The recommended daily dietary allowance for zinc among males older than age 50 y

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^bar = 12.3, and s = 6.89. Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance? (Use alpha = 0.05.) State the appropriate null and alternative hypotheses. H_0: mu = 15 H_a = mu lessthanorequalto 15 H_0: mu = 15 H_a: mu notequalto 15 H_0: mu = 15 H_a: mu > 15 H_0: mu = 15 H_a: mu

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: > 15

Alternative hypothesis: < 15

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE) and the z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.6425

z = (x - ) / SE

z = - 4.20

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of - 4.20. We use the z Distribution Calculator to find P(z < - 4.20) = 0.00003.

Interpret results. Since the P-value (0.00003) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that zinc intake in the population of males of age group 65-74 has fallen below the recommended allowance.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.