Blood samples from 10 persons were sent to each of two labs for cholesterol determination. The measurements are recorded below:
Table 1. Serum Cholesterol measurements of Lab 1 and Lab 2
Subject Lab 1 Lab 2
1 296 318
2 268 287
3 244 260
4 272 279
5 240 245
6 244 249
7 282 294
8 254 271
9 244 262
10 262 285
Is there a statistical significant difference (at α = 0.01) in the cholesterol level reports by lab 1 and lab 2? Perform the following steps:
a. State in symbolic form the null and alternative hypothesis
b. Should you use the independent t test or the paired t test to answer this question?
c. Once you choose the appropriate t test, state:
a. α
b. degrees of freedom (df)
c. Critical value
d. Calculate the t test statistic
e. Provide the exact p-value
f. Calculate 99% C.I. and interpret the confidence interval using one or two sentences.
g. State your conclusion (reject or fail to reject the null hypothesis)
h. Express your conclusion in less technical terms (“there is or there is not a significant difference……..”)
i. Based on your conclusion state what type of error you might have made (Type I or Type II error).
a. The null hypothesis (H0) can be stated as: There is no statistical significant difference between the cholesterol level reports by Lab 1 and Lab 2.
The alternative hypothesis (Ha) can be stated as: There is a statistical significant difference between the cholesterol level reports by Lab 1 and Lab 2.
b. We should use the independent t-test to answer this question because the measurements from Lab 1 and Lab 2 are independent of each other.
c. Once we choose the appropriate t-test, the following information needs to be stated:
a. α (significance level) = 0.01
b. degrees of freedom (df) = (n1 + n2) - 2 = (10 + 10) - 2 = 18 (where n1 and n2 are the sample sizes of Lab 1 and Lab 2, respectively)
c. Critical value: To find the critical value, we need to look at the t-distribution table with the significance level (α) and degrees of freedom (df). Since α is 0.01 and df is 18, the critical value is ±2.878 (for a two-tailed test).
d. Calculate the t-test statistic: The formula for the t-test statistic for independent samples is: t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2)), where x1 and x2 are the means, s1 and s2 are the standard deviations, and n1 and n2 are the sample sizes for Lab 1 and Lab 2, respectively. We will calculate this in the next step.
e. Provide the exact p-value: The p-value can be determined by looking at the t-distribution table or using statistical software to calculate it. We will calculate this in the next step.
f. Calculate 99% confidence interval: The formula for the confidence interval is: CI = (x1 - x2) ± t * sqrt((s1^2 / n1) + (s2^2 / n2)). We will calculate this in the next step.
d. To calculate the t-test statistic:
Mean of Lab 1 (x1) = (296 + 268 + 244 + 272 + 240 + 244 + 282 + 254 + 244 + 262) / 10 = 260.6
Mean of Lab 2 (x2) = (318 + 287 + 260 + 279 + 245 + 249 + 294 + 271 + 262 + 285) / 10 = 272
Standard deviation of Lab 1 (s1) = sqrt(((296 - 260.6)^2 + (268 - 260.6)^2 + (244 - 260.6)^2 + (272 - 260.6)^2 + (240 - 260.6)^2 + (244 - 260.6)^2 + (282 - 260.6)^2 + (254 - 260.6)^2 + (244 - 260.6)^2 + (262 - 260.6)^2) / (10 - 1)) = 19.46
Standard deviation of Lab 2 (s2) = sqrt(((318 - 272)^2 + (287 - 272)^2 + (260 - 272)^2 + (279 - 272)^2 + (245 - 272)^2 + (249 - 272)^2 + (294 - 272)^2 + (271 - 272)^2 + (262 - 272)^2 + (285 - 272)^2) / (10 - 1)) = 19.05
t = (260.6 - 272) / sqrt((19.46^2 / 10) + (19.05^2 / 10)) = -2.29
e. To find the exact p-value, we can use statistical software or a t-distribution table. The p-value for t = -2.29 with df = 18 is approximately 0.0334.
f. To calculate the 99% confidence interval:
CI = (260.6 - 272) ± (2.878 * sqrt((19.46^2 / 10) + (19.05^2 / 10))) = -25.8 to -4.2
Interpretation: We are 99% confident that the true difference in cholesterol levels between Lab 1 and Lab 2 falls within the range of -25.8 to -4.2.
g. Based on the critical value (±2.878), the t-test statistic (-2.29), and the p-value (0.0334), we can conclude that there is a significant difference between the cholesterol level reports by Lab 1 and Lab 2.
h. In less technical terms, we can say that there is a significant difference in the cholesterol levels reported by Lab 1 and Lab 2.
i. Based on the conclusion, a Type I error might have been made, which means we rejected the null hypothesis even though it was true.