194k views
5 votes
The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the following summary data on intake for a sample of males age 65−74 years: n = 114, x = 11.9, and s = 6.42.

Does this data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance? (Use α = 0.05.)Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

User Paleo
by
8.2k points

1 Answer

2 votes

Answer:

Data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance.

Explanation:

Let mu be the average daily zinc intake in the population of all males age 65−74


H_(0) : mu= 15 mg/day


H_(0) : mu< 15 mg/day

Test statistic can be calculated using the equation:

z=
(X-M)/((s)/(√(N) ) ) where

  • X is the sample average daily zinc intake of males age 65−74 years (11.9)
  • M is the recommended amount (15)
  • s is the sample standard deviation (6.42)
  • N is the sample size (114)

Then z=
(11.9-15)/((6.42)/(√(114) ) ) ≈ -5.15

Since p-value≈0.0001 < 0.05 is very low and smaller than significance level, we can reject the null hypothesis.

User Harshal Waghmare
by
8.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories