Statistical Significance and P Value

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The quality-control manager at a light bulb factory needs to determine whether the mean life of a large shipment of light bulbs is equal to 375 hours. The population standard deviation is 100 hours. A random sample of 64 light bulbs indicates a sample mean life of 350 hours.
a. At the 0.05 level of significance, is there evidence that the mean life is different from 375 hours?
b. Compute the p-value and interpret its meaning.
c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs.
d. Compare the results of (a) and (c). What conclusions do you reach?
 
For you to learn this stuff, it’s probably better for you to work out the answers rather than merely to read what others here write about it. So, with that in mind:
  1. Is this a 1-tail test, or a 2-tail test?
  2. Is this a small sample or a large sample?
  3. Will you be using a Z-table or a t-table?
  4. If a t-table, what is the number of degrees of freedom?
  5. What are the critical values from the table?
Answer those and we’ll continue.
 
S2000magician wrote:
For you to learn this stuff, it’s probably better for you to work out the answers rather than merely to read what others here write about it. So, with that in mind:
  1. Is this a 1-tail test, or a 2-tail test? - 2, more or less.
  2. Is this a small sample or a large sample? Large? It is more than 30.
  3. Will you be using a Z-table or a t-table? Z, N>30.
  4. If a t-table, what is the number of degrees of freedom? DF = N-1. It’s not a t though.
  5. What are the critical values from the table? I think it’s .05/2 = .025 in each tail.
Answer those and we’ll continue.
Click to expand...
  1. Is this a 1-tail test, or a 2-tail test? - 2, more or less than mean.
  2. Is this a small sample or a large sample? Large? It is more than 30.
  3. Will you be using a Z-table or a t-table? Z, N>30.
  4. If a t-table, what is the number of degrees of freedom? DF = N-1. It’s not a t though.
  5. What are the critical values from the table? I think it’s .05/2 = .025 in each tail.
Now you’ll probably tell me I’m wrong wrong wrong again, just like my g/f does daily…
 
FinanceConsultant wrote:
S2000magician wrote: For you to learn this stuff, it’s probably better for you to work out the answers rather than merely to read what others here write about it. So, with that in mind:
  1. Is this a 1-tail test, or a 2-tail test? - 2, more or less.
  2. Is this a small sample or a large sample? Large? It is more than 30.
  3. Will you be using a Z-table or a t-table? Z, N>30.
  4. If a t-table, what is the number of degrees of freedom? DF = N-1. It’s not a t though.
  5. What are the critical values from the table? I think it’s .05/2 = .025 in each tail.
Answer those and we’ll continue.
Click to expand...
  1. Is this a 1-tail test, or a 2-tail test? - 2, more or less than mean.
  2. Is this a small sample or a large sample? Large? It is more than 30.
  3. Will you be using a Z-table or a t-table? Z, N>30.
  4. If a t-table, what is the number of degrees of freedom? DF = N-1. It’s not a t though.
  5. What are the critical values from the table? I think it’s .05/2 = .025 in each tail.
Now you’ll probably tell me I’m wrong wrong wrong again, just like my g/f does daily…
Click to expand...
I cannot help with the girlfriend, but you’ve nailed almost everything.
#1: Correct! It’s a 2-tail test because the claim is that μ = 375 hours, so we could scupper that by showing that μ > 375 or μ < 375.
#2: Correct!
#3: Correct! We could use a t-table, but the values are close enough to the Z-table values that we might as well use the latter.
#4: Correct! Though I wish you’d written “63”, to let us know that you know what n is.
#5: Whoops! The critical values are the Z-values that correspond to 2.5%. What does the table tell you?
 
This is the right way to handle any question related to Hypothesis Testing. You have already answered everything but still:
1) It is a 2 tail test, as the question asks if the mean life is different from 375 or not? so if the mean life is less than 375 then its different and if it is greater than 375 it is different. So, we are looking both sides. So, it is a 2-tail test
I usually combine step 2 and 3.
2) & 3) We need to check which test is applicable. We are dealing with mean so we may apply z-test or t-test depending on the size of the sample. Our sample size 64>30, we will apply z-test
4) Its not a t though, but in case then we could have used 64 - 1 = 63 df (n-1). In our case, the values of t and z will be close
5) It is a 2-tail test @ 5% significance level. So, we will be looking at 2.5% and the value is +- 1.96
After this, we need to define Null and Alternate Hypothesis. We always give equality sign to Null. It is a 2 - tail test, so we will say:
Null, H0: Mean = 375
Alternate, HA: Mean <> 375
I will answer rest in the next post…
Hey s2000, I am new to AF, is there any way I can post pics to AF?
 
exotichedge wrote: Hey s2000, I am new to AF, is there any way I can post pics to AF?
Click to expand...
I have no idea. Some people have suggested posting the picure on some public picture-posting website and linking to it, but it’s never worked for me. Then again, I’m an idiot.
 
a. At the 0.05 level of significance, is there evidence that the mean life is different from 375 hours?

H0: mu = 375
HA: mu ≠ 375
First I calculated the critical value of z at 0.05. Since it’s a one tailed test, there is no reason to divide by 2.
=NORMSINV(0.05) = -1.64485
Then I calculated z
z = (xbar - mu)/(sigma/√n)
z = (375- 350)/(100/√64)
z = 25 / 12.5
z = 2.00
Since 2.00 is very close to the critical value of z at .05, we cannot reject the Null and conclude there may be material difference between the sample mean and population mean.

b. Compute the p-value and interpret its meaning.

p = NORMSDIST(2.0) = 0.97725
To convert this into a one-tailed p value, I subtracted from 1. That shows the “tail” of the distribution. Since this is a one tail test, there is no need to multiply it by 2 like a 2 tailed test.
1 - 0.97725 = 0.02275
Interpretation: There’s a 2.275% chance that we could see a difference this far from 350 due to random sampling error, if the true population mean was 350

c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs.

I calculated z
z = (xbar - mu)/(sigma/√n)
z = (375- 350)/(100/√64)
z = 25 / 12.5
z = 2.00
375+/- 2*(100/sqrt64) = 375 +/- 2*(12.5) = 350 < x < 400

d. Compare the results of (a) and (c). What conclusions do you reach?
We see that the Sample Mean is barely in the range of a 95% confidence interval.
 
I tried to answer by solving in on a note-book. Please take a look on the following link and see if this is helpful. I have tried to answer only part a, if helpful I will answer more.
cfa.exotichedge.com
 
You’re going about this backwards.
Start with the Z-value that corresponds to your α: 1.96.
What’s X-bar – 1.96(standard error)?
By the way: why are you using standard error instead of standard deviation?
 
S2000magician wrote:
You’re going about this backwards.
Start with the Z-value that corresponds to your α: 1.96.
What’s X-bar – 1.96(standard error)?
By the way: why are you using standard error instead of standard deviation?
Click to expand...
this for me?
 
exotichedge wrote:
S2000magician wrote: You’re going about this backwards.
Start with the Z-value that corresponds to your α: 1.96.
What’s X-bar – 1.96(standard error)?
By the way: why are you using standard error instead of standard deviation?
Click to expand...
this for me?
Click to expand...
Nope: FinancialConsultant.
I’m trying to help him, not you.
 
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