The recommended daily allowance (RDA) of calcium for women between the ages of 18 and 24 years is 1200 milligrams (mg). Researchers who were involved in a large-scale study of women’s bone health suspected that

their participants had significantly lower calcium intakes than the RDA. To test this suspicion, the researchers measured the daily calcium intake of a random sample of 36 women from the study who fell in the desired age range. The Minitab output below displays descriptive statistics for these data, along with the results of a significance test.

(a) Determine whether there are any outliers. Show your work.

(b) Interpret the P-value in context.

(c) Do these data give convincing evidence to support the researchers’ suspicion? Carry out a test to help your answer

The recommended daily allowance (RDA) of calcium for women between the ages of $18$and $24$years is $1200$ milligrams (mg). The researchers measured the daily calcium intake of a random sample of $36$ women from the study who fell in the desired age range.

we know from the tab,

$Q_1=632.5\text{and}{Q}_3=1090.5$

$1QR=Q_3-Q_1=1090.5-632.5=458$

Calculating the lower and upper boundaries,

$\begin{array}{r}\mathrm{LB}=Q_1-1.5\times 1QR=632.5-1.5\times 458=-54.5\\ \text{UB}=Q_3+1.5\times 1QR=1090.5+1.5\times 458=1777.5\end{array}$

Hence we can say that all the values lie between the boundaries and there is no outlier.

The probability of getting a random sample of $36$ladies with a typical intake $856.2$is close to zero if the typical daily calcium intake of ladies is $1200$mg. It additionally infers that the typical intake among $36$ example of ladies is under $1200$.

Calculating the null and alternative hypotheses,

$\begin{array}{r}{H}_0:\mu =1200\\ H_a:\mu <1200\end{array}$

A random selection of $36$women is done. The population is of huge size, so we can approximately expect it to be an ordinary distribution. There will be in excess of $360$ladies in U.S matured somewhere in the range of $18$and $24$, so $10\%$condition is additionally satisfied.

t test is done and we get test statistic t= $-6.73$and the corresponding p value as $0.000$. As the p-value is not exactly the degree of significance $0.005$, we have sufficient evidence at $5\%$level of significance to dismiss the null hypothesis. Hence, there is convincing evidence to support the researchers’ suspicion