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QUESTIONI New York is known as 'the city that never sleeps: Arandom sample of 25 New Yorkers was asked how much sleep they get per night: Statistical summaries...

Question

QUESTIONI New York is known as 'the city that never sleeps: Arandom sample of 25 New Yorkers was asked how much sleep they get per night: Statistical summaries of these data are shown below: The point estimate suggests New Yorkers sleep less than 8 hours night 0n average: Is the result statistically significant?minmax 9.78257.730.776.17Write the hypotheses in symbols and words: Check conditions, calculate the test statistic, T,= and the associated degrees of freedom Find the p-value and dra

QUESTIONI New York is known as 'the city that never sleeps: Arandom sample of 25 New Yorkers was asked how much sleep they get per night: Statistical summaries of these data are shown below: The point estimate suggests New Yorkers sleep less than 8 hours night 0n average: Is the result statistically significant? min max 9.78 25 7.73 0.77 6.17 Write the hypotheses in symbols and words: Check conditions, calculate the test statistic, T,= and the associated degrees of freedom Find the p-value and draw conclusion based on a significance level of a = 0.05.



Answers

Preliminary data analyses indicate that you can reasonably apply the $t$ -interval procedure (Procedure 8.2 on page 346 ). \In $1908,$ W. S. Gosset published the article "The Probable Error of a Mean" (Biometrika, Vol. 6, pp. 1-25). In this pioneering paper, written under the pseudonym "Student," Gosset introduced what later became known as Student's $t$ -distribution. Gosset used the following data set, which gives the additional sleep in hours obtained by a sample of 10 patients using laevohysocyamine hydrobromide. $$\begin{array}{ccccc} \hline 1.9 & 0.8 & 1.1 & 0.1 & -0.1 \\ 4.4 & 5.5 & 1.6 & 4.6 & 3.4 \\ \hline \end{array}$$ a. Obtain and interpret a $95 \%$ confidence interval for the additional sleep that would be obtained on average for all people using laevohysocyamine hydrobromide. (Note: $\bar{x}=2.33 \mathrm{hr}$ $s=2.002 \mathrm{hr} .$ b. Was the drug effective in increasing sleep? Explain your answer.

All right. So we have our non hypothesis of no change and an alternative hypothesis of a change. So that would indicate migration. Um Our population is the United States and the region is variable that we are looking at. Currently we're looking at four different regions which means that our sample sizes for And that are degrees of freedom is one lesson that so three At the alpha equals 5% significance level. Um we're looking to see if there is a change. So that means that our graph, it's gonna look like this fat Oh that is a beautiful, beautiful guys, great distribution. And we're gonna be chopping off each tail here and tail there. We're gonna be adding 2.5% to each tail right there. Doesn't look like a five. But yeah, and then we find out that And table 7,5%, is over here. So we get 711 and 5% over there is gonna be uh lost the page here, 5% 3° of freedom. 7.815. Yeah. So those are bounds. I used Excel to calculate the test statistic Guys Square is given by this, and Excel gives us a value of 8.163 which is going to land us to the right of that farthest rejection line. So we reject the null hypothesis in favor of the alternative. This just means that there has been a migration.

Alright, population of this problem is the usa the rarely bill being tested. It is the regions song. Yeah. Regions. And there's four of them, Meaning that there's a sample size of four. So therefore we have 3° of freedom At the 5% significance level. So alpha is equal 2.05. We got our graph that's archive square chop it off there. That area under the curve. There is 5%. We'll go to the back of book Table seven. Mhm Yeah. Yeah. And at 3° of freedom. 5%. Oh, just lost a page hold up on a second. Yeah. Mm. Mm hmm. Uh huh, wow. It's probably easier to page through it in an actual book. All right. So 3° of freedom at 5% is 7.815. So that critical value is 7.815. And may I present you the table where we have expected minus observed square divided by the expected right here. Oh, it's up there. Yeah. Mm. Keep scrolling down. Yeah, there it is. All right. So we got 8.163 which clearly lands in the rejection region. So we reject the null hypothesis, which is that there isn't movement and accept the alternative that there has been migration of the regions of people in the regions.

So the variable under consideration is hours slept. The two populations we are looking at is some a long word. Virtual. Just abbreviate to LH And the second one there is one C. Is going to be just be the differences between these populations. Okay. And can we use these differences? Yes we can. Since they are differences. Yeah. So our alternative hypothesis is going to be the thing that one of them is greater than the other. Actually it's gonna be, the first one is greater than the other. So um you want greater than youtube making that a right tailed test. Now they give us that D. bar is 2.33. So put that there & SD. is 2.02. Mhm. Yeah. And we have how many samples? 345 10. So two divided by Square root of 10. This is going to be approximately equal to let me bust out my calculator here. 2.02 Divided by Square Root of 10. 2.33 divided by 6.66 there. Yeah .63. This is about equal to 3.680. This is going to get us a p value of Oh, I gotta go all the way back the book here. It wasn't at the right page. And by analysis of critical values 3.688 lands as well, past the 1.833 for the 5% critical value And for the 2.4-1 critical value at the 1% significance level. So in both cases, we will reject the null hypothesis.

Alright for this problem we're going to verify that we can use the chi squared independence tests before we stayed out our hypotheses. So we got lawyers making bank um 250,000, to 500,000. More than 500,000 in the government, judicial, private and salaried sectors. So this is our original table right here. All good. And then uh I just copied it from the last table since we had the same dimensions and adjusted it that way. So this is our expected table right here. Um So what we're looking at right here are 12 three values that are less than five, which violates our second term per second assumption being that we can't have expected frequencies that are less than five. So with that said we cannot apply the chi squared test. Okay?


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