Hey guys,

This is my first thread on this forum so please bare with me as I try to be as concise as possible, I'll then write down my questions at the end of the thread.

I'm working on some class notes, and stumbled upon this problem that confuses me on some point:

"The drying time of a certain type of paint, under fixed environmental conditions, is known to be normally distributed with mean 75 min. and standard deviation 9 min, i.e. N(75,9). Chemists have added a new additive that is believed to decrease drying time and have obtained a sample of 35 drying times with a sample mean of 70.8 min. and wish to test their assertion."

The professor then explains how to set the null hypothesis (that the mean drying time hasn't changed) after adding the additive.

He then says that we'll be testing for the type I error.

He then sets up the normal distribution: Xbar~N(75,(9^2)/35), but then some tomfoolery happens and all of sudden it becomes (Xbar-75)/(9/sqrt(35))~N(0,

He then sets up the plugs Xbar=70.8 and gets a Z-score of -2.76.

Questions:

1. What does it mean to test for type I error, in terms of the problem? And what would it look like testing for type II errors (in terms of that specific problem again)?

2. What is the meaning of the aforementioned tomfoolery? My guess is that the professor scaled down the distribution to make it easier to read, but why is the 1 (the one in bold and underlined) standing there by itself? the second parameter of the normal distribution is (SD^2/n), is the SD and n scaled to one after the tomfoolery?

3. The T.S (test statistics) is set to be -2.76. Why is test statistic the same as the Z-score formula? Also, since the standard deviation is set to be 9 (by the problem) and that the Z-score tells you how many standard deviation a sample mean is away from the population mean, why is the T.S -2.76? 70.8 (sample mean) is clearly not 2.76 times away from 75 (population mean).

Sorry if I bombarded you guys with questions, I don't learn unless I have a rock solid understanding of first principles. Thanks again for reading!

LOVE...

EDIT: I have actually answered my own question (why is the T.S -2.76? 70.8 is clearly not 2.76 times away from 75).

I've made two mistakes: I've mixed SD and sampling distribution SD. I've also realized that hypothesis testing requires to take the null hypothesis as being originally correct.

The original problem states an SD of 9, meaning that on average, most of the values (more specifically 68%) are within +- 9 of the mean. On the other hand, the sampling distribution SD is set by SD/sqrt(n) (which is 1.52 in the problem) means that on average, sample means are +-1.52 away from the real mean.

Given that the null hypothesis is correct, that would mean that 70.8 is -2.76 SD from the mean. We've already established that most values are within 1 SD from the mean, and 95% of values are within 2SDs from the mean. If 75 was truly the mean, it would be ludicrous to think that we got a sample mean of 70.8 since 95% of the values are between (75-2(1.52),75+2(1.52)) or (71.96,78.04). We then reject the null hypothesis since it doesn't make sense from a probability perspective.

This is my first thread on this forum so please bare with me as I try to be as concise as possible, I'll then write down my questions at the end of the thread.

I'm working on some class notes, and stumbled upon this problem that confuses me on some point:

"The drying time of a certain type of paint, under fixed environmental conditions, is known to be normally distributed with mean 75 min. and standard deviation 9 min, i.e. N(75,9). Chemists have added a new additive that is believed to decrease drying time and have obtained a sample of 35 drying times with a sample mean of 70.8 min. and wish to test their assertion."

The professor then explains how to set the null hypothesis (that the mean drying time hasn't changed) after adding the additive.

He then says that we'll be testing for the type I error.

He then sets up the normal distribution: Xbar~N(75,(9^2)/35), but then some tomfoolery happens and all of sudden it becomes (Xbar-75)/(9/sqrt(35))~N(0,

**1**).He then sets up the plugs Xbar=70.8 and gets a Z-score of -2.76.

Questions:

1. What does it mean to test for type I error, in terms of the problem? And what would it look like testing for type II errors (in terms of that specific problem again)?

2. What is the meaning of the aforementioned tomfoolery? My guess is that the professor scaled down the distribution to make it easier to read, but why is the 1 (the one in bold and underlined) standing there by itself? the second parameter of the normal distribution is (SD^2/n), is the SD and n scaled to one after the tomfoolery?

3. The T.S (test statistics) is set to be -2.76. Why is test statistic the same as the Z-score formula? Also, since the standard deviation is set to be 9 (by the problem) and that the Z-score tells you how many standard deviation a sample mean is away from the population mean, why is the T.S -2.76? 70.8 (sample mean) is clearly not 2.76 times away from 75 (population mean).

Sorry if I bombarded you guys with questions, I don't learn unless I have a rock solid understanding of first principles. Thanks again for reading!

LOVE...

EDIT: I have actually answered my own question (why is the T.S -2.76? 70.8 is clearly not 2.76 times away from 75).

I've made two mistakes: I've mixed SD and sampling distribution SD. I've also realized that hypothesis testing requires to take the null hypothesis as being originally correct.

The original problem states an SD of 9, meaning that on average, most of the values (more specifically 68%) are within +- 9 of the mean. On the other hand, the sampling distribution SD is set by SD/sqrt(n) (which is 1.52 in the problem) means that on average, sample means are +-1.52 away from the real mean.

Given that the null hypothesis is correct, that would mean that 70.8 is -2.76 SD from the mean. We've already established that most values are within 1 SD from the mean, and 95% of values are within 2SDs from the mean. If 75 was truly the mean, it would be ludicrous to think that we got a sample mean of 70.8 since 95% of the values are between (75-2(1.52),75+2(1.52)) or (71.96,78.04). We then reject the null hypothesis since it doesn't make sense from a probability perspective.

Last edited: