A software company is planning for an upgrade of their software. You must charge customers $100. Are your customers willing to pay this much? You contact a random sample of 40 customers and find that 11 would pay $100 for the upgrade. If the upgrade is to be profitable, you will need to sell it to more than 20% of your customers. Do the sample data give good evidence that more than 20% are willing to buy?

a. Formulate this problem as a hypothesis test. Give the null and alternative hypotheses.

b. Carry out the significance test. Report the test statistic and the P-value.

Question

A software company is planning for an upgrade of their software. You must charge customers $100. Are your customers willing to pay this much? You contact a random sample of 40 customers and find that 11 would pay $100 for the upgrade. If the upgrade is to be profitable, you will need to sell it to more than 20% of your customers. Do the sample data give good evidence that more than 20% are willing to buy? 

a. Formulate this problem as a hypothesis test. Give the null and alternative hypotheses. 

b. Carry out the significance test. Report the test statistic and the P-value.

anonymous · Answer

c. Should you proceed with plans to produce and market the upgrade?

anonymous · Answer

@hartnn

anonymous · Answer

@matricked

anonymous · Answer

@ganeshie8

anonymous · Answer

@amistre64

amistre64 · Answer

what will you use as your hypothesises?

amistre64 · Answer

you will need to sell it to more than 20% of your customers.

Ho is by convention a statement of equality:  Ho:  p=.20

Ha defines the alternative outcome:  Ha:  p>.20  seems fair

agreed?

anonymous · Answer

Yes, and b?

amistre64 · Answer

carry out the tests of course ...

anonymous · Answer

Could you talk me through it?

amistre64 · Answer

$$z=\frac{p-.20}{\sqrt{pq/n}}$$sounds familiar, for some sample proportion.

amistre64 · Answer

what is your sample proportion?

anonymous · Answer

I'm not sure, I tried to figure it out, but can't seem to get it right.

amistre64 · Answer

how many people were sampled?

anonymous · Answer

40

amistre64 · Answer

and how many people said they would pay?

anonymous · Answer

11

amistre64 · Answer

then ... what proportion of the sample said yes?

anonymous · Answer

.275?

amistre64 · Answer

11/40 = .275  correct,  and this means the opposing proportion is?  1- .275 right?

anonymous · Answer

right

amistre64 · Answer

we are ready to compute the z:$$z=\frac{.275-.20}{\sqrt{\frac{.275(.725)}{40}}}$$

anonymous · Answer

z=1.062321348?

amistre64 · Answer

http://www.wolframalpha.com/input/?i=%5Cfrac%7B.275-.20%7D%7B%5Csqrt%7B%5Cfrac%7B.275%28.725%29%7D%7B40%7D%7D%7D not according to the wolf ...

amistre64 · Answer

hold on, i think i got some things backwards:  we use the expected value of .20 underneath

amistre64 · Answer

$$z=\frac{.275-.20}{\sqrt{\frac{.20(.80)}{40}}}$$
that is what we want ..... underneath it

amistre64 · Answer

http://www.wolframalpha.com/input/?i=%5Cfrac%7B.275-.20%7D%7B%5Csqrt%7B%5Cfrac%7B.20%28.80%29%7D%7B40%7D%7D%7D i still get someting less than 1%: .0047

anonymous · Answer

didn't your equation show that (.275-.20) was over the square root? On the Wolfram page it shows it beside the square root.

amistre64 · Answer

$$\frac{3-2}{5}=\frac{3-1}{1}*\frac{1}{5}=(3-2)*(\frac{1}{5})$$
dont fret yourself over multiplication ....

anonymous · Answer

Thats not what it looked like, could you please humor me and go to the link and look at it because I want to make sure it is correct.

amistre64 · Answer

fine fine fine ... the wolf is spose to read latex but i see that its a little confused :)

amistre64 · Answer

http://www.wolframalpha.com/input/?i=%28.275-.20%29%2F%28sqrt%7B%5Cfrac%7B.20%28.80%29%7D%7B40%7D%7D%29 thats better: z=1.18585...

anonymous · Answer

P=0.117839956714519?

amistre64 · Answer

the right tailed area from 1.185 is the P-value

amistre64 · Answer

0.117841...

so yes

amistre64 · Answer

as an overview:

we are looking to compare our sample value with a normal distribution having a mean of .20 and a standard deviation or sqrt(.20(.80)/40)

if that helps visualize this with prior work

anonymous · Answer

Thanks that helped a bunch can you help me with another question?

amistre64 · Answer

i dont read a significance value in the information, so we tend to be free to use whatever we want.  a good rule of thumb is 5%, since 5% of the data is said to occur rarely.

amistre64 · Answer

i can try ;)

anonymous · Answer

For a single proportion the margin of error of a confidence interval is largest for any given sample size n and confidence level C when p-hat = 0.5. This led us to use p-hat = 0.5 for planning purposes. Use these conservative values in the following calculations, and assume that the sample sizes of n. Calculate the margins of error of the 99% confidence intervals for the following choices of n: 10, 30, 50, 100, 200, and 500. Present the results in a table. Summarize your conclusions.

anonymous · Answer

I'm not sure how to even approach this one.

amistre64 · Answer

it amounts to reworking the z formula; given that the zscore is equal to the 99% interval mark

amistre64 · Answer

let x be the bound for the z value such that:$$\pm z_{.005}=\frac{E-\mu}{\sigma/\sqrt n}$$
solve for E

anonymous · Answer

How?:}

amistre64 · Answer

algebra of course

amistre64 · Answer

$$\pm z_{.005}=\frac{E-\mu}{\sigma/\sqrt n}$$
$$\pm \frac{z_{.005}(\sigma)}{\sqrt n}=E-\mu$$
$$\mu\pm \frac{z_{.005}(\sigma)}{\sqrt n}=E$$

anonymous · Answer

What do I substitute where is what I mean?

amistre64 · Answer

the (+-) part is the spread, or the margin of error, i believe

you know n

$\sigma=\sqrt {\hat p(1-\hat p)}$

and the zscore related to 99% about the mean leaves 1%/2 = .005 to look up

amistre64 · Answer

in this case, sigma= sqrt(p^2) = p  since p=.5 and 1-p = .5

anonymous · Answer

I'm not following...

amistre64 · Answer

$$\pm \frac{2.576(.5)}{\sqrt n}=Error_m$$

anonymous · Answer

I got it know thanks.

anonymous · Answer

now not Know