SOLUTIONS TO THE THIRD EXAM

FALL, 1999

  1. Selasphorus platycerus is a scientific name of the broad-tailed hummingbird that is commonly found throughout the western United States. It is known that the mean incubation time for eggs of this bird is 16.5 days (The Hummingbird Book b y Stokes and Little). However for higher elevations, it is thought that the average incubation time might be different from 16.5 days. A number of broad-tailed hummingbird nests were located in mountain country above 8000 feet, and incubation times for 18 eggs revealed a mean incubation time of 17.83. Assume that the population standard deviation is known to be 2.20.
  1. Construct a 90% confidence interval for the population mean of the incubations times for higher elevations.
  2. How large a sample should be to estimate the mean incubation time to within 0.5 days with 90% confidence.
  3. Does this information indicate that the population average incubation time above 8000 feet elevation is different from 16.5 days? Use a=0.05.
. Reject the null hypothesis at 0.05 level.

 

 

  1. A hospital reported that the normal death rate for patients with extensive burns (more than 40% of the skin area) has been significantly reduced by the use of new fluid plasma compress. Before the new treatment, the mortality rate for extensive bu rn patients was about 60%. Using the new compresses, the hospital found that only 40 of 90 patients with extensive burns died.
  1. Construct a 95% confidence interval for the true proportion of the patients who would die under the new treatment.
  2. Determine the sample size required so that the estimate will be within 0.05 of the true proportion with 95% confidence.
  3. Use 0.01 significance level to test the claim that the mortality rate has dropped.
Reject the null hypothesis. The mortality rate has dropped.

 

3. Suppose you want to test using a=0.05. A random sample of size 100 will be drawn from the population in question. Assume that t he population has a standard deviation equal to 1.0.

  1. If m were really equal to 9.9, what is the probability of committing Type II error.
  2. What is the power of this test for detecting the alternative
.

Power is 1-0.7422=0.2578.

 

 

 

PLEASE ANSWER EITHER ONE OF THE FOLLOWING TWO QUESTIONS

4. A new kind of typhoid shot is being developed by a medical research team. The old typhoid shot was know to protect the population for a mean of 36 months with a standard deviation of 3 months. To test the variability of the new shot, a r andom sample of 24 people was given the new shot. The regular blood tests showed that the sample standard deviation of protection time was 1.9 months.

  1. At a 0.05 significance level, test the claim that the typhoid shot has a smaller variance of protection times.
  2. The shot has a smaller variance of protection times.

  3. Set up a 90% confidence interval for the population standard deviation.

4. A group of environmentalists claims that the warm water from Virginia Power plant at Lake Anna is limiting the growth of striped bass at the lake. In a similar southeastern reservoir the average weight of striped bass is 25 pounds with a sta ndard deviation of 8 pounds. A study of 30 striped bass at the Lake Anna revealed a mean weight of 20 pounds.

  1. Do these data support the environmentalists? Use the 0.01 level of significance.
  2. Set up a 90% confidence interval for the population mean weight of striped bass.

 

 

 

 

5. A biologist suspected that males age 20-24 have a higher mean systolic blood pressure than females in the same age group. Independent random samples of males and females were selected from the Framingham Heart Study. The data are as follows:

 

n

s

Males

31

125

13.9

Females

41

117

12.1

  1. Give a 99% confidence interval for the mean difference in systolic blood pressure between the males and females.
  2. At a 1% level of significance, is there sufficient evidence to justify the biologist's suspicion that males have a higher mean systolic blood pressure than females