Chapter 17, Problem 17.44E

a.

To determine

To obtain: The 90% confidence interval for the mean blood pressure of all executives.

Answer to Problem 17.44E

The 90% confidence interval for the mean blood pressure of all executives is 123.162 to 128.978.

Explanation of Solution

Given info:

A medical director found mean blood pressure $\overline{x}=126.07$ for an SRS of 72 executives with the standard deviation of the blood pressure of all executives as $\sigma =15$.

Calculation:

The 90% confidence interval for the mean blood pressure of all executives is calculated by the formula,

$\text{Interval}=\overline{x}\pm z\cdot \frac{\sigma }{\sqrt{n}}$ (1)

For the confidence interval 90, the value of z score is 1.645.

Substitute 126.07 for $\overline{x}$, 15 for $\sigma$, 72 for n and 1.645 for z in equation (1).

$\begin{array}{c}\overline{x}\pm z\cdot \frac{\sigma }{\sqrt{n}}=126.07\pm \left(1.645\right)\left(\frac{15}{\sqrt{72}}\right)\\ =126.07\pm \left(1.645\right)\left(1.767\right)\\ =126.07\pm 2.907\\ =123.162\text{to}\text{128}\text{.978}\end{array}$

Thus, the 90% confidence interval for the mean blood pressure of all executives is 123.162 to 128.978.

b.

To determine

To obtain: The z test for $H_0:\mu =128$ against the two sided alternative.

Answer to Problem 17.44E

The z test for $H_0:\mu =128$ against the two sided alternative is $-1.0918$.

Explanation of Solution

Given Info:

The hypothesized value ${\mu }_0=128$ falls inside the confidence level.

Calculation:

The formula to calculate the z score is,

$z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$ (2)

Substitute 126.07 for $\overline{x}$, 15 for $\sigma$, 72 for n and 128 for $\mu$ in equation (2).

$\begin{array}{c}z=\frac{126.07-128}{\frac{15}{\sqrt{72}}}\\ =\frac{-1.93}{1.76}\\ =-1.0918\end{array}$

Thus the value of z score is $-1.0918$.

Since the value of z score comes out to be $-1.0918$, and by standard mean cumulative proportion table, the value lies between $z^{\ast }=1.036$ and $z^{\ast }=1.282$, that means the P-value lies between $P=0.30$ and $P=0.20$.

The level of significance is given as 10% that is $\alpha =0.1$ and since the P value is greater than the level of significance thus it is concluded that the test is not significant on the 10% level.

c.

To determine

To obtain: The z test for $H_0:\mu =129$ against the two sided alternative.

Answer to Problem 17.44E

The z test for $H_0:\mu =129$ against the two sided alternative is $-1.66$.

Explanation of Solution

Given Info:

The hypothesized value ${\mu }_0=129$ falls outside the confidence level.

Calculation:

The formula to calculate the z score is,

$z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$ (2)

Substitute 126.07 for $\overline{x}$, 15 for $\sigma$, 72 for n and 129 for $\mu$ in equation (2) of part (b).

$\begin{array}{c}z=\frac{126.07-129}{\frac{15}{\sqrt{72}}}\\ =\frac{-2.93}{1.76}\\ =-1.66\end{array}$

Thus the value of z score is $-1.66$.

Since the value of z score comes out to be $-1.66$, thus by standard mean cumulative proportion table, the value lies between $z^{\ast }=-1.645$ and $z^{\ast }=1.96$ that means the P value lies between $P=0.05$ and $P=0.10$.

The level of significance is given as 10% that is $\alpha =0.1$ and since the P value is less than the level of significance thus it is concluded that the test is significant on the 10% level.