Confidence Intervals for Normally Distributed Population: An Example - Prof. N. Phillips, Study notes of Probability and Statistics

Download Confidence Intervals for Normally Distributed Population: An Example - Prof. N. Phillips and more Study notes Probability and Statistics in PDF only on Docsity! Math 243: Lecture File 8 N. Christopher Phillips 23 April 2009 N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 1 / 36 Confidence intervals “Statistics is never having to say you are certain.” (Outside a UO statistician’s office door.) A 95% confidence interval means: I got this result using a method that gives a correct statement 95% of the time. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 2 / 36 Confidence intervals: an example Crumple-horned snorkacks have horn lengths (in centimeters) which are normally distributed with standard deviation σ = 6, but we don’t know the mean µ. We choose a simple random sample of 9 crumple-horned snorkacks, and find that the mean x of the horn lengths of the snorkacks in our sample is 43. The sampling distribution of x is N(µ, 6/ √ 9) = N(µ, 2). (Remember, we don’t know µ.) N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 3 / 36 Confidence intervals: an example Crumple-horned snorkacks have horn lengths (in centimeters) which are normally distributed with standard deviation σ = 6, but we don’t know the mean µ. We choose a simple random sample of 9 crumple-horned snorkacks, and find that the mean x of the horn lengths of the snorkacks in our sample is 43. The sampling distribution of x is N(µ, 6/ √ 9) = N(µ, 2). (Remember, we don’t know µ.) N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 3 / 36 Confidence intervals: an example Crumple-horned snorkacks have horn lengths (in centimeters) which are normally distributed with standard deviation σ = 6, but we don’t know the mean µ. We choose a simple random sample of 9 crumple-horned snorkacks, and find that the mean x of the horn lengths of the snorkacks in our sample is 43. The sampling distribution of x is N(µ, 6/ √ 9) = N(µ, 2). (Remember, we don’t know µ.) N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 3 / 36 We want a 95% confidence interval for the mean horn length µ of all crumple-horned snorkacks. For a normal distribution, we find the number z∗ such that 95% of all data is within z∗ standard deviations of the mean. That is, 95% of all data has z-scores between −z∗ and z∗. That is, for N(µ, σ), 95% of all data lies in (µ− z∗σ, µ + z∗σ). These numbers are in the 3rd last line of Table C—see the next page. We get z∗ ≈ 1.960. (The Rule of Thumb gives z∗ ≈ 2.) We apply this to the sampling distribution, which is N(µ, 2). Thus, 95% of all simple random samples of 9 crumple-horned snorkacks have mean horn length x in (µ− (1.960)(2), µ + (1.960)(2)). N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 4 / 36 We want a 95% confidence interval for the mean horn length µ of all crumple-horned snorkacks. For a normal distribution, we find the number z∗ such that 95% of all data is within z∗ standard deviations of the mean. That is, 95% of all data has z-scores between −z∗ and z∗. That is, for N(µ, σ), 95% of all data lies in (µ− z∗σ, µ + z∗σ). These numbers are in the 3rd last line of Table C—see the next page. We get z∗ ≈ 1.960. (The Rule of Thumb gives z∗ ≈ 2.) We apply this to the sampling distribution, which is N(µ, 2). Thus, 95% of all simple random samples of 9 crumple-horned snorkacks have mean horn length x in (µ− (1.960)(2), µ + (1.960)(2)). N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 4 / 36 We want a 95% confidence interval for the mean horn length µ of all crumple-horned snorkacks. For a normal distribution, we find the number z∗ such that 95% of all data is within z∗ standard deviations of the mean. That is, 95% of all data has z-scores between −z∗ and z∗. That is, for N(µ, σ), 95% of all data lies in (µ− z∗σ, µ + z∗σ). These numbers are in the 3rd last line of Table C—see the next page. We get z∗ ≈ 1.960. (The Rule of Thumb gives z∗ ≈ 2.) We apply this to the sampling distribution, which is N(µ, 2). Thus, 95% of all simple random samples of 9 crumple-horned snorkacks have mean horn length x in (µ− (1.960)(2), µ + (1.960)(2)). N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 4 / 36 We want a 95% confidence interval for the mean horn length µ of all crumple-horned snorkacks. For a normal distribution, we find the number z∗ such that 95% of all data is within z∗ standard deviations of the mean. That is, 95% of all data has z-scores between −z∗ and z∗. That is, for N(µ, σ), 95% of all data lies in (µ− z∗σ, µ + z∗σ). These numbers are in the 3rd last line of Table C—see the next page. We get z∗ ≈ 1.960. (The Rule of Thumb gives z∗ ≈ 2.) We apply this to the sampling distribution, which is N(µ, 2). Thus, 95% of all simple random samples of 9 crumple-horned snorkacks have mean horn length x in (µ− (1.960)(2), µ + (1.960)(2)). N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 4 / 36 Interruption: Table C The first row of Table C, and the third row from the bottom, sideways (so that it fits on my page): Confidence level C z∗ 50% 0.674 60% 0.841 70% 1.036 80% 1.282 90% 1.645 95% 1.960 96% 2.054 98% 2.326 99% 2.576 99.5% 2.807 99.8% 3.091 99.9% 3.291 N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 5 / 36 What the entries in Table C mean Table entries: 90% gives 1.645. -4 -3 -2 -1 0 1 2 3 4 This shows the standard normal distribution. The shaded region extends from −1.645 to 1.645, and has area very close to 0.9. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 6 / 36 95% of all simple random samples of 9 crumple-horned snorkacks have mean horn length x such that µ− (1.960)(2) ≤ x ≤ µ + (1.960)(2)). (x is within distance (1.960)(2) of µ.) This inequality is the same as x − (1.960)(2) ≤ µ ≤ x + (1.960)(2) (µ is within distance (1.960)(2) of x .) We had x = 43. So we conclude, with 95% confidence, that 43− (1.960)(2) ≤ µ ≤ 43 + (1.960)(2), that is, that µ is in the interval (39.080, 46.920). Since we used a simple random sample, we got this answer by a method which, in the long run, gives the correct answer 95% of the time. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 8 / 36 95% of all simple random samples of 9 crumple-horned snorkacks have mean horn length x such that µ− (1.960)(2) ≤ x ≤ µ + (1.960)(2)). (x is within distance (1.960)(2) of µ.) This inequality is the same as x − (1.960)(2) ≤ µ ≤ x + (1.960)(2) (µ is within distance (1.960)(2) of x .) We had x = 43. So we conclude, with 95% confidence, that 43− (1.960)(2) ≤ µ ≤ 43 + (1.960)(2), that is, that µ is in the interval (39.080, 46.920). Since we used a simple random sample, we got this answer by a method which, in the long run, gives the correct answer 95% of the time. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 8 / 36 95% of all simple random samples of 9 crumple-horned snorkacks have mean horn length x such that µ− (1.960)(2) ≤ x ≤ µ + (1.960)(2)). (x is within distance (1.960)(2) of µ.) This inequality is the same as x − (1.960)(2) ≤ µ ≤ x + (1.960)(2) (µ is within distance (1.960)(2) of x .) We had x = 43. So we conclude, with 95% confidence, that 43− (1.960)(2) ≤ µ ≤ 43 + (1.960)(2), that is, that µ is in the interval (39.080, 46.920). Since we used a simple random sample, we got this answer by a method which, in the long run, gives the correct answer 95% of the time. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 8 / 36 95% of all simple random samples of 9 crumple-horned snorkacks have mean horn length x such that µ− (1.960)(2) ≤ x ≤ µ + (1.960)(2)). (x is within distance (1.960)(2) of µ.) This inequality is the same as x − (1.960)(2) ≤ µ ≤ x + (1.960)(2) (µ is within distance (1.960)(2) of x .) We had x = 43. So we conclude, with 95% confidence, that 43− (1.960)(2) ≤ µ ≤ 43 + (1.960)(2), that is, that µ is in the interval (39.080, 46.920). Since we used a simple random sample, we got this answer by a method which, in the long run, gives the correct answer 95% of the time. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 8 / 36 General form of the confidence interval Let σ be the population standard deviation. Let n be the sample size. Then the sampling standard deviation is σ/ √ n. So our confidence interval is ( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) , for the appropriate critical value z∗ (usually found via Table C). The confidence interval is also written: x ± z∗ ( σ√ n ) . Caution: Most actual observations of individuals will not be within this range. It is an estimate for the population mean, not for any individual. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 9 / 36 General form of the confidence interval Let σ be the population standard deviation. Let n be the sample size. Then the sampling standard deviation is σ/ √ n. So our confidence interval is ( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) , for the appropriate critical value z∗ (usually found via Table C). The confidence interval is also written: x ± z∗ ( σ√ n ) . Caution: Most actual observations of individuals will not be within this range. It is an estimate for the population mean, not for any individual. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 9 / 36 General form of the confidence interval Let σ be the population standard deviation. Let n be the sample size. Then the sampling standard deviation is σ/ √ n. So our confidence interval is ( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) , for the appropriate critical value z∗ (usually found via Table C). The confidence interval is also written: x ± z∗ ( σ√ n ) . Caution: Most actual observations of individuals will not be within this range. It is an estimate for the population mean, not for any individual. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 9 / 36 Margin of error The number z∗ ( σ√ n ) is called the margin of error. (It is the distance from the middle of the confidence interval to one of the ends.) N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 10 / 36 Suppose now we want a 99% confidence interval. Recall that the population standard deviation was 6, the sample size was 9, so the sampling standard deviation was 6/ √ 9 = 2. Also, x = 43. Now Table C gives z∗ ≈ 2.576. So we conclude, with 99% confidence, that 43− (2.576)(2) ≤ µ ≤ 43 + (2.576)(2), that is, that µ is in the interval (37.848, 48.152). It is larger than the 95% confidence interval (39.080, 46.920). The margin of error was (1.960)(2) = 3.920, but is now (2.576)(2) = 5.152. If we want to be more certain that µ is in our confidence interval, the interval must be bigger. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 11 / 36 What 95% confidence really means: Pictures Population mean 64, population standard deviation 2.7, and for a simple random sample of size 100 I got x = 64.3921. My 95% confidence interval for the true mean µ is 64.3921± 0.5292. (The number 0.5292 was obtained as 1.960(2.7/ √ 100), and 1.960 came from the third row from the bottom in Table C.) That is, with 95% confidence I think the true mean µ is between 63.8629 and 64.9213. I found that the mean is between 63.8629 and 64.9213, using a method that gives a correct statement 95% of the time. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 13 / 36 Here are one hundred 95% confidence intervals obtained using simple random samples of size 100. The ones in red do not contain the true mean: those statisticians got an incorrect statement. 62.5 63.0 63.5 64.0 64.5 65.0 65.5 N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 14 / 36 Another hundred 95% confidence intervals. 62.5 63.0 63.5 64.0 64.5 65.0 65.5 N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 15 / 36 Another hundred 95% confidence intervals. 62.5 63.0 63.5 64.0 64.5 65.0 65.5 N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 18 / 36 Conditions for z confidence intervals and hypothesis tests We are using the “one sample z procedure”. We have talked about the procedure for confidence intervals; hypothesis tests are in the next chapter. Conditions: We have a simple random sample from the population in question. (In particular, no difficulties with such things as nonresponse.) The variable we measure has normal distribution N(µ, σ). We know σ. These are of course not realistic. Even though the first two may be approximately satisfied, the third is usually quite unreasonable. We will learn later how to do without it. We concentrate on this case first because it is the easiest to understand. (There are occasional situations in which we do know σ but not µ, such as for the calibration of certain scientific instruments.) N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 19 / 36 Conditions for z confidence intervals and hypothesis tests We are using the “one sample z procedure”. We have talked about the procedure for confidence intervals; hypothesis tests are in the next chapter. Conditions: We have a simple random sample from the population in question. (In particular, no difficulties with such things as nonresponse.) The variable we measure has normal distribution N(µ, σ). We know σ. These are of course not realistic. Even though the first two may be approximately satisfied, the third is usually quite unreasonable. We will learn later how to do without it. We concentrate on this case first because it is the easiest to understand. (There are occasional situations in which we do know σ but not µ, such as for the calibration of certain scientific instruments.) N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 19 / 36 Conditions for z confidence intervals and hypothesis tests We are using the “one sample z procedure”. We have talked about the procedure for confidence intervals; hypothesis tests are in the next chapter. Conditions: We have a simple random sample from the population in question. (In particular, no difficulties with such things as nonresponse.) The variable we measure has normal distribution N(µ, σ). We know σ. These are of course not realistic. Even though the first two may be approximately satisfied, the third is usually quite unreasonable. We will learn later how to do without it. We concentrate on this case first because it is the easiest to understand. (There are occasional situations in which we do know σ but not µ, such as for the calibration of certain scientific instruments.) N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 19 / 36 Conditions for z confidence intervals and hypothesis tests We are using the “one sample z procedure”. We have talked about the procedure for confidence intervals; hypothesis tests are in the next chapter. Conditions: We have a simple random sample from the population in question. (In particular, no difficulties with such things as nonresponse.) The variable we measure has normal distribution N(µ, σ). We know σ. These are of course not realistic. Even though the first two may be approximately satisfied, the third is usually quite unreasonable. We will learn later how to do without it. We concentrate on this case first because it is the easiest to understand. (There are occasional situations in which we do know σ but not µ, such as for the calibration of certain scientific instruments.) N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 19 / 36 Further examples Recall the general form of the confidence interval: Let σ be the population standard deviation. Let n be the sample size. Then the sampling standard deviation is σ/ √ n. Let z∗ be the appropriate critical value (usually found via Table C). Then the confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . It is also written: x ± z∗ ( σ√ n ) . N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 20 / 36 Further examples Spiral-horned snorkacks are much more common than crumple-horned snorkacks. J. K. Rowling has told us that their horn lengths (measured in cm) have a Normal distribution with standard deviation σ = 12. (Thus, σ is 12 cm.) Unfortunately, the mean µ is unknown. We want a 90% confidence interval for the mean horn length of spiral-horned snorkacks. (We will also find 95% and 99% confidence intervals.) Suppose we have managed to choose a simple random sample of spiral-horned snorkacks of size 100 (never mind how; with real animals, this is a significant issue). Suppose further that the sample mean horn length (the mean of the lengths of the horns of the 100 spiral-horned snorkacks in our sample) turned out to be x = 121.502 (again, measured in cm). N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 21 / 36 Further examples Spiral-horned snorkacks are much more common than crumple-horned snorkacks. J. K. Rowling has told us that their horn lengths (measured in cm) have a Normal distribution with standard deviation σ = 12. (Thus, σ is 12 cm.) Unfortunately, the mean µ is unknown. We want a 90% confidence interval for the mean horn length of spiral-horned snorkacks. (We will also find 95% and 99% confidence intervals.) Suppose we have managed to choose a simple random sample of spiral-horned snorkacks of size 100 (never mind how; with real animals, this is a significant issue). Suppose further that the sample mean horn length (the mean of the lengths of the horns of the 100 spiral-horned snorkacks in our sample) turned out to be x = 121.502 (again, measured in cm). N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 21 / 36 Further examples Spiral-horned snorkacks are much more common than crumple-horned snorkacks. J. K. Rowling has told us that their horn lengths (measured in cm) have a Normal distribution with standard deviation σ = 12. (Thus, σ is 12 cm.) Unfortunately, the mean µ is unknown. We want a 90% confidence interval for the mean horn length of spiral-horned snorkacks. (We will also find 95% and 99% confidence intervals.) Suppose we have managed to choose a simple random sample of spiral-horned snorkacks of size 100 (never mind how; with real animals, this is a significant issue). Suppose further that the sample mean horn length (the mean of the lengths of the horns of the 100 spiral-horned snorkacks in our sample) turned out to be x = 121.502 (again, measured in cm). N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 21 / 36 Further examples (continued) From the previous page, but in condensed form: Spiral-horned snorkacks have horn lengths (in cm) distributed N(µ, 12) with µ unknown. A simple random sample of 100 spiral-horned snorkacks has sample mean horn length 121.502 (again, in cm). Find a 90% confidence interval for the true mean horn length of spiral-horned snorkacks. (You should be able to solve problems stated in this condensed form.) The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . The population standard deviation is σ = 12 (given). x = 121.502. n = 100. For z∗, look at the entry under 90% in the 3rd last line of Table C, finding z∗ ≈ 1.645. Put these into the formula for the confidence interval, getting:( 121.502− (1.645) ( 12√ 100 ) , 121.502 + (1.645) ( 12√ 100 )) . N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 22 / 36 Further examples (continued) From the previous page, but in condensed form: Spiral-horned snorkacks have horn lengths (in cm) distributed N(µ, 12) with µ unknown. A simple random sample of 100 spiral-horned snorkacks has sample mean horn length 121.502 (again, in cm). Find a 90% confidence interval for the true mean horn length of spiral-horned snorkacks. (You should be able to solve problems stated in this condensed form.) The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . The population standard deviation is σ = 12 (given). x = 121.502. n = 100. For z∗, look at the entry under 90% in the 3rd last line of Table C, finding z∗ ≈ 1.645. Put these into the formula for the confidence interval, getting:( 121.502− (1.645) ( 12√ 100 ) , 121.502 + (1.645) ( 12√ 100 )) . N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 22 / 36 Further examples (continued) The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . We have σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.645. Put these into the formula for the confidence interval, getting: ( 121.502− (1.645) ( 12√ 100 ) , 121.502 + (1.645) ( 12√ 100 )) = (121.502− (1.645)(1.2), 121.502 + (1.645)(1.2)) ≈ (119.528, 123.476) . N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 23 / 36 Further examples (continued) Suppose that, instead of being asked for a confidence interval, we were asked for the margin of error for a 90% confidence interval. The margin of error is m = z∗ ( σ√ n ) . We have σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.645. (These are gotten the same way as for the confidence interval problem. However, we don’t actually need x .) Put these into the formula for the margin of error, getting: (1.645) ( 12√ 100 ) = (1.645)(1.2) ≈ 1.974. Back to the confidence interval problem: we can also write our confidence interval as x ±m (now we do need x), that is, 121.502± 1.974. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 24 / 36 Further examples (continued) Suppose that, instead of being asked for a confidence interval, we were asked for the margin of error for a 90% confidence interval. The margin of error is m = z∗ ( σ√ n ) . We have σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.645. (These are gotten the same way as for the confidence interval problem. However, we don’t actually need x .) Put these into the formula for the margin of error, getting: (1.645) ( 12√ 100 ) = (1.645)(1.2) ≈ 1.974. Back to the confidence interval problem: we can also write our confidence interval as x ±m (now we do need x), that is, 121.502± 1.974. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 24 / 36 Further examples (continued) Suppose that, instead of being asked for a confidence interval, we were asked for the margin of error for a 90% confidence interval. The margin of error is m = z∗ ( σ√ n ) . We have σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.645. (These are gotten the same way as for the confidence interval problem. However, we don’t actually need x .) Put these into the formula for the margin of error, getting: (1.645) ( 12√ 100 ) = (1.645)(1.2) ≈ 1.974. Back to the confidence interval problem: we can also write our confidence interval as x ±m (now we do need x), that is, 121.502± 1.974. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 24 / 36 Further examples (continued) Spiral-horned snorkacks have horn lengths (in cm) distributed N(µ, 12) with µ unknown. A simple random sample of 100 spiral-horned snorkacks has sample mean horn length 121.502 (again, in cm). Find a 95% confidence interval for the true mean horn length of spiral-horned snorkacks. (The only difference from before is the required confidence level, now 95% instead of 90%.) The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . We have σ = 12, x = 121.502, and n = 100. For z∗, look at the entry under 95% in the 3rd last line of Table C, finding z∗ ≈ 1.960. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 25 / 36 Further examples (continued) Spiral-horned snorkacks have horn lengths (in cm) distributed N(µ, 12) with µ unknown. A simple random sample of 100 spiral-horned snorkacks has sample mean horn length 121.502 (again, in cm). Find a 95% confidence interval for the true mean horn length of spiral-horned snorkacks. (The only difference from before is the required confidence level, now 95% instead of 90%.) The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . We have σ = 12, x = 121.502, and n = 100. For z∗, look at the entry under 95% in the 3rd last line of Table C, finding z∗ ≈ 1.960. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 25 / 36 Further examples (continued) Spiral-horned snorkacks have horn lengths (in cm) distributed N(µ, 12) with µ unknown. A simple random sample of 100 spiral-horned snorkacks has sample mean horn length 121.502 (again, in cm). Find a 95% confidence interval for the true mean horn length of spiral-horned snorkacks. (The only difference from before is the required confidence level, now 95% instead of 90%.) The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . We have σ = 12, x = 121.502, and n = 100. For z∗, look at the entry under 95% in the 3rd last line of Table C, finding z∗ ≈ 1.960. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 25 / 36 Further examples (continued) The margin of error is m = z∗ ( σ√ n ) . We have σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.960. Put these into the formula for the margin of error, getting: (1.960) ( 12√ 100 ) = (1.960)(1.2) ≈ 2.352. So we can also write our confidence interval as 121.502± 2.352. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 27 / 36 Further examples (continued) The margin of error is m = z∗ ( σ√ n ) . We have σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.960. Put these into the formula for the margin of error, getting: (1.960) ( 12√ 100 ) = (1.960)(1.2) ≈ 2.352. So we can also write our confidence interval as 121.502± 2.352. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 27 / 36 Further examples (continued) The margin of error is m = z∗ ( σ√ n ) . We have σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.960. Put these into the formula for the margin of error, getting: (1.960) ( 12√ 100 ) = (1.960)(1.2) ≈ 2.352. So we can also write our confidence interval as 121.502± 2.352. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 27 / 36 Further examples (continued) Spiral-horned snorkacks have horn lengths (in cm) distributed N(µ, 12) with µ unknown. A simple random sample of 100 spiral-horned snorkacks has sample mean horn length 121.502 (again, in cm). Find a 99% confidence interval for the true mean horn length of spiral-horned snorkacks. (The only difference from before is the required confidence level, now 99% instead of 90% or 95%.) The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . We have σ = 12, x = 121.502, and n = 100. For z∗, look at the entry under 99% in the 3rd last line of Table C, finding z∗ ≈ 2.576. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 28 / 36 Further examples (continued) Spiral-horned snorkacks have horn lengths (in cm) distributed N(µ, 12) with µ unknown. A simple random sample of 100 spiral-horned snorkacks has sample mean horn length 121.502 (again, in cm). Find a 99% confidence interval for the true mean horn length of spiral-horned snorkacks. (The only difference from before is the required confidence level, now 99% instead of 90% or 95%.) The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . We have σ = 12, x = 121.502, and n = 100. For z∗, look at the entry under 99% in the 3rd last line of Table C, finding z∗ ≈ 2.576. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 28 / 36 Further examples (continued) Spiral-horned snorkacks have horn lengths (in cm) distributed N(µ, 12) with µ unknown. A simple random sample of 100 spiral-horned snorkacks has sample mean horn length 121.502 (again, in cm). Find a 99% confidence interval for the true mean horn length of spiral-horned snorkacks. (The only difference from before is the required confidence level, now 99% instead of 90% or 95%.) The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . We have σ = 12, x = 121.502, and n = 100. For z∗, look at the entry under 99% in the 3rd last line of Table C, finding z∗ ≈ 2.576. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 28 / 36 Further examples (continued) The margin of error is m = z∗ ( σ√ n ) . We have σ = 12, x = 121.502, n = 100, and z∗ ≈ 2.576. Put these into the formula for the margin of error, getting: (2.576) ( 12√ 100 ) = (2.576)(1.2) ≈ 3.091. So we can also write our confidence interval as 121.502± 3.091. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 30 / 36 Further examples (continued) The margin of error is m = z∗ ( σ√ n ) . We have σ = 12, x = 121.502, n = 100, and z∗ ≈ 2.576. Put these into the formula for the margin of error, getting: (2.576) ( 12√ 100 ) = (2.576)(1.2) ≈ 3.091. So we can also write our confidence interval as 121.502± 3.091. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 30 / 36 Further examples (continued) The margin of error is m = z∗ ( σ√ n ) . We have σ = 12, x = 121.502, n = 100, and z∗ ≈ 2.576. Put these into the formula for the margin of error, getting: (2.576) ( 12√ 100 ) = (2.576)(1.2) ≈ 3.091. So we can also write our confidence interval as 121.502± 3.091. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 30 / 36 Compare the margins of error Our results: Confidence level z∗ Confidence interval Margin of error 90% 1.645 (119.528, 123.476) 1.974 95% 1.960 (119.150, 123.854) 2.352 99% 2.576 (118.411, 124.593) 3.091 Insisting on higher confidence requires a larger z∗, so gives a larger margin of error, and hence a larger confidence interval. If you want to be more confident that the true mean is in your interval, you must use a larger interval. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 31 / 36 A further example You have a scientific instrument which measures the amount of uranium hexaflouride in air samples. For any given air sample, the results of using this instrument vary normally with mean the true amount of uranium hexaflouride in the air sample and standard deviation 12.5 mg (milligrams). You use this instrument on 10 one liter bottles of air from a room suspected to be polluted with uranium hexaflouride. The air is well mixed, so the concentration is the same in all the bottles. The results (in mg) are: 236 254 258 269 258 261 247 251 272 239 Assuming your measurements can be treated as a simple random sample of all possible measurements of air from this room with this instrument (a common and reasonable assumption for repeated use of a measuring device), find a 98% confidence interval for the concentration of uranium hexaflouride in the air in this room. Warning: The word “sample” is used in two different senses here! One is “air sample”; the other is as in simple random sample. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 32 / 36 A further example You have a scientific instrument which measures the amount of uranium hexaflouride in air samples. For any given air sample, the results of using this instrument vary normally with mean the true amount of uranium hexaflouride in the air sample and standard deviation 12.5 mg (milligrams). You use this instrument on 10 one liter bottles of air from a room suspected to be polluted with uranium hexaflouride. The air is well mixed, so the concentration is the same in all the bottles. The results (in mg) are: 236 254 258 269 258 261 247 251 272 239 Assuming your measurements can be treated as a simple random sample of all possible measurements of air from this room with this instrument (a common and reasonable assumption for repeated use of a measuring device), find a 98% confidence interval for the concentration of uranium hexaflouride in the air in this room. Warning: The word “sample” is used in two different senses here! One is “air sample”; the other is as in simple random sample. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 32 / 36 Check for approximate normality of your data One should verify that the data do not obviously depart from normality. For this purpose, here is a stemplot: 23 6 9 24 7 25 1 4 8 8 26 1 9 27 2 The shape is somewhat irregular, but this happens with small samples from truly normal distributions. There are no outliers, and it is roughly symmetric, so we proceed. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 33 / 36 Check for approximate normality of your data One should verify that the data do not obviously depart from normality. For this purpose, here is a stemplot: 23 6 9 24 7 25 1 4 8 8 26 1 9 27 2 The shape is somewhat irregular, but this happens with small samples from truly normal distributions. There are no outliers, and it is roughly symmetric, so we proceed. N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 33 / 36 Compare with Lecture 2: 100 observations from a normal distribution should be much more regular than 10, but here is what we actually got: -4 -2 2 4 5 10 15 20 N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 34 / 36 We have σ = 12.5, x = 254.5, n = 10, and z∗ ≈ 2.326. The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . Put the numbers into the formula for the confidence interval, getting:( 254.5− (2.326) ( 12.5√ 10 ) , 254.5 + (2.326) ( 12.5√ 10 )) . ≈ (254.5− (2.326)(3.95285), 254.5 + (2.326)(3.95285)) ≈ (245.306, 263.694) . One should probably round to four significant digits: (245.3, 263.7) . N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 36 / 36 We have σ = 12.5, x = 254.5, n = 10, and z∗ ≈ 2.326. The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . Put the numbers into the formula for the confidence interval, getting: ( 254.5− (2.326) ( 12.5√ 10 ) , 254.5 + (2.326) ( 12.5√ 10 )) . ≈ (254.5− (2.326)(3.95285), 254.5 + (2.326)(3.95285)) ≈ (245.306, 263.694) . One should probably round to four significant digits: (245.3, 263.7) . N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 36 / 36 We have σ = 12.5, x = 254.5, n = 10, and z∗ ≈ 2.326. The confidence interval is( x − z∗ ( σ√ n ) , x + z∗ ( σ√ n )) . Put the numbers into the formula for the confidence interval, getting:( 254.5− (2.326) ( 12.5√ 10 ) , 254.5 + (2.326) ( 12.5√ 10 )) . ≈ (254.5− (2.326)(3.95285), 254.5 + (2.326)(3.95285)) ≈ (245.306, 263.694) . One should probably round to four significant digits: (245.3, 263.7) . N. Christopher Phillips () Math 243: Lecture File 8 23 April 2009 36 / 36