1. ## Hypothesis Test

I was wondering if I am even answering these questions right.

The Better Business Bureau has several complaints that a flour company is under-filling their 5 pound flour bags. The Bureau randomly selects 750 bags of flour and determines the weight of each bag. The sample average of the bags is 4.80 pounds with a standard deviation of .15 pounds. Is there overwhelming evidence at the .01 level that the bags are under-filled?

M = 5
x = 4.8
n = 750
s = .15

z = (x-M)/(s/n^(1/2))
z = (4.8-5)/(.15/750^(1/2))
= -36.5

The next step I am unsure of.

P = P(z>-36.5) < .005?
How do I get the z value? I have a table in the back of the book that only shows up to -3.4 but that is .0003 so -36.5 has to be much less then that so it is true? Not even sure I am doing this right.

A fuel oil company claims that one-fifth of the homes in a certain city are heated by oil. Do we have reason to doubt this claim if, in a random sample of 1000 homes in this city, it is found that 136 are heated by oil? Use a .01 level of significance.

p = .20
q = .8
n = 1000
x = 136

z = (x-np)/(npq)^(1/2)
z = (136-1000(.2))/(1000*.2*.8)^(1/2)
= -5.06

P = P(z > -5.06) < .005

The oil company is right?

Again not sure I am using right formula or doing it right. Also q is just 1-p, correct?

Thanks

2. Originally Posted by kenshinofkin
I was wondering if I am even answering these questions right.

The Better Business Bureau has several complaints that a flour company is under-filling their 5 pound flour bags. The Bureau randomly selects 750 bags of flour and determines the weight of each bag. The sample average of the bags is 4.80 pounds with a standard deviation of .15 pounds. Is there overwhelming evidence at the .01 level that the bags are under-filled?

M = 5
x = 4.8
n = 750
s = .15

z = (x-M)/(s/n^(1/2))
z = (4.8-5)/(.15/750^(1/2))
= -36.5

The next step I am unsure of.

P = P(z>-36.5) < .005?
How do I get the z value? I have a table in the back of the book that only shows up to -3.4 but that is .0003 so -36.5 has to be much less then that so it is true? Not even sure I am doing this right.

[snip]
Since the population sd is unknown you're using the sample sd. This means that you have to use the t-distribution, NOT the z-distribution. BUT ..... since n is really big there's no real difference between using z-distribution and t-distribution for 750 - 1 = 749 degrees of freedom. So you're OK this time.

You should realise that Pr(z < -36.5) is for all practical purposes equal to zero. And zero is certainly less than 0.01 ..... Overwhelming is certainly a good adjective to use here ......

Note: It's Pr(z < - 36.5) NOT Pr(z > - 36.5) ....... Capisce?

3. Originally Posted by mr fantastic
Since the population sd is unknown you're using the sample sd. This means that you have to use the t-distribution, NOT the z-distribution. BUT ..... since n is really big there's no real difference between using z-distribution and t-distribution for 750 - 1 - 749 degrees of freedom. So you're OK this time.

You should realise that Pr(z < -36.5) is for all practical purposes equal to zero. And zero is certainly less than 0.01 ..... Overwhelming is certainly a good adjective to use here ......

Note: It's Pr(z < - 36.5) NOT Pr(z > - 36.5) ....... Capisce?
Ok so I guess I got the second one right?

So the first question should be t-distributions? So

t = (x-M)/(s/n^(1/2))
t = (4.8-5)/(.15/750^(1/2))
= -36.5

So the v = 749 and a(forgot what the symbol is called) = .01 the critical value equals 2.326 or so. So I should reject the value?

4. Originally Posted by kenshinofkin
[snip]
A fuel oil company claims that one-fifth of the homes in a certain city are heated by oil. Do we have reason to doubt this claim if, in a random sample of 1000 homes in this city, it is found that 136 are heated by oil? Use a .01 level of significance.

p = .20
q = .8
n = 1000
x = 136

z = (x-np)/(npq)^(1/2)
z = (136-1000(.2))/(1000*.2*.8)^(1/2)
= -5.06

P = P(z < -5.06) < .005

The oil company is right?

Again not sure I am using right formula or doing it right. Also q is just 1-p, correct? Mr F says: Yes.

Thanks
You seem to have a general problem getting the correct direction for your inequality signs. Note my correction in red.

The P-value is 2 P(z < -5.06) < 0.01 (since it's a two-tailed test).

Do you understand that the P-value is the probability that you get the data that you got given that the null hypothesis is true. So the null hypothesis is rejected.

The null hypothesis is that the proportion is equal to 0.2 .......

5. Originally Posted by mr fantastic
Since the population sd is unknown you're using the sample sd. This means that you have to use the t-distribution, NOT the z-distribution. BUT ..... since n is really big there's no real difference between using z-distribution and t-distribution for 750 - 1 = 749 degrees of freedom. So you're OK this time.

You should realise that Pr(z < -36.5) is for all practical purposes equal to zero. And zero is certainly less than 0.01 ..... Overwhelming is certainly a good adjective to use here ......

Note: It's Pr(z < - 36.5) NOT Pr(z > - 36.5) ....... Capisce?
Originally Posted by kenshinofkin
Ok so I guess I got the second one right?

So the first question should be t-distributions? So

t = (x-M)/(s/n^(1/2))
t = (4.8-5)/(.15/750^(1/2))
= -36.5

So the v = 749 and a(forgot what the symbol is called) = .01 the critical value equals 2.326 or so. So I should reject the value?
Look again at what I said and you will see that I said that because n was so large there is no real difference between using z-distribution and t-distribution for 750 - 1 = 749 degrees of freedom. So it's OK not to use the t-distribution. Notice that z-critical = 2.33 and t-critical = 2.33 ......