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 = 115, x = 12.1, and s = 6.57. 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.)

H₀: μ = 15
H_a: μ < 15

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.)

z	=?
P-value	=?

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Solution :

Given that,

Population mean = = 15

Sample mean = = 12.1

Population standard deviation = = 6.57

Sample size = n = 115

Level of significance = = 0.05

The test statistics,

Z =( - )/ (/n)

= ( 12.1 - 15 ) / ( 6.57 / 115)

= -4.73

p-value = P(Z < z )

= P(Z < -4.73)

= 0

The p-value is p = 0, and since p = 0 < 0.05, it is concluded that the null hypothesis is rejected.