Question

Consider differences between daily calcium and iron consumption among a hypothetical sample of 300 U.S. adults, age 18 and older, who completed a nutrition assessment survey. Among the survey results: The mean difference between daily calcium and iron consumption equaled 970mg, and the standard deviation of the difference equaled 75mg.

(i) Calculate a 95% confidence interval for the average difference between daily calcium and iron amounts of U.S. adults.

(ii) Does the confidence interval suggest a genuine difference in the average amounts consumed? Why?

Among the answer choices that follow, CHOOSE THE BEST COMBINATIONS OF THE OPTIONS GIVEN IN (i) AND (ii).

Group of answer choices

a.

(i)

(950, 1000)

(ii)

No, because the value 950 is about one-half a standard deviation away from 970.

b.

(i)

(950,1000)

(ii)

Yes, because 970 is contained within the interval.

c.

(i)

(960,980)

(ii)

Yes, because this interval does not contain the value 0, which, if obtained, would suggest there is no difference, statistically speaking at least.

d.

(i)

(960,980)

(ii)

No, because the interval needs must not contain 0, given the high value on the difference (970mg).

e.

None of the above answers sets is correct.

Answer 1

Since the sample size n =300 is very large we can use z score for constructing the confidence interval.

95% confidence interval for the population mean difference is given as:

Sample mean +- z*sample SD/√n

= 970 +- 1.96*75/√300

= (960, 980)

ii) this interval does not contain 0, so it's statistically significant that is we can conclude that a genuine difference between the two groups exists.

So option c. Is correct.