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2 peer replies, must be a minumum of 80 words and include 1 direct question

I don’t know how to handle this Statistics question and need guidance.

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In your first peer response post, look at the hypothesis test results of one of your classmates and explain what a type 1 error would mean in a practical sense. Looking at your classmate’s outcome, is a type 1 error likely or not? What specific values indicated this?

Peer 1: Lauren

Please refer to the Excel spreadsheet attached below for this forum post.

For this Hypothesis Test (AKA “Test of Significance”) we will be using a t-test. We are not using a z-test because z-tests relate to problems focused on proportions, which this problem is not. The 4-steps I will be taking to test the statement posed in the forum prompt are:

1. Stating the null hypothesis (H0) and alternative hypothesis (Ha or H1)

2. Finding the T-test state

3. Finding the P-value

4. Making a conclusion regarding the statement

First, I must determine what the 40th percentile of the data set is. To do this, I will use “=PERCENTILE.INC(E2:E11,0.4)”. This resulted in the value $29,280. The statement, “more than the 40th percentile” coincides with an upper-tailed (one-tailed) test in the form f$bar{x} data-verified=29280f$ ” src=”https://edge.apus.edu/cgi-bin/latex.cgi?%5cinline%20%5Cbar%7Bx%7D%3E29280″>. This is the alternative hypothesis. The null hypothesis is thus f$bar{x}=29280f$ .

Next, using the previously calculated mean ($31,930) and standard deviation ($6315.7739), I will calculate the T-Test State using f$frac{bar{x}-c}{frac{SD}{sqrt{n}}}f$ . Let c=29280, n=10, f$bar{x}f$ =31930, and SD=6315.7739. Excel calculates the T-Stat to be 1.33.

Next, I will calculate the p-value using “=T.DIST.RT(J4,9)” where J4 is my T-State and 9 is the Degrees of Freedom (n-1). Excel calculated the p-value to be 0.108619.

Since alpha=0.05, the p-value=0.108619, and 0.05<0.108619, the p-value is greater than alpha. Thus, I do not reject H0. This means that I cannot reject the claim that the average sale price for a car in the town’s area is $29,280. However, I have proved the town official wrong because I could not accept Ha, which was their claim that the average vehicle sells for more than $29,280.

:

Peer 2: Martine

In this week’s scenario, a town official claims that the average vehicle in their area sells for more than the 40th percentile of your data set. Using my week 1 data, we’ll run a hypothesis test to determine if the claim can be supported.

Here are my important values:

Mean: $41,117

SD: $16,398

n: 10

40th Percentile: $39,183.40 (I found this by using =PERCENTILE.INC(E2:E11,0.4)

Therefore, the statement made by the town official is that their average vehicles sells for more than $39,183.40.

Our next step is to calculate the T-Test State using this formula: ̅− / (/√)

x̅ = 41,117

c = 39,183.40

SD = 16,398

n = 10

After plugging our data into the formula, we receive an answer of 0.37288572.

We then calculate the P-Value by using the following formula in Excel: =T.DIST.RT(J4,9), giving us a result of 0.358970.

Because my P-Value of 0.358970 is greater than the Alpha of .05, I cannot reject the claim made by the town official.

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