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3 years ago

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A manufacturer produces bolts of a fabric with a fixed width. A quantity q of this fabric (measured in yards) that is sold is a

function of the selling price p(in dollars per yard), so we can write q=f(p). Then, the total revenue earned with selling price p is R(p)=pf(p).
Find R′(20), given f(20)=13000 and f′(20)=−550. Also explain, what these equations mean.

1 answer:

BigorU [14]3 years ago

7 0

Answer:

$R'(20) = 2000$

Step-by-step explanation:

We are given the following in the question:

Quantity, q

Selling price in dollars per yard, p

$q=f(p)$

Total revenue earned =

$R(p)=pf(p)$

f(20)=13000

This means that 13000 yards of fabric is sold when the selling price is 20 dollars per yard.

f′(20)=−550

This means that increasing the selling price by 1 dollar per yards there is a decrease in fabric sales by 550.

We have to find R'(20)

Differentiating the above expression, we have,

$R'(p) = \displaystyle\frac{d(R(p))}{dp} = \frac{d(pf(p))}{dp} = f(p) + pf'(p)$

Putting the values, we get,

$R'(p) = f(p) + pf'(p)\\\\R'(20) = f(20) + 20(f'(20))\\\\R'(20) = 13000 + 20(-550) = 2000$

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a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

$\hat p=\frac{64}{100}=0.64$ estimated proportion for were in favor of firing the coach

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

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Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

$0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719$

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

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The statistic is calculated with the following formula:

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$p_v =P(z>2.8)=1-P(z$

Part c

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2 years ago