Question

Sample Proportions: Can someone please help me?

Answer 1

I might be wrong, but part B looks like a hypothesis testing problem in disguise. You have to compute the test statistic and determine the \(p\) value from that.

If I happen to be right, you are essentially testing the following:
\[\text{Null hypothesis: }p_0=0.95\\
\text{Alt. hypothesis: }p_0<0.95\]
Your test statistic would likely be
\[Z=\frac{\hat{p}-p_0}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}\]
where \(\hat{p}=0.91\) is the sample proportion and \(n=100\) is the sample size.

Answer 2

Okay. Like I did find the standard deviation which is Square root of .95 times (1-.95)/100=Sq root of .000475=.02179

Answer 3

Now I was told in order to find probability use the normalcdf function on a graphing calculator but what numbers would I use?

Answer 4

I've never used a calculator's built in functions to find \(p\) values, so I'd refer you to this page for the proper syntax: http://tibasicdev.wikidot.com/normalcdf

From what I can tell, you need to manually input the mean and standard deviation, so that's just \(n\hat{p}\) and \(\sqrt{n\hat{p}(1-\hat{p})}\), respectively. I'm not sure about the limits just yet... My first guess would be \(-\infty\) to the test statistic, which I believe is around \(-1.4\) ? Of course, you can't plug in \(-\infty\), so you should just use a large negative number, like \(-100,000\).

Answer 5

Just so you have an idea of why I did what I did: the question asks for the probability that any SRS would get a proportion of \(0.91\) or less; that is, \(P(\hat{p}\le0.91)\). To use the normal distribution as an approximation, we transform to the standard normal distribution:
\[P(\hat{p}\le0.91)\approx P\left(\frac{\hat{p}-p_0}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}\le\frac{0.91-0.95}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}\right)=P(Z\le-1.4)\]

Answer 6

Oh I see. The reason why I asked because I do not want to use something that wasn't taught such as the equation you used up above.

Answer 7

I know that I would use the normalcdf function but I just don't know what numbers to insert for the limits.

Answer 8

normalcdf should give you the probability above. I think you would type something like

`normalcdf( -1000000, -1.4, 95, 2.92 )`

Answer 9

Alternatively, you could use

`normalcdf( (-1000000-95)/2.92, (-1.4-95)/2.92 )`

which, according to the page I linked above, should give the same value.

Answer 10

Okay, why is there a -1.4 in normalcdf?

Answer 11

Something's not sitting right with me... Let me think about this for a minute or two.

Anyway, you can think of \(P(Z\le-1.4)\) as the area under the curve of the probability density function (the standard normal bell curve). The `normalcdf` function computes the area over the interval `a` to `b` in the command `normalcdf( a, b )`. In this case, we want the entire area under the curve to the left of \(Z=-1.4\).

Answer 12

Im sorry for being difficult. I am just a bit confused

Answer 13

I think I know what the problem is. The syntax should be `normalcdf( -10000, -1.4)`, as in the screenshot below.

Answer 14

Second opinion? @amistre64 @kropot @perl

Answer 15

Oops, @kropot72

Answer 16

lol, I hate reading ... give me a moment

Answer 17

Your mail-order company advertises that it ships 95% of its orders within three working days. Ho, and po = .95


You select an SRS of 100 of the 5000 orders received in the past week for an audit. The audit reveals that only 91 of these orders were shipped on time.

A) What is the sample proportion of orders shipped on time?
p^ = 91/100 = .91


B) If the company really ships 95% of its orders on time, what is the probability that the proportion in an SRS of 100 orders is as small as the proportion in your sample or smaller? (Use a normal approximation, but be sure to justify the standard deviation and approximation first.)

np and nq > 5 is required for norm approx

standard error is used in place of standard deviation for comparing samples to pops




C) A critic says, "Aha! You claim 95% but in your sample the on-time percentage is lower than that. So the 95% claim is wrong." Does your probability calculation in (b) support or refute the 95% claim? Explain.

gonna have to do a cal

Answer 18

also, normal approx adjusts the X by +- .5 depending on if we include or exclude it

Answer 19

are we spose to do a hypt test?

Answer 20

B) I know that np>10 and n(1-p)>10 must be true.

Answer 21

I'm not so sure about the hyp testing. It seemed like it at first, but it looks like we're just computing a probability. The actual hyp test might be more related to part C.

Answer 22

Yeah it's more of finding the probability and I was taught by my stats teacher to use normalcdf. But I do not know which numbers to plug in for the limits. I believe the mean is .95 and the SD is .02179

Answer 23

there are 2 versions of the test that tend to get me twisted about

sample compared to pop; and pop compared to sample.

one used the sample data in its cals, the other uses the pop data in its calc

Answer 24

https://onlinecourses.science.psu.edu/stat200/node/53

Answer 25

sample compared to null data is what that site seems to infer

Answer 26

and the courses seem to be inconsistent in their np nq qualifications .. >5 was when i took it

Answer 27

normalcdf(high, low, mean, sd) is the ti83 syntax

Answer 28

low, high? mabe

Answer 29

I know that normalcdf is (lower,upper,mean,SD)

Answer 30

so the question is, do we use np and sqrt(npq) using sample or claim values?

Answer 31

It could very well be that
\[Z=\frac{\hat{p}-p_0}{\sqrt{\dfrac{\color{red}{p_0}(1-\color{red}{p_0})}{n}}}\]
I often mess up the formulas. I'll check my textbook.

Answer 32

I've never even seen that formula before. My teacher never taught that one. Is that Ap Stats?

Answer 33

If the company really ships 95% of its orders on time, then we sould be comparing po stuff i beleive

Answer 34

given that the distribution of the po stuff is actual, then we want to compare the sample to the actual distribution

Answer 35

i never took AP stats so im not sure if its in it or not

Answer 36

Same here, I took a stats course just last year. @amistre64 your formula is correct, we're using \(p_0\), not \(\hat{p}\). We're considering the distribution of \(Z\) under the null hypothesis.

Answer 37

\[P(\hat{p}\le0.91)\approx P\left(\frac{\hat{p}-p_0}{\sqrt{\dfrac{\color{red}{p_0}(1-\color{red}{p_0})}{n}}}\le\frac{0.91-0.95}{\sqrt{\dfrac{\color{red}{p_0}(1-\color{red}{p_0})}{n}}}\right)=P(Z\le\color{red}{-1.84})\]

Answer 38

\[\frac{\frac{x}{n}-p_o}{\sqrt{\frac{p_oq_o}{n}}}\]
\[\frac{x-np_o}{n\sqrt{\frac{p_oq_o}{n}}}\]
\[\frac{x-np_o}{\sqrt{\frac{n^2}{n}p_oq_o}}\]
\[\frac{x-np_o}{\sqrt{n~p_oq_o}}\]

since normal approx for "as small as 91" includes 91, X = 91.5

Answer 39

notice the binomial equaivalents for mean = np and sd = sqrt(npq) :)

Answer 40

Oh alright. I understand! That makes sense

Answer 41

Thank you to you both!

Answer 42

You're welcome! Glad we could clear things up.

Answer 43

good luck :)

Answer 44

where is the complete directions to the question