MAS183 Statistical Data Analysis
Assessment 2
Student name Student number
- -
Question 1
a. • p̂ =
276
1912
= 0.1443
p̂(1−p̂)
• μp̂ = p̂ and SE(p̂) = σp̂ = √
n
• The sampling distribution is approximately normal, so for a 90% CI we have k = 1.645
276 276
276 (1− )
• p̂ ± k x SE(p̂) = ± 1.645√1912 1912
= 0.1443 ± 1.645(0.0012) = 0.1423 to 0.1463
1912 1912
b. We can be 90% sure that between 14.23% and 14.63% of people in this population have
experienced a flu-like illness.
Question 2
a. i. Graph pictured right
ii. • H0: μ0 = 6.3, the population mean pH of
river water is 6.3
• H1: μ0 ≠ 6.3, the population mean pH of
river water is not equal to 6.3
• n = 29, μˆ = 5.759, μ = 6.3, s = 0.5973
5.759 − 6.3
• Test statistic = 0.5973 = -5.18
√29
• p-value: 2 x P(Z < -5.18) = >0.0001
• Assumptions: sample is random,
population follows normal distribution, and
the sample size is large enough for CLT
• Conclusion: the small p-value provides strong evidence against the H0. There is
sufficient evidence to suggest that the pH changed from the usual value of 6.3.
iii. • Standard error =
0.5973
√29
= 0.1104
•
Critical value = 1.96 for a 95% CI
•
CI = 5.759 ± 0.1104 x 1.96
o Upper bound: 5.759 - 1.96 x 0.1104 = 5.543
o Lower bound: 5.759 + 1.96 x 0.1104 = 5.975
• The 95% confidence interval is 5.543 to 5.975. We can be 95% confident that the
mean pH falls within this range.
b. The p-value is very small which provides strong evidence against the null hypothesis. It is
rejected because the p-value is smaller than the significance level. There is enough
evidence to suggest that the average pH of water in this river system has changed from
6.3. The 95% confidence interval for the mean pH in this river system is approximately
5.543 to 5.975, which means we can be 95% confident that the mean pH in the river
system falls within this range.
Question 3
a. The data should be analysed as paired data. The same subjects are measured more than
once under different conditions; in this case, the subjects are measured twice – once after
taking oligofructose and once after taking a placebo.
b. i. Graph pictured on next page
Assessment 2
Student name Student number
- -
Question 1
a. • p̂ =
276
1912
= 0.1443
p̂(1−p̂)
• μp̂ = p̂ and SE(p̂) = σp̂ = √
n
• The sampling distribution is approximately normal, so for a 90% CI we have k = 1.645
276 276
276 (1− )
• p̂ ± k x SE(p̂) = ± 1.645√1912 1912
= 0.1443 ± 1.645(0.0012) = 0.1423 to 0.1463
1912 1912
b. We can be 90% sure that between 14.23% and 14.63% of people in this population have
experienced a flu-like illness.
Question 2
a. i. Graph pictured right
ii. • H0: μ0 = 6.3, the population mean pH of
river water is 6.3
• H1: μ0 ≠ 6.3, the population mean pH of
river water is not equal to 6.3
• n = 29, μˆ = 5.759, μ = 6.3, s = 0.5973
5.759 − 6.3
• Test statistic = 0.5973 = -5.18
√29
• p-value: 2 x P(Z < -5.18) = >0.0001
• Assumptions: sample is random,
population follows normal distribution, and
the sample size is large enough for CLT
• Conclusion: the small p-value provides strong evidence against the H0. There is
sufficient evidence to suggest that the pH changed from the usual value of 6.3.
iii. • Standard error =
0.5973
√29
= 0.1104
•
Critical value = 1.96 for a 95% CI
•
CI = 5.759 ± 0.1104 x 1.96
o Upper bound: 5.759 - 1.96 x 0.1104 = 5.543
o Lower bound: 5.759 + 1.96 x 0.1104 = 5.975
• The 95% confidence interval is 5.543 to 5.975. We can be 95% confident that the
mean pH falls within this range.
b. The p-value is very small which provides strong evidence against the null hypothesis. It is
rejected because the p-value is smaller than the significance level. There is enough
evidence to suggest that the average pH of water in this river system has changed from
6.3. The 95% confidence interval for the mean pH in this river system is approximately
5.543 to 5.975, which means we can be 95% confident that the mean pH in the river
system falls within this range.
Question 3
a. The data should be analysed as paired data. The same subjects are measured more than
once under different conditions; in this case, the subjects are measured twice – once after
taking oligofructose and once after taking a placebo.
b. i. Graph pictured on next page
