Question

Use the same sample of 51 boys in problem. The standard deviation of daily iron intake in the larger population of 9 to 11-year-old boys was 5.56mg. a. We wish to test if the standard deviation from the low-income group is lower than that of the general population. Use -0.10. : H: ii. Summary Sample Statistics: iii. Significance level: iv. p-value: v. Conclusion: vi. Is the standard deviation of iron intake among the low-income group lower than that of the general population? Explain. b. Compute a 99% confidence interval for the underlying variance of daily iron intake in the low- income group i. Point estimate: ii. Confidence Interval: iii. What does the confidence interval mean?

Answer 1

Answer)

2)

i) Ho : u = 12.5

H1 : u > 12.5

ii)

Sample mean = 14.44 mg

Sample s.d = 4.75 mg

iii)

Significance level is 0.01

iv)

As the population s.d is unknown, we will use t table to conduct the test and construct the interval as well.

Test statistics t = (sample mean - population mean)/(s.d/√n)

N = sample size = 51

t = (14.44-12.5)/(4.75/√51)

t = 2.917

Degrees of freedom is = n-1, 50

For df 50 and test statistics 2.917, P-Value is = 0.00264

V)

As the obtained P-Value is less than the given significance level

We reject the null hypothesis

We have enough evidence to support the claim

Vi)

Yes, as we have rejected the null hypothesis that, u = 12.5, and colcluded that we have enough evidence to support the claim that mean iron intake among the low-income group is higher than that of the general population.

B)

i)

Point estimate is = sample mean = 14.44

ii)

For df 50 and 95% confidence level, critical value t is = 2.009 (from t table)

Margin of error (MOE) = t*(s.d/√n) = 2.009*4.75/√51 = 1.33625227718

Confidence interval is given by

(Mean - MOE, Mean + MOE)

(13.1037477228, 15.7762522771)

C)

Confidence interval is the range in which population parameter lies.

For example here

We are 95% confident that true population mean lies in between 13.104 and 15.776