Statistical Inferences on Means: Confidence Intervals and Hypothesis Tests
1. One category of lymphoma is non-Hodgkin lymphoma, which is the classification of many different types. Of interest is to estimate m = the mean survival time, in years, of a patient diagnosed with non-Hodgkin lymphoma. For this problem only, assume that the standard deviation of the survival times for all patients diagnosed with non-Hodgkin lymphoma is 9.2 years. What is the minimum number of non-Hodgkin lymphoma patients that would need to be selected to allow the calculation of a 95% confidence interval with margin of error no greater than 3.5 years? Please circle your final answer.
n =
= (5.152)2 = 26.54; Round up to 27 patients
2. A simple random sample of 51 non-Hodgkin lymphoma patients was selected and their survival time, from initial diagnosis to death, was determined for each. The mean survival time for this sample of 51 patients was 17.8 years with a standard deviation of 4.6 years, and the distribution was heavily skewed to the right since a few patients had lived many years between initial diagnosis and death. If appropriate, use this information to calculate and interpret a 95% confidence
We have a simple random sample, and the sample size is large enough for the Central Limit Theorem to apply (n = 51 > 40), so the assumptions are satisfied. The population standard deviation 
is unknown, so we must use the t-distribution.
95% CI implies 
= 2.009.
So 
= (16.51, 19.09)
[Calculator gives (16.506, 19.094)]
We have 95% confidence that the mean survival time of all non-Hodgkin lymphoma patients is between 16.51 years and 19.09 years
For questions 3 and 4 the answer choices for both multiple choice questions are as follows.
(D) Both the margin of error and the width would increase.
_____ 3. In question 2 a confidence interval was computed based on a sample of 51 patients. If the number of patients in the sample were decreased to 41, what impact would this have on the margin of error and width of the confidence interval?
_____ 4. In question 2 a 95% confidence interval was computed. If the confidence level were increased from 95% to 99% confidence, what impact would this have on the margin of error and the width of the confidence interval?
5. If diagnosed early lymphoma is often curable, and it is usually diagnosed in older people and not the young. Suppose that the ages of patients when diagnosed with lymphoma are skewed heavily to the left (since diagnosis in the young is rare) with a mean of 61 years of age and a standard deviation of 10.4 years of age. If a simple random sample of 64 lymphoma patients is selected and the age of each at diagnosis recorded, describe completely the sampling distribution of 
, the resulting mean age at diagnosis of this sample of 64 lymphoma patients.
We have a simple random sample, and the sample size is large enough to apply the Central Limit Theorem (n = 64 > 40). So 
= 61; 
1.3; since the CLT applies, the shape is normal. Hence 
(61, 1.3).
6. For patients diagnosed with non-Hodgkin lymphoma it is known that most patients diagnosed are older and hence the distribution of the ages of patients when diagnosed with non-Hodgkin lymphoma is skewed heavily to the left. However, m = the mean age at diagnosis of all people diagnosed with non-Hodgkin lymphoma is not known. It is conjectured that this mean age is 55 years, and researchers want to test this versus the alternative that the mean age at diagnosis of all people diagnosed with non-Hodgkin lymphoma is different from 55 years. State the appropriate null and alternative hypotheses that should be tested. 6. H0: 
= 55 versus Ha: 
55
7. Consider all the information in question 6, and assume that the standard deviation of the ages at diagnosis for all people diagnosed with non-Hodgkin lymphoma is 10.4 years. To test the hypotheses in question 6, data for the 12 patients most recently diagnosed with non-Hodgkin lymphoma at VCU-MCV Hospitals was recorded. The mean age of these 12 patients was 58.75 years. If appropriate, use this information to test the hypotheses stated in question 6 at the a = .05 level of significance.
The assumptions are not met – the sample was not a simple random sample, and the population was heavily skewed to the left, so the sample size of 12 is too small (n = 12
11. If a low-grade lymphoma is becoming symptomatic, radiotherapy or chemotherapy are the treatments of choice; although they do not cure the lymphoma, they can alleviate the symptoms, particularly painful lymphadenopathy. Patients with these types of lymphoma can live near-normal lifespans, but the disease is incurable. Treatment of some other, more aggressive, forms of lymphoma can result in a cure in the majority of cases, but the prognosis for patients with a poor response to therapy is worse. Treatment for these types of lymphoma typically consists of aggressive chemotherapy. For patients treated with chemotherapy, of interest is m = the mean number of days between the first and last treatment. It is conjectured that the mean number of days between the first and last chemotherapy treatment is 42 days, and of interest is to test this conjecture versus the alternative that the mean number of days between the first and last chemotherapy treatment is less than 42 days. State the appropriate null and alternative hypotheses that should be tested
. H0:
= 42 versus Ha:
12. Consider the information and hypotheses specified in question 11. A simple random sample of 28 lymphoma patients is selected and the number of days between the first and last chemotherapy treatment was recorded for each. The mean for this sample of 28 observations was 38.25 days, the standard deviation for this sample of 28 observations was 7.75, and the distribution of the data was fairly symmetrical. If appropriate, use this information to test the hypotheses stated in question 11 at the a = .10 level of significance.
(1) Hypotheses were stated in question 11. 
= .10
(2) We had a simple random sample, and the sample size of 28 is large enough for the Central Limit Theorem to apply (n = 28 > 15). Therefore, the assumptions are satisfied. The population standard deviation
is unknown, so we must use the t-distribution.
t =
= –2.560
(3) p-value = P(t27 ≤ –2.560) = P(t27 ≥ 2.560) = .005
(calculator gives .0082)
(4) Since p-value
(5) There is sufficient evidence that the mean number of days between the first and last chemotherapy treatment is less than 42 days.
2. (8 points) In the current chapter covering statistical inferences on means, all of the procedures we have discussed (sampling distributions, confidence intervals and statistical tests) have involved two assumptions. What are those two assumptions?
1. Simple random sample
2. Normal population or large enough sample for the Central Limit Theorem to apply.
_____ 8. Is the mean amount of money paid per month on rent by all renters in the state of New York in the confidence interval computed in question 7?
- The mean amount of money paid per month on rent by all renters in the state of New York is not known and hence we do not know if it is in the interval or not.
_____ 9. If all other things are held constant, what impact would increasing the confidence level have on the margin of error and width of a confidence interval?
(A) Both the margin of error and the width would increase.