Question 635:

Asked on February 22, 2009

Tags: Confidence Intervals , 1-sample t-test

1

Answer:

No answer provided yet.

The average of the sample above is the sum of all roadmarking values divided by the total = 4692/17 = 267. 

 

The next thing we need to do is determine if the mean of this sample is above the minimum acceptable average reflectivity of 100. To do so we can compute a 1-sample t-test against the mean of 100. We will also need the standard deviation of the sample which can be found using excels' STDEV() function. We get 108.8761.

 

The null hypothesis is that mean reflectivity is less than 100. The alternative hypothesis is the mean reflectivity is greater than 100.

 

Our test statistic is made up of the difference between the sample mean and test mean divided by the standard error of the difference (SE). The difference is 176. The standard error is the standard deviation divided by the square root of the sample size = 95.35619/SQRT(17) = 26.40633 

 

This makes our test statistic t = 176/26.40633  = 6.665

 

To find the p-value associated with this statistic we use the excel function =TDIST(6.665,16,1) or the T distribution calculator http://www.usablestats.com/calcs/tdist and get a p-value of less than .0001.

 

With a p-value this low we'd reject the Null Hypothesis and conclude the mean reflectivity is greater than 100.

 

 

We can also generate a 95% confidence interval around the mean. A confidence interval is made up of a margin or error, which is added and subtracted from the mean. The margin of error is found as the standard error of the mean(SEM) times a critical value from the t-distribution corresponding to the level of confidence. We are using the t, instead of z (normal) distribution because the sample size is relatively small and we are not given the population standard deviation. To find the critical t-value we can use the excel function =TINV(.05,17) = 2.10982 or use the Inverse t calculator here http://www.usablestats.com/calcs/tinv. The parameters are 1-confidence level and the degrees of freedom, or n-1.

 

Next to find the standard error of the mean, we divide the standard deviation by the square root of the sample size = 108.876 /SQRT(17) =26.4063.  Now we multiply this value times the critical value 26.4063*2.10982 = 55.97891, which becomes our margin of error.

 

To build the interval we add and subtract this to the mean: 267-55.97891  and 267+55.97891  equals a 95% confidence interval between 220 and 332.

 

The next part of the question asks us to generate an estimate of the percent of roadmarkings that are below the minimum standard of 80. In looking at the sample, we see there is only 1 value (56) out of the 17 that is below the specification minimum of 80. This gives us an observed percentage of 1/17= 5.88%. To compute a confidence interval around the proportion for a small sample size we should use the adjusted wald method which is explained here: http://www.measuringusability.com/wald.htm.

 

We adjust our proportion by adding ½ the critical z-value squared to the numerator and 1 squared z-value to the denominator. This gives us a new proportion of 2.9207/20.84146 or an adjusted p of .14014

 

The standard error of the mean is found using these adjusted values SQRT(pq/n) = .076038. We multiply this times the z-critical value of 1.96 and get a margin of error of .149032. We add and subtract this to the adjusted p to get a 95% confidence interval between <.00001 and .289172. In other words, we can be 95% confident the true proportion of roadmarkings is between .001% and 28.9172%.