Chapter 9.3, Problem 73E

a)

To determine

To verify whether there are any outliers in given data.

Explanation of Solution

Given:The Minitab output of data of daily calcium intake of a random sample of 36 women of age between 18 and 24 is shown below

  

Concept used:The boundaries for outliers are shown below

  $\begin{array}{l}Lowerboundary=Q_1-1.5*IQR\\ Upperboundary=Q_3+1.5*IQR\end{array}$

Any values which does not lie between these boundaries are considered as outliers in a data distribution.

Calculation:From the given information

  $\begin{array}{l}{Q}_1=632.5\\ Q_3=1090.5\\ IQR=Q_3-Q_1=1090.5-632.5=458\end{array}$

We get Inter Quartile Range as 458, using this we calculate the boundaries as shown below

  $\begin{array}{l}Lowerboundary=Q_1-1.5*IQR\\ =632.5-1.5*458=-54.5\\ Upperboundary=Q_3+1.5*IQR\\ =1090.5+1.5*458=1777.5\end{array}$

Interpretation:

Here the minimum value is 374 and maximum value is 1425. Hence all values do lie between the boundaries of outliers. Hence there are no outliers in the data.

b)

To determine

To interpret the p value from Minitab output.

Explanation of Solution

From the Minitab output the p value is 0.000

If the average daily calcium intake of women is 1200 mg, then the probability of getting a random sample of 36 women with an average intake 856.2 is approximately 0. It also concludes that the average intake among 36 sample of women is less than 1200.

b)

To determine

To verify if the data provide convincing evidence to support researchers suspicion.

Explanation of Solution

Concept used:The conditions to perform a one sample t testfor µ.The sample has to be randomly selected.

The population should follow a normal distribution.

The sample should be less than 10% of the population.

Calculation:

To test the claim that recommended daily allowance of calcium of women between ages 18 and 24 years µ is less than the true value 1200.

The null and alternate hypothesis are shown below

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

The sample of 36 women are selected randomly. The population is of large size, so we can approximately assume it to be a normal distribution. There will be more than 360 women in U.S aged between 18 and 24, so 10% condition is also satisfied.

A one sample t test is run and we get test statistic $t=-6.73$ and the corresponding p value as 0.000. As the p value is less than the level of significance (0.05), we have sufficient evidence at 5% level of significance to reject the null hypothesis.

Conclusion:At 5% level of significance there is evidence to conclude that the women are getting calcium less than 1200 mg.