HSCI 2117: Problem Set 5 - Hypothesis Tests Part 1 Solution & Analysis

This document provides a comprehensive solution to the HSCI 2117 Problem Set 5 assignment, focusing on hypothesis tests using the Smart Pill dataset. The solution includes detailed calculations and explanations for each question. Question 1 calculates the probability of 30 or fewer patients with a stomach pH below 2.5 using binomial distribution, resulting in a p-value of 0.002506. Question 2 concludes that the null hypothesis is rejected as the p-value is less than alpha=0.01. Question 3 computes rate difference and rate ratio. Question 4 determines the 95% confidence interval. Question 5 provides the mean, standard deviation, and sample size for ill patients and healthy volunteers. Questions 6 to 8 calculate the t-statistic, p-value, and confidence intervals for the difference in means using the t-test. Question 9 rejects the null hypothesis based on the p-value and suggests that the median may be a better measure than the mean if the data is not normally distributed.

Question 1
In order to get the probability of 30 or fewer patients, we will fast compute the probability of success for exactly
x and then find the cumulative success from 1 to 30 (in excel). We find that P ( X ≤30 )=0.002506 as indicated
below. The overall formula can be indicated as P ( X ≤30 ) =∑
i=1
30
( 87 Ci )∗Pi∗( 1−P ) ( 87−i )
x Proability of exactly X Probability of X or less
1 5.62224E-25 5.62224E-25
2 2.41756E-23 2.47379E-23
3 6.84977E-22 7.09715E-22
4 1.43845E-20 1.50942E-20
5 2.38783E-19 2.53877E-19
6 3.26337E-18 3.51724E-18
7 3.77618E-17 4.1279E-17
8 3.77618E-16 4.18897E-16
9 3.31465E-15 3.73354E-15
10 2.58542E-14 2.95878E-14
11 1.8098E-13 2.10568E-13
12 1.1462E-12 1.35677E-12
13 6.61272E-12 7.96949E-12
14 3.4953E-11 4.29224E-11
15 1.70104E-10 2.13027E-10
16 7.6547E-10 9.78497E-10
17 3.19696E-09 4.17546E-09
18 1.24326E-08 1.66081E-08
19 4.51501E-08 6.17582E-08
20 1.5351E-07 2.15268E-07
21 4.89771E-07 7.05039E-07
22 1.46931E-06 2.17435E-06
23 4.1524E-06 6.32676E-06
24 1.10731E-05 1.73998E-05
25 2.79042E-05 4.5304E-05
26 6.65407E-05 0.000111845
27 0.000150333 0.000262177
28 0.000322141 0.000584319
29 0.000655391 0.00123971
30 0.00126709 0.002506799
We multiply our probability from question 1 i.e. 0.0025 by 2; hence, the P-value=0.005.
Question 2
No, this is because the p-value is less than alpha=0.01 which means we will reject the null hypothesis and
conclude that sampling variability alone is not sufficient to example why the sample mean is far away from
50%.

Question 3
Rate difference= 34%-50%=-16%
Rate ratio=34%/50%=0.68
Based on the sizable nature of the results from the two computations, we can conclude that there is a significant
difference between the sample proportion and the hypothesized proportion.
Question 4
(a). With regard to the null hypothesis of 50% the 95% confidence interval for health patients is (39.83%,
60.17%). This means that the researcher should expect with 95% confidence that at least 39.83% of patients
and at most 60.17% of patients to be healthy. This confidence interval increases the assessment scope for
researcher.
(b). With regard to a numerator of 30 and denominator of 87 the 95% confidence interval for health patients is
(24.62%, 45.44%). This means that the researcher should expect with 95% confidence that at least 24.62% of
patients and at most 45.44% of patients to be healthy.
Question 5
The table contains the mean, standard deviation and sample size for both samples
Ill Patients healthy Volunteers
Mean 7.114 4.059
Standard Deviation 3.618 1.613
Sample Size 7 83
Question 6
Using the following formula we can compute the t-statistic
ttest= x1−x2
√ s1
2
n1
+ s2
2
n2
ttest= 7.114−4.059
√ (3.618)2
7 + (1.613)2
83
ttest= 3.055
1.37889
t−statistic=2.2156 at d . f .=7+ 83−2=8 8

Question 7
The p-value for t-statistic=2.2156 at d.f.=88 (2 tailed test) is given as
Formula in excel 2007 is “=TDIST(2.2156,88,2)”
P-value= 0.0292997
Question 8
Difference in means confidence interval= (x ¿¿ 1−x2)± tCI ( √ S1
n1
+ S2
n2 )¿
tCI is calculated in excel 2007 “=TINV(0.05,88)”
tCI=1.9873
95% confidence intervals is
( 7.114−4.059 ) ± 1.9873∗( √ ( 3.618 )2
7 + (1.613 )2
83 )
3.055 ±2.7403
95% confidence interval= (0.314, 5.795)
Question 9
Given that our p-value is less than alpha=0.05 we will reject the null hypothesis and conclude that the means are
different for healthy volunteers and ill patients. The usage of median by the research may be an improved if we
assume that our data is not normally distributed. Under the assumption of normality the median and mean are
equal and the usage of either measure will yield the same results. However, if the data is not normally
distributed then the median is the true measure of center and can be used in place of mean to improve the
confidence interval computation and hypothesis testing.