# I have a 50-point Algebra 2 project and I want to make sure it looks good and it all checks out. I think I understood everything and did it correctly, but math has never been my strongest point, so...

I have a 50-point Algebra 2 project and I want to make sure it looks good and it all checks out. I think I understood everything and did it correctly, but math has never been my strongest point, so any help/corrections (and explanations if needed, please!) would be greatly appreciated.

According to the Library of Congress, the House of Representatives has 137 legislative work-days per year.

(Null Hypothesis) Ho: μ =137

(Alternative Hypothesis) Ha: μ ≠ 137

2013: 159, 2012: 153, 2011: 175, 2010: 127, 2009: 159, 2008: 119, 2007: 164, 2006: 101, 2005: 120, 2004: 110, 2003: 133, 2002: 123, 2001: 143

Mean: 137. 38

Standard Deviation: 23.05

95% confidence interval: 124.85 ≥ μ ≤ 149.91

Because the claim the Library of Congress made (Ho: μ =137) falls within this confidence interval, we fail to reject our null hypothesis. Because the null hypothesis is the Library s claim, there is sufficient evidence to accept, or at least fail to reject, the claim that the HoR works 137 legislative days per year.

So, do my numbers all check out and make sense? This assignment is really important to my grade, and could potentially boost me from a B+ to an A if I get a good score, so please help me out! Thank you!

violy | High School Teacher | (Level 1) Associate Educator

Posted on

The null hypothesis is correct, as well as the sample mean and the sample standard deviation.

For the confidence interval, we are not given with the population standard deviation, and `n lt 30` , therefore, we will use the formula:

`barx +- t_(alpha/2)*(s/sqrt(n))`

where `bara` is the sample mean,` s` is the sample standard deviation,` n` is

the sample size and `t_(alpha/2) ` is the critical value.

We will refer on the t-table for the critical value.

Kindly refer on the attachment on how to find the value of the` t_(alpha/2)` .

So, the value of `t_(alpha/2) = 2.179` .

Applying the formula we will have:

`137.38 +- 2.179*(23.05/sqrt(13))`

`137.38 +- 2.179*6.392919761`

`137.38+- 13.93017216`

Therefore, we are 95% confidence that the population mean falls

within the interval 123.45, 151.31.

The conclusion will be the same as what you have above though, since the `137` can be found on the interval above.

That is it!

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