Question

Calcium is the most abundant mineral in the body and also one of the most important. According to the Food and Nutrition Board of the National Academy of Sciences, the recommended daily allowance (RDA) of calcium for adults is 800 millgrams (mg). A simple random sample of 18 people with incomes below the poverty level gives the daily calcium intakes shown below.

620	647	634	574	1113	850
641	775	686	858	743	609
879	734	672	993	433	992

Given the daily calcium intake is normally distributed with population standard deviation 188 mg.

a) Find the sample mean and sample standard deviation.

b) Do the data provide sufficient evidence to conclude that, at 5% significance level, the mean calcium intake of all people with incomes below the poverty level is less than the 800mg? (Use Critical-Value approach)   

c) What is the type I error of the test in part (b)?   

d) Determine a 95% confidence interval for the mean daily calcium intakes.   

Answer 1

a) The sample mean = 747.4

sample sd =   172.0  

b) Hypothesis :

Test statistic

Here we will be using population sd . n = sample size = 18

So,

at 5% significance level ,

Critical region : Reject null if zobs  < z0.05

Here . zobs = -1.187 > -1.645

So,not enough evidence to reject null .

i.e we cannot conclude that  at 5% significance level, the mean calcium intake of all people with incomes below the poverty level is less than the 800mg.

c) P[type 1 error ] = P[ rejecting True Null hypothesis ]

=P[ z < -1.187] = 0.118

d) 95% confidence interval for mean is given by :