Question

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.)

Answer 1

Answer:

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

Step-by-step 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=$\frac{X-M}{\frac{s}{\sqrt{N} } }$ 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=$\frac{11.9-15}{\frac{6.42}{\sqrt{114} } }$ ≈ -5.15

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