimate of the mean pulse rate for males. 59.9<μ<79.3 b. Construct a 95​% confidence interval estimate of the mean pulse rate for females. 74.5<μ<102.3 c. Compare the preceding results. Can we conclude that the population means for males and females are​ different? No, because the two confidence intervals​ overlap, we cannot conclude that the two population means are different.

A genetic experiment with peas resulted in one sample of offspring that consisted of 447 green peas and 174 yellow peas. a. Construct a 95​% confidence interval to estimate of the percentage of yellow peas. b. It was expected that​ 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?

a. use equation: p̂± Zα/2 (√ p̂q̂/n) since you are given a proportion to find!!! p̂= 174/174+447=0.280 q̂= 1-0.280=0.720 Zα/2= from table for 95% confidence (last row!!) (Very large Degrees of freedom)= 1.960 n=174+447=621 plug: 0.280±1.960(√(0.280)(0.720)/621)= 0.245 & 0.316 = 0.245<p<0.316 b. Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations? No, the confidence interval includes​ 0.25, so the true percentage could easily equal​ 25%

In the week before and the week after a​ holiday, there were 10 comma 000 total​ deaths, and 4936 of them occurred in the week before the holiday. a. Construct a 95​% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday. b. Based on the​ result, does there appear to be any indication that people can temporarily postpone their death to survive the​ holiday?

a.) 0.484<p<0.5064 b.)No​, because the proportion could easily equal 0.5. The interval is not less than 0.5 the week before the holiday.

Find the sample size needed to estimate the percentage of residents of one region of a country who are​ left-handed. Use a margin of error of one percentage point​, and use a confidence level of 90​%. Complete parts​ (a) through​ (c) below

a.) Assume that ModifyingAbove p hat and q hat are unknown Use equation: n= ((Zα/2)/E)^2 p̂q̂ Zα/2= from table use 90% confidence so look at 2 tail for a very large sample row in column for 0.10 (10%). Zα/2=1.645 E= 1%= 0.01 p̂=q̂= 0.50 (assume since not known) plug n=(1.645/0.01)^2 (0.5)(0.5) n=6765 The sample size needed is 6765 b.) Assume that based on prior​ studies, about 8% of residents of the region are​ left-handed Zα/2=1.645 E= 1%= 0.01 p̂= 0.08 q̂= 1-0.08=0.92 plug n=(1.645/0.01)^2 (0.08)(0.92) The sample size needed is 1992 c).How do the results from parts​ (a) and​ (b) change if the entire country is used instead of the one region of that​ country? If the entire country is used instead of the one region of that​ country, the result from part​ (a) does not change and the result from part​ (b) does not change.

John wanted to estimate the average literacy rate for all countries. He randomly selected 35 countries and researched the literacy rate for the 35 countries.

a.) The statistical unit for​ John's study is A country b.) The population John is interested in calculating an estimate for is all countries in world c.) Suppose John pluged his data into SPSS and calculated a sample mean of x bar equals 82 % and a​ 90% confidence interval of​ (78%, 86%). Find the center of the confidence interval by adding the two end points and dividing by 2. What value did you​ calculate? ​ 82% (The same value as his sample​ mean). ------------------------------------------------------ No mater how many times we calculate our sample mean and confidence​ interval, the the sample mean will always be exactly in the middle of our confidence interval. That is because we use the sample mean as the center point of the confidence interval as seen in the formula​ below: x̅- tα/2(s/√ ̅n) <μ< x̅+ tα/2(s/√ ̅n) Notice we are taking our sample mean ​(x overbar​) and subtracting and adding the same amount to get the two endpoints of our confidence interval. Notice we are taking our sample mean ​(x overbar​) and subtracting and adding the same amount to get the two endpoints of our confidence interval. --------------------------------------------------- d.) This means we are ​100% confident that our sample mean is exactly in the middle of our confidence interval. e.) So what is John only​ 90% confident​ of??? John is​ 90% confident that his confidence interval​ (78%, 86%) captures the population mean literacy rate for all countries. ----------------------------------------------------- Why are the other answers not​ correct? ​"John is​ 90% confident that all countries have literacy rates between​ 78% and​ 86%." is not correct because a confidence interval is calculated to serve as an estimate for a population mean. If all the values in your population were contained within the​ interval, the interval would be too wide to be of any value. ​"John is​ 90% confident that his confidence interval​ (78%, 86%) captures the population mean literacy rate for his 35​ countries." is not correct because of the​ ".....for his 35​ countries." He is calculating a confidence interval that serves as an estimate for the population mean which is the mean literacy rate for all countries and not just the 35 countries he​ selected! ​"John is​ 90% confident that the average literacy rate for all countries falls in his confidence interval​ (78%, 86%)." is not correct because the word​ "falls" implies that the average literacy rate for all countries is what varies. Population parameters are fixed​ values, so it is the end points of the confidence interval that will vary and not the average literacy rate for all countries. ​"John is​ 90% confident that his confidence interval​ (78%, 86%) captures the population mean literacy rate for a​ country." is not correct because of the​ ".......a country." He is not estimating the population mean for a country​, but for all countries. Make sure you understand these concepts

In a test of the effectiveness of garlic for lowering​ cholesterol, 47 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.4 and a standard deviation of 19.7. Complete parts​ (a) and​ (b) below.

a.) What is the best point estimate of the population mean net change in LDL cholesterol after the garlic​ treatment? THE BEST POINT ESTIMATE WILL ALWAYS EQUAL THE SAMPLE MEAN!! The best point estimate is 3.4 ​mg/dL b.) 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? What is the confidence interval estimate of the population mean μ​? use equation: x̅ ±tα/2(s/√ ̅n) Since you are given Mean and standard dev!!! x̅ =3.4 tα/2= for 95% confidence and 46 degrees of freedom =2.0129 s=19.7 n=47 plug: 3.4±2.0129(19.7/√ ̅47)= -2.39 & 9.19 -2.39 mg/dl <μ<9.19 mg/dl What does the confidence interval suggest about the effectiveness of the​ treatment? The confidence interval limits contain ​0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.

In a sample of seven​ cars, each car was tested for​ nitrogen-oxide emissions​ (in grams per​ mile) and the following results were​ obtained: 0.08​, 0.18​, 0.06​, 0.15​, 0.13​, 0.15​, 0.15 . Assuming that this sample is representative of the cars in​ use, construct a 98​% confidence interval estimate of the mean amount of​ nitrogen-oxide emissions for all cars. If the EPA requires that​ nitrogen-oxide emissions be less than 0.165 g divided by mi​, can we safely conclude that this requirement is being​ met?

confidence interval: 0.078 g/mi <μ<0.180 g/mi can we safely conclude that the requirement that​ nitrogen-oxide emissions be less than 0.165 g divided by mi is being​ met? No, it is possible that the requirement is being​ met, but it is also very possible that the mean is not less than 0.165 g/mi

A table to get values for confidence intervals based on degreed of freedom. (Two-tailed)

http://davidmlane.com/hyperstat/t_table.html

The​ _______ is the best point estimate of the population mean.

sample mean

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

​Percentage, Probability, and Proportion

Listed below are the amounts of mercury​ (in parts per​ million, or​ ppm) found in tuna sushi sampled at different stores. The sample mean is 1.053 ppm and the sample standard deviation is 0.267 ppm. Use technology to construct a 90​% confidence interval estimate of the mean amount of mercury in the population. use spss to find values: 1.43 0.89 0.75 1.13 0.73 1.22 1.22

0.857ppm<μ<1.249 ppm

Refer to the accompanying data set and construct a 90​% confidence interval estimate of the mean pulse rate of adult​ females; then do the same for adult males. Compare the results. Male/Females 81 82 77 94 53 60 59 66 53 53 60 81 54 78 76 83 52 87 64 53 73 34 57 64 65 83 78 74 79 81 66 66 69 65 94 76 45 61 89 64 71 82 66 80 70 71 74 77 52 88 68 90 56 87 79 91 75 89 62 93 66 68 96 87 60 83 65 81 55 74 57 56 70 101 70 71 83 74 57 77

Construct a 90​% confidence interval of the mean pulse rate for adult females: 72.0 bpm <μ<79.2 bpm Construct a 90​% confidence interval of the mean pulse rate for adult males: 64.2 bpm<μ<70.6 bpm (values from spss) The confidence intervals do not​ overlap, so it appears that adult females have a significantly higher mean pulse rate than adult males.

A data set includes 105 body temperatures of healthy adult humans having a mean of 98.7degreesF and a standard deviation of 0.64degreesF. Construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6degreesF as the mean body​ temperature?

The degree of freedom is df=n-1=105-1=104 . For 99% confidence level, the critical value of t is t^*=2.624 . A 99 % confidence interval estimate of the mean body temperature of all healthy humans is: x̅ ± s/√ ̅n= 98.7± 2.624 (0.64/√ ̅105)= 98.7± 0.16 or (98.536,98.864) This suggests that the mean body temperature could very possibly be 98.6degreesF

Which of the following would be a correct interpretation of a​ 99% confidence interval such as 4.1 <μ<5.6?

We are​ 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of μ

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 550 babies were​ born, and 275 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective?

a. 0.445<p<0.555 using equation= p hat plus or minus z alpha/2 times the square root of (p hat)(q hat)/n b. No​, the proportion of girls is not significantly different from 0.5