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 = 111, x = 12.4, and s = 6.69. Does this data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance? (Use ? = 0.05.) State the appropriate null and alternative hypotheses. H0: ? = 15 Ha: ? > 15H0: ? = 15 Ha: ? < 15 H0: ? = 15 Ha: ? ≤ 15H0: ? = 15 Ha: ? ≠ 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 = State the conclusion in the problem context. Do not reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/day.Reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/day. Do not reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day.Reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day.
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 = 111, x = 12.4, and s = 6.69. Does this data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance? (Use ? = 0.05.)
State the appropriate null and alternative hypotheses.
Ha: ? > 15H0: ? = 15
Ha: ? < 15 H0: ? = 15
Ha: ? ≤ 15H0: ? = 15
Ha: ? ≠ 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 | = |
State the conclusion in the problem context.
The random variable daily zinc intake follows normal distribution.
We have to test whether average daily zinc intake falls below 15 mg/day or not.
This is One-sample z-test because the sample size is large (n > 30).
We have to assume the sample standard deviation as population standard deviation.
Trending now
This is a popular solution!
Step by step
Solved in 4 steps with 3 images
