IHP 525 Module Six Problem Set

1. A pharmacist reads that a 95% confidence interval for the average price of a particular prescription drug is $30.50 to $35.50. Asked to explain the meaning of this, the pharmacist says “95% of all pharmacies sell the drug for between $30.50 and $35.50.” Is the pharmacist correct? Explain your response.

1. A pharmacist reads that a 95% confidence interval for the average price of a particular prescription drug is $30.50 to $35.50. Asked to explain the meaning of this, the pharmacist says “95% of all pharmacies sell the drug for between $30.50 and $35.50.” Is the pharmacist correct? Explain your response.
   1. The pharmacist is wrong because given a 95% confidence interval is applicable to the population mean. The confidence interval is not applicable to distribution of individual observations though.

 

2. Hemoglobin levels in 11-year-old boys vary according to a normal distribution with σ=1.2 g/dL.
3. How large a sample is needed to estimate µ with 95% confidence so the margin of error is no greater than 0.5 g/dL?
   1. n = (Z_{(1 -a/2)}^x (σ/m) )² = [ (1.2 / 0.5) x 1.96 ]² = 22.1 = 22
4. How large a sample is needed to estimate µ with margin of error 0.5 g/dL with 99% confidence?
   1. n = (Z_{(1 -a/2)}^x (σ/m) )² = [ (1.2 / 0.5) x 2.58 ]² = 38.3 = 38

 

3. A t-test calculates t_stat = 6.60. Assuming the study had more than just a few observations, you do not need a t table or software utility to draw a conclusion about the test. What is this conclusion, and why is a look-up table unnecessary?
   1. The conclusion with the given information is that the p-value will be low because of the t_stat and the mean values.

 

4. A researcher fails to find a significant different in mean blood pressure in 36 matched pairs. The standard deviation of the difference was 5 mmHg. What was the power of the test to find a mean difference of 2.5 mmHg at α = 0.05 (two-sided)?
   1. n = 36
   2. ∂ = 5
   3. X₁ -X₂= 2.5
   4. H₀: µ₁ = µ₂
   5. H₁: µ₁҂ µ₂
   6. Z_{stat =} (X_{1 –} X₂) / [∂ / (√n)  = 2.5 / [5 / (√36) ] = 2.5 / 0.833 = 3.000

 

5. Would you use a one-sample, paired-sample, or independent-sample t procedure in the following situations?
6. A lab technician obtains a specimen of known concentration from a reference lab. He/she tests the specimen 10 times using an assay kit and compares the calculated mean to that of the known standard.
   1. One-sample t procedure should be used in this situation. The data tested is from a single group and it is also compared to the known standard of data.
7. A different technician compares the concentration of 10 specimens using 2 different assay kits. Ten measurements are taken with each kit. Results are then compared.
   1. Paired-sample t procedure should be used in this situation. There are two samples of data being used.

 

6. In a study of maternal cigarette smoking and bone density in newborns, 77 infants of mothers who smoked has a mean bone mineral content of 0.098 g/cm³ (s₁ = 0.026 g/cm³). The 161 infants whose mothers did not smoke has a mean bone mineral content of 0.095 g/cm³ (s₂ = 0.025 g/cm³).

Sample size of smoking mothers:                      Sample size of non-smoking mothers:

n₁ = 77                                                             n₂ =161

x̅ ₁= 0.098                                                        x̅_{2 =} 0.095

S₁ 0.026                                                          S₂ =0.025

df – n₁-1= 76                                                    df – n₂-1= 160

a = 0.05

 

(x̅₁ – x̅₂) =0.003 df = 76

2. Calculate the 95% confidence interval for µ₁ – µ₂.
   1. SE² _x̅1 = S₁²/n₁ = 0.026²/ 77  = 8.779 x 10^-6                      
   2. SE _{x̅1 – x̅2} = √ S₁²/n_{1 +} S₂²/n₂ = √(8.779 x 10^-6) + (3.881 x 10^-6)  = 0.003558
   3. SE² _x̅2 = S₂²/n₂ = 0.025²/ 161 = 3.881 x 10^-6
   4. t_df – a/2 ₌ t₁ – 0.05/2 1.99
   5. (x̅₁ – x̅₂) ± t_{df – a/2} * SE _{x̅1 – x̅2} = 003 ± 1.99 * 0.199 * 0.003558 = (0.003 -0.007) -0.004 / (0.003 + 0.007) = 0.01
3. Based on the confidence interval you just calculated, would you reject or retain H₀: µ₁ – µ₂ = 0 at α=0.05?
   1. The null hypothesis is not rejected.

 

7. A randomized, double-blind, placebo-controlled study evaluated the effect of the herbal remedy Echinacea purpurea in treating upper respiratory tract infections in 2- to 11-year olds. Each time a child had an upper respiratory tract infection, treatment with either echinacea or a placebo was given for the duration of the illness. One of the outcomes studied was “severity of symptoms.” A severity scale based on four symptoms was monitored and recorded by the parents of subjects for each instance of upper respiratory infection. The peak severity of symptoms in the 337 cases treated with echinacea had a mean score of 6.0 (standard deviation 2.3). The peak severity of symptoms in the placebo group (n2 = 370) had a mean score of 6.1 (standard deviation 2.4). Test the mean difference for significance. Discuss your findings.

Control Group:                         Treatment Group:

n₁ = 337                                                                                                                                             n₂  = 370

x̅₁ = 6.0                                                x̅₂ = 6.1

S₁ =2.3                                                 S₂ =2.4

df = 336                                               df =369

 

(x̅₁ – x̅₂) = .1

2. SE² _x̅1 = S₁²/n₁ = 2.3² / 337 = 0.016
3. SE _{x̅1 – x̅2} = √ S₁²/n_{1 +} S₂²/n₂ = √ 0.0156 + 0.0155 = 0.176
4. SE² _x̅2 = S₂²/n₂ =2.4² /370 =0.0156
5. t_stat = (x̅₁ – x̅₂) / SE _{x̅1 – x̅2} = (6.1 – 6) / 0.176 = 0.568
6. 568 > 0.05
7. Since the p-value is greater than the value of a, the conclusion is that there is no significant difference in the mean severity between the control group and the treatment group.