Question

The mass of a standard loaf of white bread is, by law meant to be 700g with a population standard deviation of 21g. The Lauren Bakery, which supplies outlets throughout the Eastern Cape, regularly checks the masses of its standard loaf of bread white bread. If their bread is underweight, on average, they are liable for a fine by the Provincial Department of Health, whose inspectors undertake random checks. If the bread is overweight, on average, the bakery is wasting its ingredients. On a given day, a random sample of 64 loaves is selected and weighed. The sample mean mass was found to be 695g. Assume that the mass of bread is approximately normally distributed 3.2 Polokwane Butchery supplies Vienna sausages in the entire of KZN province. An Inspector from the CCSA has been sceptical of the mass of the Vienna sausage packs after receiving numerous complaints. He wants to investigate the marked mass shown on Vienna sausages. A pilot study showed a mean of 11.8kg per pack and a variance of 0.49kg

Answer 1

Given the information:

For a standard loaf of white bread,

Hypothesized mean $\mu_o=700$

Population standard deviation $\sigma=21$

Sample size $n=64$

Sample mean $\overline{x}=695$

We have given, mass of bread is approximately normally distributed.

Here we assume the level of significance $\alpha=0.05$

Here to check if breads are overweight or not. We use z test statistic

Hypothesis to test

$H_o:\mu=700$

$H_1:\mu>700$

Region rejection:

$Z_c=Z_\alpha=Critical$ $Value$

Where $\alpha=0.05$

$Z_c$ is the critical value from Z - table at $\alpha=0.05$

Therefore, $Z_c=Z_\alpha=1.645$

Critical rejection region

$R=\left\{Z:Z>1.645\right\}$

Test statistic:

$z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}=\frac{695-700}{\frac{21}{\sqrt{64}}}=-1.905$

Decision rule:

$z=-1.905<Z_c=1.645$

Hence we accept $H_o$ at $\alpha=0.05$

and we can conclude that there is no overweight production of bread.

Now, from $z-table$ we find out $p-value$ for $z=-1.905$

$p-value=p(z>-1.905)$

$=0.9716$

If $p-value>level$ $of$ $significance(\alpha)$

$p-value=0.9716>\alpha=0.05$

Hence, we accept $H_o$ at $\alpha=0.05$

Conclusion

We can conclude that there is no overweight production of bread.

3.2

Given:

Sample mean $(\overline{x})=11.8$

Sample variation $(s)^2=0.49$

Sample standard deviation $(s)=\sqrt{s^2}$

$=\sqrt{0.49}$

$s=0.7$

Margin error $=0.2$

Confidence level(CI) $=99$%

We have to find out how many packs should the inspector sample out

In short we have to find out sample size.

We know that,

$Confidence\:Interval=\overline{x}\:\pm margin\:of\:error$

$=\overline{x}\:\pm Z\alpha \left(\frac{s}{\sqrt{n}}\right)$

Here we have given margin of error

Margin of error $=Z\alpha \times(\frac{s}{\sqrt{n}})$

$0.2=Z\alpha \times \left(\frac{0.7}{\sqrt{n}}\right)$

Where, $Z\alpha$ is a critical value derived from $z-statistical$ $table$

For confidence level $=99$%

$(1-\alpha)=0.99$

$\alpha=0.01$

$Z\alpha=0.01=2.576$

$0.2=2.576\times \left(\frac{0.7}{\sqrt{n}}\right)$

$n=\left(\frac{2.576\times 0.7}{0.2}\right)^2$

$n=80.72\approx 81$

Therefore, we can say that 81 samples are required to find out 99% confidence that sample mean will differ by 0.2g.