Section 7.2: Confidence Intervals for a Population Mean, σ Unknown

An IQ test is designed to have scores that have a standard deviation of σ= 15. A simple random sample of students at a large university will be given the test in order to construct a 98% confidence interval for the mean IQ of all students at the university. How many students must be tested so that the margin of error will be equal to 3 points?

98% Z*=2.326 m=3 σ=15 n=(2.326(15)/3)^2= 135.3 -> 136 students

The Critical Value tα/2 (or t*)

The critical value tα/2 can be found in Table A.3, in the row corresponding to the number of degrees of freedom and the column corresponding to the desired confidence level or by technology.

company has developed a new type of lightbulb and wants to estimate its mean lifetime. A simple random sample of 100 bulbs had a sample mean lifetime of 750.2 hours with a sample standard deviation of 30 hours. a) Verify that the assumptions are satisfied. b) Construct and interpret a 95% confidence interval for the population mean lifetime of all bulbs manufactured by this new process.

a) n=100 (random, ~normal, n>30) x=750.2 s=30 b) TInt-> (744.3, 756.2) We are 95% confident that mean lifetime for these bulbs is between 744.3 hours and 756.2 hours

Comparing Confidence Levels: 70% 95% 99.7%

although the MOE is smaller, they cover the pop mean only 70% of the time this represents a good compromise between reliability and MOE for many purposes they almost always succeed in covering the pop mean, but their MOE is large

A food chemist analyzed the calorie content for a popular type of chocolate cookie. Following are the numbers of calories in a sample of eight cookies. 113, 114, 111, 116, 115, 120, 118, 116 Find and interpret a 98% confidence interval for the mean number of calories in this type of cookie.

construct boxplot -> normal t*=invT(.01,7)= 2.998 (7=df) x=115.375 s=2.825 115.375+/-2.998(2.825/sqrr(8)= (112.4, 118.4) -> round final result we are 98% confident the mean number of calories is between 112.4 and 118.4 cal.

If we want to be more confident that our interval contains the true value, we must

increase the critical value, which increases the margin of error. There is a trade-off. We would rather have a higher level of confidence than a lower level, but we would also rather have a smaller margin of error than a larger one.

We would rather have a smaller margin of error than a larger one in order for the confidence interval to be more useful. We can make the margin of error smaller if we are willing to reduce our ________________________________.We can also reduce the margin of error by increasing the ___________________

level of confidence, sample size *If the population standard deviation is known, the margin of error is m=Zα/2⋅σ/sqrr(n), which may be rewritten as n=(Zα/2⋅σ/m)^2

Let m represent the margin of error. • If the population standard deviation is known, the necessary sample size is • In cases where the population standard deviation is not known, sis reasonably close to σfor large samples. Given a large sample (n> 30), the necessary sample size is • If the value of nis not a whole number,

n=(Zα/2⋅σ/m)^2 n=(Zα/2⋅S/m)^2 round it up to the nearest whole number.

A simple random sample of size 10 is drawn from a normal population. Find the critical value tα/2 for a 95% confidence interval.

n=10 df= 10-1=9

A potato chip company wants to evaluate the accuracy of its potato chip bag-filling machine. Bags are labeled as containing 8 oz. of chips. A SRS of 12 bags has a mean weight of 8.12 oz. with a standard deviation of 0.1 oz. Construct and interpret a 99% CI for the population mean weight of bags of chips.

n=12 (don't know if ~normal, SRS) x=8.12 s=.1 99% TInt-> (8.03, 8.21) We are 99% confident the mean weight is between 8.03 ounces and 8.21 ounces. The machine appears to be overfilling.

A sample of 123 people aged 18-22 reported the number of hours they spent on the internet in an average week. The sample mean was 8.20 hours, with a sample standard deviation of 9.84 hours. Assume this is a simple random sample from the population. Construct and interpret a 95% confidence interval for μ, the population mean number of hours per week spent on the internet in that age group.

n=123 (n>30 and SRS) x=8.2 s=9.84 95% TInt->(6.44,9.96) round same decimal place as the stats (2) we are 95% confident the mean number of hours per week spent on the internet between 6.44 and 9.96 hours per week

A machine used to fill beverage cans is supposed to put exactly 12 ounces of beverage in each can, but the actual amount varies randomly from can to can. In a sample of 50 cans, the standard deviation of the amount was s= 0.05 ounces. A simple random sample of filled cans will have their volumes measured, and a 95% confidence interval for the mean fill volume will be constructed. Estimate the number of cans that must be sampled for the margin of error to be equal to 0.01 ounces.

n=50 s=.05 95% Z*=1.96 m=.01 n=(1.96(.05)/.01)^2= 96.04 -> 97 cans

Scientists want to estimate the mean weight of mice after they have been fed a special diet. A sample of 60 mice have been weighed, and the standard deviation was s= 3 grams. Estimate the number of mice that must be weighed for a 95% confidence interval to have a margin of error of 0.5 grams.

n=60 s=3 m=.5 95% Z*=1.96 n=(1.96(3)/.5)^2= 138.3 -> 139 mice

Following are interest rates for a 30-yr fixed-rate mortgage in Macon GA. It is reasonable to assume that the population is approximately normal. Construct and interpret a 98% CI for the population mean mortgage rate. 4.750 4.375 4.176 4.679 4.426 4.2274.125 4.250 3.950 4.191 4.299 4.415 One week earlier, the mean rate was 4.050%. A broker claims that the mean rate is now higher. Is this a reasonable claim? explain

random? -> ~normal TInt-> (4.1444, 4.4995) round 1 more dec place than the data (4) We are 98% confident the main mortgage rate is between 4.1444% and 4.4995% yes, the entire CI is above 4.05%

95%, sample size of 15 99%, sample size of 22 90%, sample size of 63 95%, sample size of 2

t*=invT(.025,14)= 2.145 t*=invT(.005,21)= 2.831 t*=invT(.05,62)= 1.670 t*=invT(.025,1)= 12.706

When constructing a confidence interval where the population standard deviation σ is known. It is rare that we would know the value of σ and not know μ. When we don't know the value of σ, we replace it with

the sample standard deviation s. However, we cannot then use z* as the critical value because the quantity x-μ/(s/sqrr(n) does not have a normal distribution. The distribution of this quantity is called the Student's t distribution. *Degrees of Freedom for the Student's t Distribution: n-1

What we do know is that our confidence interval was constructed by a method that succeeds 95% of the time. The confidence level describes

the success rate of the method used to construct a confidence interval, not the success of any specific interval

Student's t distributions are _______________ and _______________, just like the normal distribution. However, they are more ____________ because sis, on average, a bit smaller than σ. The distribution looks almost normal but has ____________ ____________.

unimodal, symmetric, spread out, fatter tails *The larger the df, the more "normal" the distribution becomes.

Decide in each case whether the methods of this section should be used to construct a confidence interval for the population mean. Assume that σ is unknown. A simple random sample of size 8 is drawn from a distribution that is approximately normal. A simple random sample of size 15 is drawn from a distribution not known to be normal. A simple random sample of size 150 is drawn from a distribution not known to be normal. A nonrandom sample is drawn of size 212. A simple random sample with the following dotplot.

yes (random, ~normal) no, n<30 yes, n>30 no (nonrandom) no, not ~norm and outlier

Confidence Intervals Using the TI-Plus84

~The Tinterval command constructs confidence intervals when the population standard deviation σis not known. This command is accessed by pressing STAT and highlighting the TESTS menu. ~If the summary statistics are given the Stats option should be selected for the input option. ~If the raw sample data are given, the Data option should be selected.

Finding a Critical Value Using Technology:

~Various forms of technology can be used to find critical values from a Student's t distribution. ~The following screenshots demonstrate finding the critical value for a 95% confidence interval from a Student's t distribution with 122 degrees of freedom. ~Recall that α/2=0.0252 with a 95% confidence level. ~invT(area to left, df)

If the desired number of degrees of freedom isn't listed in Table A.3, then

• If the desired number is less than 200, use the next smaller number that is in the table. • If the desired number is greater than 200, use the Z-value found in the last row of Table A.3, or use Table A.2.

The assumptions for constructing a confidence interval for a population mean are:

• We have a simple random sample. • The sample size is large (n> 30), or the population is approximately normal. *note: When the sample size is small (n≤ 30), we must check to determine whether the sample comes from a population that is approximately normal, Section 6.5.