Given the equation \[\simplify[std]{{a} * x + {b} = {f}/{g}({c} * x + {d})}\] we first multiply both sides by $\var{g}$ to get

\[\simplify[std]{{g}*({a} * x + {b} )= {f}*({c} * x + {d})}.\]

Then expand both sides of this equation to get:

\[\simplify[std]{{g*a} x + {g*b} = {f*c}x + {f*d}}.\]

and then collect together all the constant terms on the right hand-side, and collect together all the terms in $x$ on the left-hand side of the equation.

The equation can then be written as:
\[\simplify[std]{({g*a}-{f*c})x=({f*d}+{-g*b})}\] i.e.
\[\simplify{{g*a-f*c}x={f*d-b*g}}\]
which gives \[x =\simplify[std]{{(f*d-b*g)}/{(g*a-f*c)}}\] as the solution.

Check the answer

You can check that this is the correct solution by inputting this solution back into the equation to see if it satisfies the equation. 

Using the method given by Show steps we have:

\[\begin{eqnarray*}\simplify[std]{ ({a}x+{b})({c}x+{d})}&=&\simplify[std]{{a}x*({c}x+{d})+{b}({c}x+{d})}\\&=&\simplify[std]{{a*c}x^2+{a*d}x+{b*c}x+{b*d}}\\&=&\simplify[std]{{a*c}x^2+{(a*d+b*c)}x+{b*d}}\end{eqnarray*}\]

The equation of the line is of the form $y=mx+c$.

You are given the slope or gradient $\displaystyle m= \simplify{{b-d}/{a-c}}$ and we can calculate the constant term $c$ by noting that $y=\var{b}$ when $x=\var{a}$.

Using this we get:
\[ \begin{eqnarray} \var{b}&=&\simplify[std]{({b-d}/{a-c}){a}+c} \Rightarrow\\ c&=&\simplify[std]{{b}-({b-d}/{a-c}){a}={(b*c-a*d)}/{(c-a)}} \end{eqnarray} \]

Hence the equation of the line is
\[y = \simplify[std]{({b-d}/{a-c})x+{b*c-a*d}/{c-a}}\]

\[ \begin{eqnarray} \simplify[std]{{a}x+{b}y}&=&\var{c}&\mbox{ ........(1)}\\ \simplify[std]{{a1}x+{b1}y}&=&\var{c1}&\mbox{ ........(2)} \end{eqnarray} \]
To get a solution for $x$ multiply equation (1) by 7 and equation (2) by 1

This gives:
\[ \begin{eqnarray} \simplify[std]{{a*this}x+{b*this}y}&=&\var{this*c}&\mbox{ ........(3)}\\ \simplify[std]{{a1*that}x+{b1*that}y}&=&\var{that*c1}&\mbox{ ........(4)} \end{eqnarray} \]
Now take away the equation (4) from equation (3) to get
\[\simplify[std]{({a*this}+{s6*a1*that})x={this*c}+{s6*that*c1}}\]
And so we get the solution for $x$:
\[x = \simplify{{c*b1-b*c1}/{b1*a-a1*b}}\]
Substituting this value into any of the equations (1) and (2) gives:
\[y = \simplify{{c*a1-a*c1}/{b*a1-a*b1}}\]
You can check that these solutions are correct by seeing if they satisfy both equations (1) and (2) by substituting these values into the equations.

Direct Factorisation

If you can spot a direct factorisation then this is the quickest way to do this question.

For this example we have the factorisation

\[\simplify{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d} = ({a} * x + { -c}) * ({b} * x + { -d})}\]

Hence we find the roots:
\[\begin{eqnarray} x&=& \simplify{{n1-n4}/{2*a*b}}\\ x&=& \simplify{{n1+n4}/{2*a*b}} \end{eqnarray} \]

Other Methods.

There are several methods of finding the roots – here are the main methods.

Finding the roots of a quadratic using the standard formula.

We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are

\[ x = \frac{-b +\sqrt{b^2-4ac}}{2a}\mbox{ and } x = \frac{-b -\sqrt{b^2-4ac}}{2a}\]
there are three main types of solutions depending upon the value of the discriminant $\Delta=b^2-4ac$

1. $\Delta \gt 0$. The roots are real and distinct

2. $\Delta=0$. The roots are real and equal. Their common value is $\displaystyle -\frac{b}{2a}$

3. $\Delta \lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

For this question the discriminant of $\simplify{{a*b}x^2+{-b*c-a*d}x+{c*d}}$ is $\Delta = \simplify[std]{{-n1}^2-4*{a*b*c*d}}=\var{disc}$

The roots exist and are distinct. .

So the roots are:

\[\begin{eqnarray} x = \frac{\var{n1} - \sqrt{\var{disc}}}{\var{n3}} &=& \frac{\var{n1} - \var{n4} }{\var{n3}} &=& \simplify{{n1 - n4}/ {n3}}\\ x = \frac{\var{n1} + \sqrt{\var{disc}}}{\var{n3}} &=& \frac{\var{n1} + \var{n4} }{\var{n3}} &=& \simplify{{n1 + n4}/ {n3}} \end{eqnarray}\]

Completing the square.

First we complete the square for the quadratic expression $\simplify{{a*b}x^2+{-n1}x+{c*d}}$
\[\begin{eqnarray} \simplify{{a*b}x^2+{-n1}x+{c*d}}&=&\var{n5}\left(\simplify{x^2+({-n1}/{a*b})x+ {c*d}/{a*b}}\right)\\ &=&\var{n5}\left(\left(\simplify{x+({-n1}/{2*a*b})}\right)^2+ \simplify{{c*d}/{a*b}-({-n1}/({2*a*b}))^2}\right)\\ &=&\var{n5}\left(\left(\simplify{x+({-n1}/{2*a*b})}\right)^2 -\simplify{ {n2^2}/{4*(a*b)^2}}\right) \end{eqnarray} \]
So to solve $\simplify{{a*b}x^2+{-n1}x+{c*d}}=0$ we have to solve:
\[\begin{eqnarray} \left(\simplify{x+({-n1}/{2*a*b})}\right)^2&\phantom{{}}& -\simplify{ {n2^2}/{4*(a*b)^2}}=0\Rightarrow\\ \left(\simplify{x+({-n1}/{2*a*b})}\right)^2&\phantom{{}}&=\simplify{ {n2^2}/{4*(a*b)^2}=({abs(n2)}/{2*a*b})^2} \end{eqnarray}\]
So we get the two solutions:
\[\begin{eqnarray} \simplify{x+({-n1}/{2*a*b})}&=&\simplify{-{abs(n2)}/{2*a*b}} \Rightarrow &x& = \simplify{({-abs(n2)+n1}/{2*a*b})}\\ \simplify{x+({-n1}/{2*a*b})}&=&\simplify{({abs(n2)}/{2*a*b})} \Rightarrow &x& = \simplify{({n1+abs(n2)}/{2*a*b})} \end{eqnarray}\]

Rearrange the equation by adding -7 to both sides to get:
\[\simplify{{t} / ({a} * x + {b}) = {d} + { -c} = {d -c}}\]
This gives \[\simplify{({a} * x + {b}) / {t} = 1 / {d -c}}\] (this is because if $\displaystyle \frac{a}{b}=c$ then $\displaystyle \frac{b}{a}=\frac{1}{c}$ on turning the fraction round the other way)
and so \[\simplify{({a} * x + {b}) = {t} / {d -c}}\] on multiplying both sides by 2.
Hence \[\simplify{{a} * x = {t} / {d -c} -{b} = ({a * an1} / {an2})}\]
and so \[\simplify{x={an1}/{an2}}\] is the solution on dividing both sides by 8.

a)

The mean of a set of numbers $\left(x_1,\ldots,x_N\right)$ is given by

\[\mu = \frac{1}{N}\sum_{i=1}^N x_i.\]

b)

The median is the middle value when the set of numbers is ordered.

c)

The mode is the most common value in the set.

d)

The range is the difference between the maximum value and the minimum value in the set.

e)

The standard deviation is given by

\[\sigma = \sqrt{ \frac{1}{N-1} \left[ \sum_{i=1}^N x_i^2 \right] - \mu^2}.\]

a)

1. $X \sim \operatorname{bin}(\var{number1},\var{prob})$, so $n= \var{number1},\;\;p=\var{prob}$.

2. The expectation is given by $\operatorname{E}[X]=n\times p=\var{number1}\times \var{prob}=\var{number1*prob}$

3. $\operatorname{stdev}(X)=\sqrt{n\times p \times (1-p)}=\sqrt{\var{number1}\times \var{prob} \times \var{1-prob}}=\var{sd}$ to 3 decimal places.

b)

1. \[ \begin{eqnarray*}\operatorname{P}(X = \var{thisnumber}) &=& \dbinom{\var{number1}}{\var{thisnumber}}\times\var{prob}^{\var{thisnumber}}\times(1-\var{prob})^{\var{number1-thisnumber}}\\& =& \var{comb(number1,thisnumber)} \times\var{prob}^{\var{thisnumber}}\times\var{1-prob}^{\var{number1-thisnumber}}\\&=&\var{prob1}\end{eqnarray*} \] to 3 decimal places.

2. 

\[ \begin{eqnarray*}\operatorname{P}(X \leq \var{thatnumber})& =& \simplify[all,!collectNumbers]{P(X = 0) + P(X = 1) + {v}*P(X = 2)}\\& =& \simplify[zeroFactor,zeroTerm,unitFactor]{{1 -prob} ^ {number1}+ {number1} *{prob} *{1 -prob} ^ {number1 -1} + {v} * ({number1} * {number1 -1}/2)* {prob} ^ 2 *( {1 -prob} ^ {number1 -2})}\\& =& \var{prob2}\end{eqnarray*} \]

to 3 decimal places.

a) The number of loans less than £$\var{u1}$ is $\var{accumdisp(n,t)}=\var{sum(n[0..t+1])}$

Since there are $\var{thismany}$ loans the probability of choosing one of these loans is  $\displaystyle \frac{\var{sum(n[0..t+1])}}{\var{thismany}}=\var{ans1}$ to 2 decimal places.

b) The number of loans greater than £$\var{o1}$ is $\var{accumdisp(n[v+1..abs(n)],abs(n)-v-2)}=\var{sum(n[v+1..abs(n)])}$.

Since there are $\var{thismany}$ loans the probability of choosing one of these loans is  $\displaystyle \frac{\var{sum(n[v+1..abs(n)])}}{\var{thismany}}=\var{ans2}$ to 2 decimal places.

c) There are $\var{accumdisp(n[p+1..q+1],q-p-1)}=\var{sum(n[p+1..q+1])}$ loans between  £$\var{a[p]}$ and £$\var{a[q]-1}$.

Hence the probability of selecting one of these loans in this range for review is $\displaystyle \frac{\var{sum(n[p+1..q+1])}}{\var{thismany}}=\var{ans3}$ to 2 decimal places.

Ordering the data gives:

Splitting into the groups of 10s gives

Then putting this into stem-and-leaf plot gives

We show how to calculate the relative percentage frequency for one range of values for  $\var{a[r]} \le X \lt \var{a[r+1]}$    - you can then check the rest.

Note that there were $\var{daysopen}$ days  in the year when sales took place. 

There were $\var{norm1[r]}$ days out of the  $\var{daysopen}$ when there were between $\var{a[r]}$ and $\var{a[r+1]}$ thousand pounds worth of sales (including  $\var{a[r]}$ thousand but not $\var{a[r+1]}$ thousand) .

Hence the relative frequency percentage for such sales is given by \[100 \times \frac{\var{norm1[r]}}{\var{daysopen}}\%=\var{rel[r]}\%\] to one decimal place.

a)

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \simplify{{2a1}x+{b1}}$.

To find the $x$-coordinate of the stationary point, solve $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ for $x$:

\[ \begin{align} \frac{\mathrm{d}y}{\mathrm{d}x} = \simplify{{2a1}x + {b1}} &= 0 \\ x &= \simplify[all,fractionNumbers]{{sx1}} \end{align} \]

Find the $y$-coordinate by substituting this value of $x$ into the definition of $y(x)$:

\[\begin{align} \simplify[fractionnumbers]{y({sx1})} &= \simplify[basic,fractionnumbers]{{a1}{sx1}^2+{b1}{sx1}+{c1}} \\ &= \simplify[fractionnumbers]{{sy1}} \end{align}\]

Finally, to determine the nature of the stationary point, look at $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ at $x = \simplify[fractionnumbers]{{sx1}}$.

\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \simplify{{2*a1}} \]

This is positive, so the stationary point is a minimum.

b)

$\dfrac{\mathrm{d}z}{\mathrm{d}x} = \simplify{{2a2}x+{b2}}$.

To find the $x$-coordinate of the stationary point, solve $\frac{\mathrm{d}z}{\mathrm{d}x} = 0$ for $x$:

\[ \begin{align} \frac{\mathrm{d}z}{\mathrm{d}x} = \simplify{{2a2}x + {b2}} &= 0 \\ x &= \simplify[all,fractionNumbers]{{sx2}} \end{align} \]

Find the $y$-coordinate by substituting this value of $x$ into the definition of $z(x)$:

\[\begin{align} \simplify[fractionnumbers]{z({sx1})} &= \simplify[basic,fractionnumbers]{{a2}{sx2}^2+{b2}{sx2}+{c2}} \\ &= \simplify[fractionnumbers]{{sy2}} \end{align}\]

Finally, to determine the nature of the stationary point, look at $\frac{\mathrm{d}^2z}{\mathrm{d}x^2}$ at $x = \simplify[fractionnumbers]{{sx2}}$.

\[ \frac{\mathrm{d}^2z}{\mathrm{d}x^2} = \simplify{{2*a2}} \]

This is negative, so the stationary point is a maximum.

c)

$\dfrac{\mathrm{d}t}{\mathrm{d}x} = \simplify{{3m}x^2}$.

To find the $x$-coordinate of the stationary point, solve $\frac{\mathrm{d}t}{\mathrm{d}x} = 0$ for $x$:

\[ \begin{align} \frac{\mathrm{d}t}{\mathrm{d}x} = \simplify{{3m}x^2} &= 0 \\ x &= 0 \end{align} \]

Find the $y$-coordinate by substituting this value of $x$ into the definition of $t(x)$:

\[\begin{align} z(0) &= \simplify[basic,fractionnumbers]{{m}*0^3 + {q}} \\ &= \var{q} \end{align}\]

Finally, to determine the nature of the stationary point, look at $\frac{\mathrm{d}^2t}{\mathrm{d}x^2}$ at $x = 0$.

\[ \frac{\mathrm{d}^2t}{\mathrm{d}x^2} = \simplify{{6m}x} \]

\[ \left.\frac{\mathrm{d}^2t}{\mathrm{d}x^2} \right\rvert_{x=0} = \var{6m} \times 0 = 0 \]

This is zero, so the stationary point is a point of inflection.

1. Converting to $\operatorname{N}(0,1)$

$\simplify[all,!collectNumbers]{P(X < {lower}) = P(Z < ({lower} -{m}) / {s}) =1 -P(Z < {m-lower}/{s})} = 1-P(z<\var{zlower})=1 -\var{p} = \var{prob1}$ to 2 decimal places.

2. Converting to $\operatorname{N}(0,1)$

$\simplify[all,!collectNumbers]{P(X > {upper}) = P(Z > ({upper} -{m}) / {s}) = 1 -P(Z < {upper-m}/{s})} = 1-P(z<\var{zupper})=1-\var{p1} = \var{prob2}$ to 2 decimal places.

a)

1. $X \sim \operatorname{bin}(\var{number1},\var{prob})$, so $n= \var{number1},\;\;p=\var{prob}$.

2. The expectation is given by $\operatorname{E}[X]=n\times p=\var{number1}\times \var{prob}=\var{number1*prob}$

3. $\operatorname{stdev}(X)=\sqrt{n\times p \times (1-p)}=\sqrt{\var{number1}\times \var{prob} \times \var{1-prob}}=\var{sd}$ to 3 decimal places.

b)

1. \[ \begin{eqnarray*}\operatorname{P}(X = \var{thisnumber}) &=& \dbinom{\var{number1}}{\var{thisnumber}}\times\var{prob}^{\var{thisnumber}}\times(1-\var{prob})^{\var{number1-thisnumber}}\\& =& \var{comb(number1,thisnumber)} \times\var{prob}^{\var{thisnumber}}\times\var{1-prob}^{\var{number1-thisnumber}}\\&=&\var{prob1}\end{eqnarray*} \] to 3 decimal places.

2. 

\[ \begin{eqnarray*}\operatorname{P}(X \leq \var{thatnumber})& =& \simplify[all,!collectNumbers]{P(X = 0) + P(X = 1) + {v}*P(X = 2)}\\& =& \simplify[zeroFactor,zeroTerm,unitFactor]{{1 -prob} ^ {number1}+ {number1} *{prob} *{1 -prob} ^ {number1 -1} + {v} * ({number1} * {number1 -1}/2)* {prob} ^ 2 *( {1 -prob} ^ {number1 -2})}\\& =& \var{prob2}\end{eqnarray*} \]

to 3 decimal places.

a)

1. $X \sim \operatorname{Poisson}(\var{thismany})$, so $\lambda = \var{thismany}$.

2. The expectation is given by $\operatorname{E}[X]=\lambda=\var{thismany}$

3. $\operatorname{stdev}(X)=\sqrt{\lambda}=\sqrt{\var{thismany}}=\var{sd}$ to 3 decimal places.

b)

1. \[ \begin{eqnarray*}\operatorname{P}(X = \var{thisnumber}) &=& \frac{e ^ { -\var{thismany}}\var{thismany} ^ {\var{thisnumber}}} {\var{thisnumber}!}\\& =& \var{prob1} \end{eqnarray*} \] to 3 decimal places.

2. If an employee receives a warning then he or she must have sold less than 3.

Hence we need to find :

\[ \begin{eqnarray*}\operatorname{P}(X < \var{number1})& =& \simplify[all,!collectNumbers]{P(X = 0) + P(X = 1) + {v}*P(X = 2)}\\& =& \simplify[all,!collectNumbers]{e ^ { -thismany} + {thismany} * e ^ { -thismany} + {v} * (({thismany} ^ 2 * e ^ { -thismany}) / 2)} \\&=& \var{prob2} \end{eqnarray*} \]

to 3 decimal places.

If $X \sim \operatorname{exp}(\lambda)$ then $\displaystyle \operatorname{E}[X] =\frac{1}{\lambda}$ and  $\displaystyle \operatorname{Var}(X)=\frac{1}{\lambda^2}$.

Also $P(X \lt a)=1-e^{-\lambda a}$.

a)

If $X \sim \operatorname{exp}(\var{ra})$ then:

$\displaystyle \operatorname{E}[X] =\frac{1}{\lambda}=\frac{1}{\var{ra}}=\var{ans1}$ to 3 decimal places.

$\displaystyle \operatorname{Var}(X) =\frac{1}{\lambda^2}=\frac{1}{\var{ra}^2}=\var{ans2}$ to 3 decimal places.

b)

$P(X \lt \var{thistime}) = 1 -(e ^ {-\var{ ra} \times \var{thistime}}) = 1 -(e ^ { -\var{ra * thistime}}) = \var{ans3}$ to 3 decimal places.

a) For a Uniform distribution \[X \sim \operatorname{U}(\var{lower},\var{upper})\] we have:

$\displaystyle \operatorname{E}[X] = \frac{\var{lower}+\var{upper}}{2}=\var{ans1}$m

$\displaystyle \operatorname{Var}(X) = \frac{(\var{upper}-\var{lower})^2}{12}=\frac{(\var{upper-lower})^2}{12}=\var{ans2}$ to 3 decimal places.

b)

$\displaystyle P(X \le \var{thisdis}\textrm{km})=\frac{\var{thisdis}\times 1000 -\var{lower}}{\var{upper}-\var{lower}}=\var{ans3}$ to 3 decimal places.

1.

The population variance is unknown. So we have to use the t tables to find the confidence interval.

2.

We now calculate the $\var{confl}$ confidence interval:

As we have $\var{n}-1=\var{n-1}$ degrees of freedom, the interval is given by:

\[ \var{m[s]} \pm t_{\var{n-1}}\sqrt{\frac{\var{sd[s]}^2}{\var{n}}}\]

Looking up the t tables for $\var{confl}$% we see that $t_{\var{n-1}}=\var{invt}$ to 3 decimal places.

Hence:

Lower value of the confidence interval $=\;\displaystyle \var{m[s]} -\var{invt} \sqrt{\frac{\var{sd[s]} ^ 2} {\var{n}}} =\var{p} \var{lci}$ to 2 decimal places.

Upper value of the confidence interval $=\;\displaystyle \var{m[s]} +\var{invt} \sqrt{\frac{\var{sd[s]} ^ 2} {\var{n}}} = \var{p}\var{uci}$ to 2 decimal places.

a)

We use the z tables to find the confidence interval as we know the population variance.

We now calculate the $\var{confl}$% confidence interval.

Note that $z_{\var{confl/100}}=\var{zval}$ and the confidence interval is given by:

\[ \var{m[s]} \pm z_{\var{confl/100}}\sqrt{\frac{\var{sd2}}{\var{n}}}\]

Hence:

Lower value of the confidence interval $=\;\displaystyle \var{m[s]} -\var{zval} \sqrt{\frac{\var{sd2}} {\var{n}}} = \var{lci}$pounds to 2 decimal places.

Upper value of the confidence interval $=\;\displaystyle \var{m[s]} +\var{zval} \sqrt{\frac{\var{sd2}} {\var{n}}} = \var{uci}$pounds to 2 decimal places.

b)

Since $\var{aim}$ does not lie in the confidence interval the answer is no.

For part a) you calculate $r$ using:

\[r=\frac{S_{XY}}{\sqrt{S_{XX} \times S_{YY}}}\] where :

$\displaystyle S_{XY}=\sum xy - n\overline{x}\overline{y}$

$\displaystyle S_{XX}=\sum x^2 - n\overline{x}^2$

$\displaystyle S_{YY}=\sum y^2 - n\overline{y}^2$

For part b): The regression line has equation:

$\simplify[all,!collectNumbers]{Y={a}+{b}X}$ and this is displayed below:  

For part c):

Predicted sales whem $X=\var{thisval}^{\small o}$C:

\[\begin{align} Y&=\simplify[all,!collectNumbers]{{a}+{b}* {thisval}}\\
&=\var{{a+b*thisval}}\\
&=\var{prediction}
\end{align}\] to nearest whole number of hundreds of pounds.

b)

Step 3 : Test statistic

We should use the   t statistic as the population variance is unknown.

The pooled standard deviation  is given by :

\[s = \sqrt{\frac{\var{n1 -1} \times \var{sd} ^ 2 + \var{n2 -1} \times \var{sd1} ^ 2 }{\var{n1} + \var{n2} -2}} = \var{tpsd} = \var{psd}\] to 3 decimal places.

The test statistic is given by \[t = \frac{|\var{m} -\var{m1}|}{s \sqrt{\frac{1}{ \var{n1} }+\frac{1}{ \var{n2}}}} = \var{tval}\] to 3 decimal places.

(Using $s=\var{tpsd}$ in this formula.)

c)

Step 4: p value range.

As  the degree of freedom is $\var{n1}+\var{n2}-2=\var{n-1}$ we use the $t_{\var{n-1}}$ tables.  We have the following data from the tables:

We see that the $p$ value is greater than 10%.

d)

Step 5: Conclusion

Hence there is no evidence against $\operatorname{H}_0$ and so we retain $\operatorname{H}_0$.

There is insufficent evidence to suggest that average call times for male and female employees differ.

a)

Step 1: Null Hypothesis

$\operatorname{H}_0\;: \; \mu=\;\var{thisamount}$

Step 2: Alternative Hypothesis

$\operatorname{H}_1\;: \; \mu \neq\;\var{thisamount}$

b)

We should use the t statistic as the population variance is unknown.

The test statistic:

\[t =\frac{ |\var{m} -\var{thisamount}|} {\sqrt{\frac{\var{stand} ^ 2 }{\var{n}}}} = \var{tval}\]

to 3 decimal places.

c)

As  $n=\var{n}$ we use the $t_{\var{n-1}}$ tables.  We have the following data from the tables:

We see that the $p$ value is less than 1%.

d)

Hence there is strong evidence against $\operatorname{H}_0$ and so we reject $\operatorname{H}_0$.

There is sufficient evidence against the claim of the flight company.

a)

Step 1: Null Hypothesis

$\operatorname{H}_0\;: \; \mu=\;\var{thismuch}$

Step 2: Alternative Hypothesis

$\operatorname{H}_1\;: \; \mu \lt\;\var{thismuch}$

b)

We should use the z statistic as the population variance is known.

The test statistic:

\[z =\frac{ |\var{m} -\var{thismuch}|} {\sqrt{\frac{\var{thisvar}}{\var{n}}}} = \var{zval}\]

to 3 decimal places.

c)

We see that the $p$ value is less than 1%.

d)

Hence there is strong evidence against $\operatorname{H}_0$ and so we reject $\operatorname{H}_0$.

There is sufficient evidence against the claim of the vending company.