Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Blood coagulation and clot formation is a complicated process with many differen

ID: 3176928 • Letter: B

Question

Blood coagulation and clot formation is a complicated process with many different plasma coagulation factors involved. Individuals with vitamin K deficiencies are thought to have low levels of (an important factor in coagulation). In an area where the population is thought to have vitamin K deficiencies, a sample of 15 people was taken and the mean level was found to be 0.2 mg/ml. The data is normally distributed and free of outliers. Do not round intermediate calculations. Round only your final answers to the questions below to the number of specified decimal places. If the population standard deviation is known to be 0.11 mg/ml, construct a 99% confidence interval for the mean population level. Enter the smaller value first, as usual. The confidence interval is [, ] mg/mL. Using the interval calculated in part (a), if the normal (healthy) level of is 0.25 mg/ml. is there evidence that the population question has a vitamin K deficiency (low levels of)? Select the correct choice below. Yes, this population has low levels of because one of the Z-scores were negative. Yes, there is evidence that this evidence that this population has low levels of because the confidence interval contains values that are less than 0.25 mg/mL. No, the population has normal levels of because one of the z-scores was positive. No, the population has normal levels of because the population mean is greater than 0.25 mg/mL.

Explanation / Answer

a) mean = 0.2

    sd = 0.11

    n =15

Z at 99% is =2.58

   For Upper confidence interval:

        = mean + Z * sd / sqrt(n)

        = 0.2 = 2.58 * sd / sqrt(15)

       = 0.273

Lower Confidence interval:

   = mean - Z * sd / sqrt(n)

        = 0.2 - 2.58 * sd / sqrt(15)

        = 0.126

interval = ( 0.126, 0.273 )

b) X = 0.25

    Z = (0.25 - 0.2) / .11/sqrt(15)

       = 1.76

   Since the caulated value 1.76 is less than 2.58 , Hence

Ans : No, the population has a normal level of prothrombin because one of the z score was positive.