Question

A health organization collects data on an individuals daily intake of selected nutrients and on their income level. The protein intake (in grams) over a 24hr period was determined for independent random samples random samples of 40 adults with income below the poverty level and 50 adults with income above the poverty level. The results:

Income Level	Sample Mean Protein intake (grams)	Sample Stardard Deviation	Number of Adults
Above Poverty (A)	73.2	15.5	50
Above Poverty (B)	65.7	15.0	40

a) Complete the following to construct a 95% confidence interval for the amount by which true mean protein intake for adults having income above the poverty level exceeds that of adults having income below the poverty level

Parameter of interest: _____________

Degrees of freedom:

Standard error (to 5 decimal places):

Critical value t*:

Confidence interval (to 3 decimal places):

Conclusion:

b) Assuming that all of the results remained unchanged, for what level of confidence C would a margin of error of 5 grams be obtained when calculating a symmetric 100(1-α)% confidence interval? Express your final answer to 2 decimal places.

Answer 1

a)

Parameter of interest: where are the true mean protein intake for adults having income above the poverty level and true mean protein intake for adults having income below the poverty level respectively.

Degrees of freedom: Using conservative approach, df = min(n1 - 1, n2 - 1) = min(50-1, 40 - 1) = 39

Standard error (to 5 decimal places):

s = = 3.22955

Critical value t*:

At df = 39 and 95% confidence interval , critical value of t is  2.023

Confidence interval (to 3 decimal places):

Point estimate of mean differences = 73.2 - 65.7 = 7.5

(7.5 - 2.023 * 3.22955 , 7.5 + 2.023 * 3.22955)

(0.96662 , 14.03338)

Conclusion:

Since the confidence interval does not contain the value 0, there is significant evidence of differences between true mean protein intake for adults having income above the poverty level and true mean protein intake for adults having income below the poverty level.

b)

Margin of error = t * Std error

5 = t * 3.22955

t = 5/3.22955 = 1.548203

For symmetric 100(1-α)% confidence interval, = 2 * P(t > 1.5482, df = 39) = 0.13

Confidence level = 1 - = 1 - 0.13 = 0.87 = 87%