Hypothesis Test – Population Median Fictitious Striped Lizard The Fictitious Striped Lizard (FSL) is a native species of Lizard Island, and is noteworthy for the both the quantity and quality of its stripes. Consider a random sample of FSL, in which the number of stripes per lizard is noted: 0, 0, 0, 1, 2, 3, 5, 6, 10, 10, 12, 14, 14, 15, 25, 33, 39, 40, 43, 50, 56, 56, 56, 60, 63, 65, 66, 67, 98, 105 Test the following: null (H 0 ): The median number of stripes per FSL is 15 (  = 15) against the alternative (H 1 ):  < 15.

Show your work. Completely discuss and interpret your test results, as indicated in class and case study summaries. Fully discuss the testing procedure and results. This discussion must include a clear discussion of the population and the null hypothesis, the family of samples, the family of errors and the interpretation of the p-value.

Numbers n = 30 (H 0 ):  = 15 and (H 1 ):  < 15, so error = {sample points < 15} = 13 p-value from row: p-value .8192

from the median test table… n computed error base p-value

Population: Fictitious Striped Lizards Population Median: Median Number of Stripes per Lizard Family of Samples: Each member is a single random sample of 30 Fictitious Spotted Lizards For each member of the FoS, compute the number of sample lizards with less than 15 stripes. Doing so for every member of the FoS yields the Family of Errors. If the true population median number of stripes is 15, then the probability of getting a sample as bad or worse than our sample is approximately 81.92%. This sample does not seem to present significant evidence against the Null Hypothesis. We do not reject the Null Hypothesis at either 5% or 1% significance.

Median Test Training Program Success Rates WidgetCorps  trains employees in a program designed to help them make better widgets. These classes are ongoing, and the company wants to know if the classes are successfully improving employee performance. A random sample of classes is scored by the percentage of employees in the class who show marked improvement in performance after the class. The percentage of employees showing markedly improved performance after the class is listed below: 22, 24, 25, 30, 40, 43, 45, 46, 47, 50, 51, 52, 55, 56, 57, 58, 60, 60, 63, 65, 65, 66, 67, 67, 70, 71, 72, 80, 85, 90 Test the following Null Hypothesis:  =  0 = 50 percent employees showing marked improvement after the class against the Alternative Hypothesis that  > 50 percent. Show your work. Completely discuss and interpret your results, as indicated in class and case study summaries.

Test the following Null Hypothesis:  =  0 = 50 percent employees showing marked improvement after the class against the Alternative Hypothesis that  > 50 percent. Show your work. Completely discuss and interpret your results, as indicated in class and case study summaries.

n=30 (H 0 ):  = 50 and (H 1 ):  > 50, so error = {sample points > 50} = of 30 classes had more than 50% of students with marked improvement. Using the table, we find this row: , which suggest an approximate p- value of.04937, or 4.937%.

from the median test table… n computed error base p-value

Discussion Population: WidgetCorps  classes. Population Median: Population median proportion of class members showing marked improvement. Family of Samples: Each member of the FoS is a single random sample of n=30 classes. Family of Errors: An error is computed from each member of the FoS as follows: FoS → Compute number of classes in sample with more than 50% of class members showing marked improvement Repeating this process for every member of the FoS yields a family of errors (FoE). If the median proportion of class members showing marked improvement is 50%, then approximately 4.937% of the Family of Samples will show errors equal to or exceeding our single computed error of 20. Since this p-value is less than.05 or 5%, we can say that this sample seems to present significant evidence against the null hypothesis.

Hypothesis Test – Population Median Generic Fictitious Spiders We have a sample of Generic Fictitious Spiders. Each spider's diameter (maximum length, in cm, from leg tip to leg tip). The spider diameters are listed below: 16.5, , 22.8, 23.5, , 28.3, , , , 31.7, 32.4, , 35.0, 35.8, 36.7, 38.4, 38.4, , , , 52.4 Let  denote the population median spider diameter. Test the null hypothesis that  = 25 centimeters against   25.

Show your work. Completely discuss and interpret your test results, as indicated in class and case study summaries. Fully discuss the testing procedure and results. This discussion must include a clear discussion of the population and the null hypothesis, the family of samples, the family of errors and the interpretation of the p-value.

n=30 (H 0 ):  = 25 and (H 1 ):   25, so Error = max of {#{spiders in sample with diam < 25}, #{spiders in sample with diam > 25} 16.5, 21.9, 22.0, 22.8, 23.5, 23.7, 24.4, 28.3, , , , 31.7, 32.4, 33.1, 34.1, 35.0, 35.8, 36.7, 38.4, 38.4, , 39.5, 40.2, 41.9, 43.3, 52.4 #{spiders in sample with diam < 25} = 7 #{spiders in sample with diam > 25} = 23 Our error = max{7,23} = 23, with n=30. Referring a p-value: ; p  2*  .522%

from the median test table… n computed error base p-value

Our population of interest consists of cases of Generic Fictitious Spiders. Each member of our Family of Samples is a single random sample of 30 spiders. Our Family of Samples consists of all such samples. An error is obtained from each member of the FoS as: max{#{spiders in sample with diam < 25}, #{spiders in sample with diam > 25}} If the true population median diameter for the Generic Fictitious Spider population is 25 cm, then approximately.522% of the Family of Samples yield errors as bad as or worse than our sample. We reject the null median of 25 centimeters at 1% and 5% significance.