# The purpose of this assignment is to help you understand how to (1) construct and interpret confidence and (2) perform sample size calculations. Study 1 To assess the accuracy of a laboratory...

The purpose of this assignment is to help you understand how to (1) construct and interpret confidence and (2) perform sample size calculations.

*Study 1*

To assess the accuracy of a laboratory scale, a standard weight known to weigh 10 grams is weighed repeatedly. The scale readings are normally distributed with unknown mean (this mean is 10 grams if the scale has no bias). The standard deviation of the scale readings is known to be 0.0002 grams.

(a) The weight is weighed 5 times. The mean result is 10.0023 grams. Give a 98% confidence interval for the mean of repeated measurements of the weight. Use the 4-step method for confidence intervals. (See handout from class that is also posted on Blackboard for the steps of the method).

(b) How many measurements must be averaged to get a margin of error of ±0.0001 with 99% confidence?

*Study 2*

A random sample of 1000 high schools students grains an average of 22 points in their second attempt at the SAT mathematics exam. The change in score has a normal distribution with standard deviation 50.

(a) Give a 95% confidence interval for the mean score gain in the population of all students. Use the 4-step method for confidence intervals.

(b) Scores from how many students must be averaged to get a margin of error of ±3 with 90% confidence?

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### 1 Answer

(1)(a) We are given a sample of size 5 with a sample mean of `bar(x)=10.0023 ` and a population standard deviation of `sigma=.0002 ` . We want a 98% confidence interval of the population mean.

The interval is given by the sample mean plus or minus the product of the confidence and the standard error.

Since we know the population standard deviation we use a z-interval. With `alpha=.02 ` we find `z_(alpha/2)=2.33 ` . (With the confidence at 98%, 1% of the sample means will lie below and 1% above the interval. Find .9900 in the standard normal table (or the opposite of the z associated with .0100) to get 2.33)

The standard error is the quotient of the standard deviation and the square root of the sample size: `sigma/sqrt(n)=.0002/sqrt(5)~~.00009 `

The interval is given by:

`bar(x)-z_(alpha/2)sigma/sqrt(n)<mu<bar(x)+z_(alpha/2)sigma/sqrt(n) `

`10.0023-(2.33)*.0002/sqrt(5)<mu<10.0023+2.33*.0002/sqrt(5) `

`10.00209<mu<10.00251 `

(b) The error is given by `E=((z_(alpha/2)sigma)/sqrt(n)) ` . Solving for n we get:

`n=((z_(alpha/2)sigma)/E)^2 ` Here `z_(alpha/2)=2.576,sigma=.0002,E=.0001 `

Note that `z_(alpha/2) ` changed from 2.33 to 2.576 as the confidence level changed from 98% 99%.

`n=(2.576*.0002/.0001)^2~~26.543 `

So we need at least 27 measurements.

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