According to the US Department of Agriculture 588 of males b

According to the U.S. Department of Agriculture, 58.8% of males between 20 and 39 years old consume the minimum daily requirement of calcium. After an aggressive “Got milk” advertising campaign, the USDA conducted a survey of 55 randomly selected males between the ages of 20 and 39 and found that 36 of them consume the recommended daily allowance of calcium.

a) Construct a 99% confidence interval for the above data. Show your work using the formulas.

b) At the = 0.01, is there evidence to conclude that the percentage of males between the ages of 20 and 39 who consume the recommended daily allowance of calcium has increased? Conduct a full hypothesis test by following the steps below.

i. State the null and alternative hypotheses.

ii. State the significance level for this problem.

iii. Check the conditions that allow you to use the test statistic, and, if appropriate, calculate the test statistic.

iv. Calculate the p-value and include the probability notation statement.

v. State whether you reject or do not reject the null hypothesis.

vi. State your conclusion in context of the problem (i.e. interpret your results).

c) Explain the connection between the confidence interval and the hypothesis test in this problem (discuss in relation to the decision made from your hypothesis test).

Solution

a)
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=36
Sample Size(n)=55
Sample proportion = x/n =0.655
Confidence Interval = [ 0.655 ±Z a/2 ( Sqrt ( 0.655*0.345) /55)]
= [ 0.655 - 2.58* Sqrt(0.004) , 0.655 + 2.58* Sqrt(0.004) ]
= [ 0.49,0.82]

b)
Set Up Hypothesis
Null, H0:P<0.588
Alternate, H1: P>0.588
Test Statistic
No. Of Success chances Observed (x)=36
Number of objects in a sample provided(n)=55
No. Of Success Rate ( P )= x/n = 0.6545
Success Probability ( Po )=0.588
Failure Probability ( Qo) = 0.412
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.65455-0.588/(Sqrt(0.242256)/55)
Zo =1.0027
| Zo | =1.0027
Critical Value
The Value of |Z | at LOS 0.01% is 2.33
We got |Zo| =1.003 & | Z | =2.33
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Right Tail - Ha : ( P > 1.00268 ) = 0.15801
Hence Value of P0.01 < 0.15801,Here We Do not Reject Ho

We have evidence to indicate that daily allowance of calcium has increased