# Intro to stats question

#### Statdummy1

##### New Member
I want to verify that I'm doing things correctly. My result is to reject the null. Am I on the right track??
1. The following data was obtained from a randomly selected sample of 20-year-old, 6-foot-1-inch males from Minnesota. This sample of 25 males had a mean weight of 177 lb. Based on this sample, are Minnesota males heavier than males in general? The mean weight of all U.S. males is 170 lb., and the population standard deviation is 16.
Null Hypothesis: Ho: µ=170
Alt Hypothesis: H1: µ≠170
Using a significance level of .05
I'm using a one-sample z test and my results were z=2.19.
The mean weight of Minnesota males are heavier than males in general.

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#### Stats fan

##### Member
Since you are asked to test whether the mean weight of Minnesota males is larger than the mean weight of all U.S. males, the alternative hypothesis is that µ > 170. So you need to test right sided, not two sided. Other than that, your answer looks correct to me!

#### Statdummy1

##### New Member
Since you are asked to test whether the mean weight of Minnesota males is larger than the mean weight of all U.S. males, the alternative hypothesis is that µ > 170. So you need to test right sided, not two sided. Other than that, your answer looks correct to me!
Thank you!
I have another one! What is throwing me is the give "variance" instead of standard deviation. Would I use a two-tailed t-test?

1. The mean hemoglobin of 16 Caucasian women is 13.7 with a variance of 2.3. The mean hemoglobin of a sample of 20 African American women is 12.5 with a variance of 2.1. Is there a significant difference between Caucasian and African American women’s hemoglobin, at a significance level of 0.05?
Two-tailed t-test.

#### Stats fan

##### Member
Two tailed t test seems good to me. Why are you confused by the variances?

#### Statdummy1

##### New Member
I thought it should be called standard deviation instead of "variances".
Can you check my work below??

Appropriate Test: Independent, non-directional (two-tailed) t-test
Null hypothesis: H 0: μ 1 = μ 2
There is no significant difference between the mean hemoglobin of Caucasian women and the mean hemoglobin of African American woman.
Alternative hypothesis: H 0: μ 1 ≠ μ 2
There is a significant difference between the mean hemoglobin of Caucasian women and the mean hemoglobin of African American women.
Calculate using:

"sum of pooled variance" = 170.12

Critical value = between 3.2 – 4.3
t value = .27
Decision rule: If t is less than -4.3 or greater than 4.3, there is not significant difference to reject the null hypothesis.
Conclusion: There is not enough of a difference between the mean hemoglobin of Caucasian women and the mean hemoglobin of African American woman to reject the null hypothesis.

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#### Stats fan

##### Member
The standard deviation is just the square root of the variance: sd = sqrt(var), so var = sd^2. Your confusion caused you to compute the pooled variance s2_p wrong. Can you retry now that you know that var = sd^2? Also, remember that the pooled variance can be considered a weighed average of the sample variances, so finding a pooled variance of 170.12 based on two sample variances of 2.3 and 2.1 would be weird ;-)

#### emrah1004

##### New Member
Hi,

I would like to compare different rating methods statistically. I will have 200 students' essays to be rated in two different methods (double raters + discussion in case of a difference VS double-blind raters + third rater in case of a difference) and I would like to compare the results.

Can I calculate z scores and compare the z scores to suggest that one method is better than the other?

Can you suggest an alternative analysis?

#### Statdummy1

##### New Member
Okay Statfan!
I tried again.
• The mean hemoglobin of 16 Caucasian women is 13.7 with a variance of 2.3. The mean hemoglobin of a sample of 20 African American women is 12.5 with a variance of 2.1. Is there a significant difference between Caucasian and African American women’s hemoglobin, at a significance level of 0.05?
I revised the sum of pooled variance (it would help if I didn't use the mean where the standard deviation goes). There were all kinds of fails in my first attempt.
How does this look??
(2.3)(15) = 34.50
(2.1)(19) = 39.90

t = 13.7-12.5
2.19 + 2.19
16 20
square root of denominator

1.2 = 2.4
.50

t = 2.4
With a critical value of -2.03 and 2.03

Decision rule: If t is less than -2.03 or greater than 2.03, there IS significant difference to reject the null hypothesis.
Conclusion: There IS a significant difference between the mean hemoglobin of Caucasian women and the mean hemoglobin of African American woman to reject the null hypothesis.

#### Stats fan

##### Member
Your critical t value and decision rule are now correct. I do get a different observed t value though. Your computations look a bit unclear due to some formatting issues (I think), so it's hard for me to see what went wrong. Here is my computation of t:

t = (13.7 - 12.5) / sqrt(2.19^2/16 + 2.19^2/20) = 1.63

#### Statdummy1

##### New Member
Yeah - I got it! I wasn't squaring the 2.19 (SS1 and SS2) One more question - My conclusion is wrong because 1.63 is lower than the critical value, 2.03. So I can't reject the null - correct?

#### Stats fan

##### Member
Woops sorry, your t of 2.4 was correct... 2.19 is the pooled variance, I misused it as the pooled standard deviataion...

#### Stats fan

##### Member
So your initial conclusion that H0 can be rejected is also correct!

#### Statdummy1

##### New Member
Oh. Okay! So I don't square the SS1 and SS2?

#### Stats fan

##### Member
The t value is equal to

t = (xbar1 - xbar2) / sqrt(sp^2 / n1 + sp^2 / n2)

The pooled variance sp^2 is equal to

sp^2 = (SS1 + SS2) / (n1 - 1 + n2 - 1)

= ((n1 - 1)*var1 + (n2 - 1)*var2) / (n1 - 1 + n2 - 1)

= ((n1 - 1)*sd1^2 + (n2 - 1)*sd2^2) / (n1 - 1 + n2 - 1)

So you are not squaring SS1 and SS2.