91Ó°ÊÓ

This is a check of a single population mean, due to the fact the hassle is set imply modifications inventory price.

We need to test

$H_0:\mathrm{μ}=5$versus $H_a:\mathrm{μ}<5$

using $\mathrm{Î\pm }=0.05$

Investors do not believe stock prices will grow that fast. Therefore, the$<$says that it is left.

Random variable $\stackrel{¯}{X}$is the mean stock price. Distribution for the test is normal because we know the standard deviation i.e.

$\stackrel{¯}{X}:\mathcal{N}\left(\mathrm{μ},\frac{\mathrm{σ}}{\sqrt{n}}\right)=\mathcal{N}\left(5,\frac{1}{\sqrt{10}}\right)$

Now we calculate the p-value using the normal distribution for a mean:

$P$-value$=P(\stackrel{¯}{x}<2.6)≈0$

where the sample mean of the problem is given by

$\frac{2+3+⋯+1+2}{10}=2.6$

The $P$-value denotes the probability to the left of the sample mean in the normal distribution.

We can consider that,

$\mathrm{Î\pm }=0.05>0=\mathrm{P}$-value

Therefore, we reject $H_0:\mathrm{μ}=5$. In other words, we do not think a stock price for a particular company will grow at a rate of $5$per week but believes the stock won't grow as quickly.

At the $5\%$significance level, we conclude that is not sufficient evidence to conclude that a stock price for a particular company will grow at a rate of $5$per week.

Rejecting the null hypothesis $H_0$when it is true is defined as a type $\text{I}$error.

Suppose the null hypothesis ${\mathrm{H}}_0$is as follows:" The stock price of a particular company rises once a week. "

So, Type $\text{I}$error is -. Reject the null hypothesis that a stock price for a particular company will not grow at a rate of $5$per week when is actually will grow at a rate of $5$per week.

Failing to reject the null hypothesis when it is false is defined as a type $\text{II}$error.

Therefore, Type $\text{II}$error is

Not to reject the null hypothesis that a stock price for a particular company will grow at a rate of $5$per week when is actually will not grow at a rate of $5$per week.

The graph for this problem is: