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Construct confidence interval for P1 Pz al the given level of confidonce X=27,0 264.*2 - 31,0_ 292, 99* conlixdence The researchers are [99]% confident Ihe differen...

Question

Construct confidence interval for P1 Pz al the given level of confidonce X=27,0 264.*2 - 31,0_ 292, 99* conlixdence The researchers are [99]% confident Ihe differencc belween Ihe JnJ population proparllons. (Use ascending order Type an integer decimal rounded Ihree decimnal JCos a5 needed }benwcenand

Construct confidence interval for P1 Pz al the given level of confidonce X=27,0 264.*2 - 31,0_ 292, 99* conlixdence The researchers are [99]% confident Ihe differencc belween Ihe JnJ population proparllons. (Use ascending order Type an integer decimal rounded Ihree decimnal JCos a5 needed } benwcen and



Answers

Construct the confidence interval for $p_{1}-p_{2}$ for the level of confidence and the data given. (The samples are sufficiently large.) a. $80 \%$ confidence, $$ \begin{array}{l} \mathbf{n}_{1}-300, \hat{p}_{1}-0.255 \\ \approx_{2}-400, \hat{p}_{2}-0.1 Q 3 \end{array} $$ b. $95 \%$ confidence, $$ \mathrm{w}_{1}-3500, \hat{p}_{1}-0.147 $$ $$ \mathrm{w}_{2}-3750, \hat{p}_{2}-0.131 $$

The following is a solution in # nine, and this says that we have a random sample of 40. Now, it doesn't say the population distribution, but it doesn't matter because that sample size is big enough to assume normality and assume the central limit theorem. So, had this been a Sample of like size 20 or 25, Then we would actually have to say, but it comes from a normal population, but since that sample sizes 40, it doesn't matter how skewed or or whatever the population is, 40 is big enough. And the means of that sample is 120.5, and the standard deviation of that sample is 12.9, and were asked to find the 99% confidence interval for the population mean, μ So first off, let's decide what method to use. We're gonna use the tea interval here. So the T interval. The reason why we use the tea interval is because we're estimating the population mean, So it's either going to be the Z. Or the tea interval. And the reason why we can't use the Z. Interval is because we don't know what sigma is, we don't know that population standard deviation. We only know the sample standard deviation S. So whenever you don't know that sigma, you don't know that population standard deviation, you have to use the T and a rule. So you can use the tea interval formula or you can just use technology, I'm going to use the calculator. The T. IATA four calculate because it works out pretty nice. So if you go to stat and then tests it's his eighth option down here at the T interval, so click eight and then summary stats should be highlighted there and then we can start punching in our numbers, so X. Bar for number nine was 1 20.5, and then the sample standard deviation was 12 9, And then the sample size was 40. And again we were asked to find the 99% confidence and it was a .99 would be our confidence level and then we calculate, and then this top band here, that's going to be Our confidence interval. So 1 14.98-1 26.02. So let's go and write that down. So 1 14 9, 8 To 1 26.02. Okay, so it doesn't say to interpret this, but if you were to interpret even though there's no context to this problem, we would just say we are 99% confident that the true population mean is between 1 14.98 and 1 26.2

This time I see is 0.8. My C is 0.8. I will have n sigma and exper What is my N 100? Sigma's 4.7 inches 100 That is my sample sizes 100 Sigma is 4.7 and my ex bar is 20.6 20.6 What is my Alfa? My Alfa is 0.2 are Alfa by two is 0.1 If I want to calculate my confidence interval by hand This is the Formula Sigma by route N But this term is known as the margin off error E All right now I can either use this formula in which I am going to need to find my critical value z Alfa by two and then put in the values and get my lower limit comma upper limit Or I can simply use an online website or a calculator. So let's do that. My confidence level is 80%. What is my standard deviation my populations and the deviation is 4.7. So this is 4.7 okay in this 100 expertise 20.6. So my sample means 20.6, 20.6 and my n sample sizes 100 sample sizes 100. Now if I had calculate if I had calculated this is going to give me my answer as 19.998 or I can write This is 2022 21.2 This is 20 to 21.2. This is my confidence in trouble. I am 80% confident that my two population mean will lie in this region.

We have the data set given below here and using it, we want to find the sample mean X. Bar, the sample standard deviation S. And constructed 99% confidence interval for the population we knew, assuming the population is normally distributed. So since we have this data we can find expire and s using the appropriate definitions expire, is that some of the data divided by n 2.353 and s. Is the square root of the sum of deviations of the mean square divided by n minus one. In this case 1.3 Now with X. Bar and sample standard deviation S, we can use a student's t distribution to construct this confidence interval. So we can identify the appropriate T score. So for the degree of freedom and minus one equals 14 and critical values um providing probability 140.99 for the confidence interval, we can use the students to table either on google or in a textbook to identify T C equals 2.977 Now, with this T. C, we can identify the margin of error E, which is given by this formula plugging in our T C. S and N. We obtain E equals 0.572 And now plugging into our confidence interval formula, which is that new falls between sample mean minus C, and sampling plus E gives us our confidence interval. That new is between 1.781 and 2.9 to 5 with 99% confidence.

The following is a solution to number two, and we have two different data sets. The first one we're going to find a 95% confidence interval for the difference of two population means where we're giving these summary stats. So in one, basically, I'm just looking at the sample sizes here, the sample sizes are greater than 30 and the fact that they're greater than 30 that means we can use the Z interval. So we're gonna use a two samples. The interval, you could use a T interval, but Once the sample size gets larger than 30, the Z and the t kind of converged to each other. So um here's the data for the first sample and I'm gonna use a T i 84. So on the T I T i 84 if you go to stat and then tests and we're gonna go down to this ninth option down here, the two Samp's E ent so I got a nine and then just make summary stats is highlighted there and then it's asked for sigma. So we're given actually the sample standard deviation, but the Z interval uses this population standard deviation in the end again and it really isn't gonna matter. But the sigma's one is 15, sigma two is 21. The samples mean, for the first sample of 77, that sample size is 110, The sample mean was 79 for the second sample and that sample size was 85 and then the confidence of those .95. So when we calculate, we get these numbers up here. So negative seven Three all the way up to 3.3. So let's go to write those down. So parentheses -7 271 all the way up to three 2714. So I didn't round Shouldn't make a whole lot of difference but that's the answer. Okay? And then the next one is a 90% confidence interval but just using different data value. So we're going to go back to stat and then test and it's the ninth option. So we just need to change this stuffy Signal one or S one in this case would be 12. The S two would be eight and then the X one bar is negative negative 83. And that sample size was 65 And then the X two bar was negative 74 And then that sample size was 65 And then we're asked to find the 90% of .9 and then whenever we calculate we get this negative 11.94 all the way up to negative 6.058. So let's write those down negative 11.94. All the way up to negative 6.0 5 8. And that's our 90% confidence interval given those summary stats


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