Step 1

It is provided that,

The null and the alternative hypothesis are:

\[\begin{array}{l} {H_0}:\mu = 75\\ {H_a}:\mu < 75 \end{array}\]

Mean, $\bar x = 72.3$

Standard deviation, $\sigma = 9$

Sample size, n = 25

a.

Based on the type of alternative hypothesis, the test is a left-tailed test.

Test statistic:

Assuming the null hypothesis true, the test statistic value is obtained as:

\[\begin{array}{c} z = \frac{{\bar x - \mu }}{{\frac{\sigma }{{\sqrt n }}}}\\ = \frac{{72.3 - 75}}{{\frac{9}{{\sqrt {25} }}}}\\ = - 1.5 \end{array}\]

Since the test statistic value is negative; therefore, the value will be below mean value.

 
Step 2

b.

Level of significance is provided as $\alpha = 0.002$

P-value:

For a left tailed test, the p-value using z-table is obtained as:

\[\begin{array}{c} P - {\rm{value}} = \phi \left( { - 1.5} \right)\\ = 0.0668 \end{array}\]

Decision:

If $P - {\rm{value}} \le \alpha$, Reject the null hypothesis.

If $P - {\rm{value}} > \alpha$, Do not reject the null hypothesis.

Conclusion:

Since p-value (0.0668) is not less than the significance level of 0.002; therefore, the null hypothesis is not rejected.

Interpretation:

It can be concluded that there is no sufficient evidence to reject the null hypothesis.

 
Step 3

c.

It is required to obtain the value of $\beta \left( {70} \right)$ .

The type II error is given by:

\[\beta = \phi \left( {{z_\alpha } + \frac{{{\mu _0} - \mu '}}{{\frac{\sigma }{{\sqrt n }}}}} \right)\]

For a significance level of 0.002, the left tailed critical value of z is obtained using z-table:

\[{z_\alpha } = - 2.88\]

Substituting the required values,

\[\begin{array}{c} \beta \left( {70} \right) = \phi \left( { - 2.88 + \frac{{75 - 70}}{{\frac{9}{{\sqrt {25} }}}}} \right)\\ = \phi \left( { - 0.10} \right)\\ = 0.46\;\;\;\;\left( {{\rm{Using }}z{\rm{ - score table}}} \right) \end{array}\]

But for a left-tailed test,

\[\begin{array}{c} \beta \left( {70} \right) = 1 - \phi \left( { - 0.10} \right)\\ = 1 - 0.46\\ = 0.54 \end{array}\]

Therefore, $\beta \left( {70} \right) = 0.54$

 
Step 4

d.

It is required to compute the sample size.

For one tailed test, the sample size at a $\alpha $ level of significance and $\beta \left( \mu \right) = {\mu '}$ is estimated by:

\[n = {\left[ {\frac{{\sigma \left( {{z_\alpha } + {z_\beta }} \right)}}{{{\mu _0} - {\mu '}}}} \right]^2}\]

For a significance level of 0.002, the critical value of z is obtained using z-table:

\[{z_\alpha } = 2.88\]

For $\beta = 0.01$ , the critical value of z is obtained using z-table:

\[{z_\beta } = 2.33\]

Substituting the required values,

\[\begin{array}{c} n = {\left[ {\frac{{9\left( {2.88 + 2.33} \right)}}{{75 - 70}}} \right]^2}\\ \approx 88 \end{array}\]

Therefore, the required sample size is 88.

 
Step 5

e.

The type I error is defined as the error arise due to rejecting the null hypothesis when it is actually true.

Concerning the problem, type I error will occur if the null hypothesis is true. But for the case the null hypothesis H0: $\mu = 76$, which is not true.

This results in the type I error to be 0.