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Question 633:



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The average of the sample above is the sum of all roadmarking values divided by the total = 6151/23 = 267.4348. 


The next thing we need to do is determine if the mean of this sample is above the minimum acceptable average reflectivity of 150. To do so we can compute a 1-sample t-test against the mean of 150. We will also need the standard deviation of the sample which can be found using excels STDEV() function. We get 95.35619.


The null hypothesis is that mean reflectivity is less than 150. The alternative hypothesis is the mean reflectivity is greater than 150.


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 117.44. The standard error is the standard deviation divided by the square root of the sample size = 95.35619/SQRT(23) = 19.883. 


This makes our test statistic t = 117.44/19.883 = 5.906.


To find the p-value associated with this statistic we use the excel function =TDIST(5.906,22,1) 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 wed reject the Null Hypothesis and conclude the mean reflectivity is greater than 150.



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(.05,22) = 2.0738 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 = 95.35619/23 = 19.88314.  Now we multiply this value times the critical value 19.88314*2.0738 = 41.235, which becomes our margin of error.


To build the interval we add and subtract this to the mean: 267.4348-41.235 and 267.4348+41.235 equals a 95% confidence interval between 226.1997 and 308.6699.


The next part of the question asks us to generate an estimate of the percent of roadmarkings that are below the minimum standard of 120. In looking at the sample, we see there is only 1 value (101) out of the 23 that is below the specification minimum of 120. This gives us an observed percentage of 1/23 = 4.3%. 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 the critical z-value squared to the numerator and 1 squared z-value to the denominator. This gives us a new proportion of 2.921/26.8415 or an adjusted p of .108814.


The standard error of the mean is found using these adjusted values SQRT(pq/n) = .06017. We multiply this times the z-critical value of 1.96 and get a margin of error of .1178. We add and subtract this to the adjusted p to get a 95% confidence interval between <.00001 and .2266. In other words, we can be 95% confident the true proportion of roadmarkings is between .01% and 22.66%.

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