Hi, Could anyone help me please, because I'm not sure with my answer, Thank you so much for your help and value time.
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.
Answer: H0: μ = 0
H1: μ ≠0
b. Should you use the independent t test or the paired t test to answer this question?
Answer: (The paired t test)
c. Once you choose the appropriate t test, state:
a. α = (0.01 Answer)
b. degrees of freedom (df)=( 8 Answer)
c. Critical value=( ±3.2498 Answer)
d. Calculate the t test statistic = (-6.7270 Answer)
e. Provide the exact p-value = (0.0001 Answer)
f. Calculate 99% C.I. and interpret the confidence interval using one or two sentences.
Answer: -21.35663 < µd < -7.443366
Reject the Null Hypothesis
Sample provides evidence to reject the claim
g. State your conclusion (reject or fail to reject the null hypothesis)
Answer: Reject the null hypothesis
h. Express your conclusion in less technical terms (“there is or there is not a significant difference……..”)
Answer: Am not sure how to answer this one
i. Based on your conclusion state what type of error you might have made (Type I or Type II error).
Answer: Type I Error because it reject Ho
Thank you for your help.
You're welcome! I'm glad I could help you. In response to your question h, here's how you can express your conclusion in less technical terms:
- "There is a significant difference in the cholesterol level reports between lab 1 and lab 2."
Regarding question i, I would like to mention a correction. It should be Type II Error because you reject the null hypothesis when it is actually true. So, the correct answer is:
- "Based on your conclusion, you might have made a Type II Error."
If you have any more questions, feel free to ask.
You're welcome! I'm glad I could help you with your question. Let's go through each step to understand how to arrive at the answers.
a. The null hypothesis (H0) states that the mean cholesterol level reported by Lab 1 is equal to the mean cholesterol level reported by Lab 2. The alternative hypothesis (H1) suggests that there is a significant difference between the two means.
- H0: μ1 = μ2
- H1: μ1 ≠ μ2
b. In this case, we would use the paired t-test because the blood samples from the same 10 persons were sent to each lab. Paired t-tests are used when the same participants or objects are measured twice under different conditions.
c. Once we choose the paired t-test, we need to determine the significance level, degrees of freedom, and the critical value.
a. The significance level (α) is given as 0.01.
b. The degrees of freedom (df) can be calculated using the formula: df = n - 1, where n is the number of pairs (in this case, 10 pairs). Therefore, df = 10 - 1 = 9.
c. The critical value can be found using a t-distribution table or a statistical software. At a significance level of 0.01 and 9 degrees of freedom, the critical value (±) is 3.2498.
d. To calculate the t-test statistic, we subtract the mean difference between Lab 1 and Lab 2 from the hypothesized value of 0, and then divide it by the standard error of the mean difference. The calculated t-statistic is -6.7270.
e. Now, to find the exact p-value, we compare the t-test statistic to the t-distribution with 9 degrees of freedom. The p-value is the probability of obtaining a t value as extreme as the one observed, assuming the null hypothesis is true. In this case, the p-value is 0.0001.
f. To calculate the 99% confidence interval (C.I.), we can use the formula: Mean Difference ± (t* * Standard Error). Here, t* represents the critical value at a 99% confidence level. The confidence interval is -21.35663 < μd < -7.443366. This means that we are 99% confident that the true mean difference falls within this interval. Since the interval does not include zero, we can conclude that there is a significant difference between the cholesterol level reports of Lab 1 and Lab 2.
g. Based on the results, we reject the null hypothesis.
h. In less technical terms, we can say that there is a significant difference in the cholesterol level reports between Lab 1 and Lab 2.
i. Based on the conclusion, the type of error made in this case is a Type I Error because we rejected the null hypothesis when it was actually true.
I hope this explanation helps! Let me know if you have any further questions.