Question

A survey collects demographic, socioeconomic, dietary, and health-related information on an annual basis. Here is a sample of 20 observations on HDL cholesterol level (mg/dl) obtained from the survey (HDL is good cholesterol: the higher ts value, the lower the risk for heart disease) 35 49 51 54 65 51 52 47 87 37 46 33 39 44 39 64 94 34 30 48 (a) Calculate a point estimate of the population mean HDL cholesterol level (b) Making no assumptions about the shape of the population distribution, calculate a point estimate of the value that separates the largest 50% of HDL levels from the smallest 50%. (c) Calculate a point estimate of the population standard deviation. (Round your answer to three decimal places.) (b) An HDL level of at least 60 is considered desirable as it corresponds to a significantly lower risk of heart disease. Making no assumptions about the shape of the population distribution, estimate the proportion p of the population having an HDL level of at least 60. Need Help? Talk to a Tutor

Answer 1

Answer:

Step-by-step explanation:

35 49 51 54 65 51 52 47 87 37 46 33 39 44 39 64 94 34 30 48

is the sample given

From the above we can find

Mean =sum of all observations /20

= $\frac{999}{20} =49.95$

Variance = $\frac{1}{20} [\Sigma_1^{20} (x_i-49.95)^2 ]\\=269.7475$

Std deviation= square root of variance

= 16.424

Population estimate for mean = 49.95

Population estimate for std dev =$\frac{s}{\sqrt{n} } \\=\frac{16,.424}{\sqrt{20} }$

=3.673

The value that separate 50% top form 50% bottom is the median

Median is the middle value after arranging in ascending order

=average of 47 and 48 = 47.5

Proportion of atleast 60 = 2/20 = 0.1