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According to the Vivino website, suppose the mean price for abottle of red wine that scores 4.0 or higher on the Vivino RatingSystem is $32.48. A New England-based ...

Question

According to the Vivino website, suppose the mean price for abottle of red wine that scores 4.0 or higher on the Vivino RatingSystem is $32.48. A New England-based lifestyle magazine wants todetermine if red wines of the same quality are less expensive inProvidence, and it has collected prices for 61 randomlyselected red wines of similar quality from wine stores throughoutProvidence. The mean and standard deviation for this sample are$30.15 and $12, respectively.(a)Develop appropriate hypotheses

According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is $32.48. A New England-based lifestyle magazine wants to determine if red wines of the same quality are less expensive in Providence, and it has collected prices for 61 randomly selected red wines of similar quality from wine stores throughout Providence. The mean and standard deviation for this sample are $30.15 and $12, respectively. (a) Develop appropriate hypotheses for a test to determine whether the sample data support the conclusion that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48. (Enter != for ≠ as needed.) H0: Ha: (b) Using the sample from the 61 bottles, what is the test statistic? (Round your answer to three decimal places.) Using the sample from the 61 bottles, what is the p-value? (Round your answer to four decimal places.) p-value = (c) At 𝛼 = 0.05, what is your conclusion? Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.Reject H0. We can conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48. Reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.Do not reject H0. We can conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48. (d) Repeat the preceding hypothesis test using the critical value approach. State the null and alternative hypotheses. (Enter != for ≠ as needed.) H0: Ha: Find the value of the test statistic. (Round your answer to three decimal places.) State the critical values for the rejection rule. Use 𝛼 = 0.05. (Round your answers to three decimal places. If the test is one-tailed, enter NONE for the unused tail.) test statistic≤test statistic≥ State your conclusion. Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.Reject H0. We can conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48. Reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.Do not reject H0. We can conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.



Answers

This is a technique to break down the variation of a random variable into useful components (called stratum) in order to decrease experimental variation and increase accuracy of results. It has been found that a more accurate estimate of population mean $\mu$ can often be obtained by taking measurements from naturally occurring subpopulations and combining the results using weighted averages. For example, suppose an accurate estimate of the mean weight of sixth grade students is desired for a large school system. Suppose (for cost reasons) we can only take a random sample of $m=100$ students, Instead of taking a simple random sample of 100 students from the entire population of all sixth grade students, we use stratified sampling as follows. The school system under study consists three large schools. School A has $N_{1}=310$ sixth grade students, School B has $N_{2}=420$ sixth grade students, and School C has $N_{3}=516$ sixth grade students. This is a total population of 1246 sixth grade students in our study and we have strata consisting of the 3 schools. A preliminary study in each school with relatively small sample size has given estimates for the sample standard deviation $s$ of sixth grade student weights in each school. These are shown in the following table. $$\begin{array}{lll} \text { School A } & \text { School B } & \text { School C } \\ N_{1}=310 & N_{2}=420 & N_{3}=516 \\ s_{1}=3 \mathrm{lb} & s_{2}=12 \mathrm{lb} & s_{3}=6 \mathrm{lb} \end{array}$$ How many students should we randomly choose from each school for a best estimate $\mu$ for the population mean weight? A lot of mathematics goes into the answer. Fortunately, Bill Williams of Bell Laboratories wrote a book called A Sampler on Sampling (John Wiley and Sons, publisher), which provides an answer. Let $n_{1}$ be the number of students randomly chosen from School $\mathrm{A}$, $n_{2}$ be the number chosen from School $\mathrm{B},$ and $n_{3}$ be the number chosen from School C. This means our total sample size will be $m=n_{1}+n_{2}+n_{3} .$ What is the formula for $n_{i} ?$ A popular and widely used technique is the following. $$n_{i}=\left[\frac{N_{i} s_{i}}{N_{1} s_{1}+N_{2} s_{2}+N_{3} s_{3}}\right] m$$ The $n_{i}$ are usually not whole numbers, so we need to round to the nearest whole number. This formula allocates more students to schools that have a larger population of sixth graders and/or have larger sample standard deviations. Remember, this is a popular and widely used technique for stratified sampling. It is not an absolute rule. There are other methods of stratified sampling also in use. In general practice, according to Bill Williams, the use of naturally occurring strata seems to reduce overall variability in measurements by about $20 \%$ compared to simple random samples taken from the entire (unstratified) population. Now suppose you have taken a random sample size $n_{i}$ from each appropriate school and you got a sample mean weight $\bar{x}_{i}$ from each school. How do you get the best estimate for population mean weight $\mu$ of the all 1246 students? The answer is, we use a weighted average. $$\mu \approx \frac{n_{1}}{m} \overline{x_{1}}+\frac{n_{2}}{m} \overline{x_{2}}+\frac{n_{3}}{m} \overline{x_{3}}$$ This is an example with three strata. Applications with any number of strata can be solved in a similar way with obvious extensions of formulas. (a) Compute the size of the random samples $n_{1}, n_{2}, n_{3}$ to be taken from each school. Round each sample size to the nearest whole number and make sure they add up to $m=100$. (b) Suppose you took the appropriate random sample from each school and you got the following average student weights: $\overline{x_{1}}=82 l b, \overline{x_{2}}=115 l b, \overline{x_{3}}=90 l b$. Compute your best estimate for the population mean weight $\mu$.

Part one out of 990 students in the sample. The number that were never awarded a voucher. R 468. And you will find that number in our by something. All the cases, this is his name of the data set dollar sign. And select years equal equal zero. You are counting the total number of cases where select years is zero. How many students had a voucher available for four years. So similarly, you find the total number of cases where selected ears double equal four. And you will get 108 students. Then how many students actually attended a choice school for four years counting all the cases were variable choice years equal equal for you get 56 party. You you run a simple regression of choice ears on select years. This is what I get. I get there slope coefficient to be minus 1.837 And this one is significant. The minus sign is not what I expect. These regression is telling us that, mm, mm hmm. Uh huh. All right. This is actually part three. All right, sir. In Part three, you run a simple regression on uh of math percentile on choice years. And this is not what I expected. Because this result is telling me being able to attend choice years actually decreases math scores. And the variable um the estimate on choice ears does not change when I add the demographic variables black, hispanic and female. Okay, so I actually accidentally skipped part two. Yeah. Let's move down to the end. In part to you. You are examining whether select ears is a good instrument for choice ears and the answer is yes. This is a regression result. This equation is telling us being selected for a voucher increases the chance of the students attending the school of choice. And the relationship is strong. Almost 1 to 1 and also highly significant. Select years can be a good instrument, assuming that it does not correlate with the error terms in the choice years. Um Now in the mass score equation let's continue with Part four. Why might choice years be endogenous in the given equation, I could think of two reasons. First one admitted bearable bias. It is possible that we have not accounted for some factors that affect both mass scores and the number of years students attending the school of choices, for example, intellectual ability and the second reason is reverse causality. The problem does not give more context on this, but it is possible that mass scores effect the ability to attend the schools of Choice, Part 56 N seven and eight. Um So for the last four parts, I put the results together and they are here in one table. This is part five, Part six and part seven. Yeah. In part five you estimate the equation in part four by instrumental variables. And Selected Years is an instrument for Choice Years does using I. V. Produced a positive effect of attending a charter school? The answer is no because the sign of the main estimate is negative. What do you make of the coefficients of the other explanatory variables? We have black, hispanic and female. As you can see the first few have negative estimates meaning students that are back and of hispanic origin tend to have significantly lower math scores. The standard error is way smaller than the estimate for variable female female dummy. This estimate is telling us being female does not affect mass scores controlling for race origin and whether attending choice schools. I wouldn't write that down about six. We at M n c E 90 the math score in 1992 the equation and we will estimate the equation by L. S. N. I. V. Now we have a positive as expected estimate on choice ears. The I. V. Estimate has a greater magnitude comparing to the old LS estimate and is actually statistically significant. I. Ve estimate is practically large each year in a school uh in a child's school is worth almost two on the math percentile score. Okay again, I will write that down about seven. So the results which has got is not entirely convincing because we have a smaller sample. So you can see from your statistical package when they run the regression in part six, they would say that 662 observations deleted due to missing. And we know that it is the missing values of the new variable M N C E 90 on a part part eight. We replaced choice years with four dummies choice here one, 23 and four. And we generate its instruments. They're instruments select years 1 to 4. We estimate the equation by ivy. This is what I get. Yeah. So we don't have choice years anymore. We don't have a constant anymore. Choice here's one is not significant. Do not significant. Three not significant. Only choice. Air Force is significant and that means being in only being in the choice school for the whole four years. The maximum amount of time matters for the math scores. So that's it and I think it's an interesting result. And you should read the paper if you have a chance to.

And this particular problem were asked to use a specific formula in order to calculate the random sample size for these three different situations. So we're asked to find the sample size for Lower case and one lower case and to And lower case and three. Then we need to make sure that the sample sizes all add up to 100. So I've got ahead and written out the formula here. And just to save a little bit of time, I'm gonna go ahead and calculate the denominator. So with my calculator my denominator will be 310 which is n one times three. That's £3 plus 4. 20 times 12 plus 516 times six. So 3 10. Mhm. Times three plus 420 times 12 plus 516 times six. So for all three of these the denominator will be 9066. So I'm gonna go ahead and use my eraser. You raised that part of the formula And the race because I know that has to be a 100 So m needs to be a 100 And what was this? 9000. So now what I need to do is I simply need to go through and for N1 I'm going to multiply these two and that will be my numerator. So I'm going to use 310 times three and then I'm going to divide that by our 9066. So I multiply the numerator And the one. I'm all done. I'll multiply that by 100. So 310 times three, divide that by 9066. And then I can either multiply it by 100 or I can just in my head move the decimal point to places and I can see I get a little over 10. I'm gonna go ahead and write the decimals down. I'm guessing it's going to be 10 But I'm going to go ahead and write down 2.54.25. Just in case I need to do some rounding. All right. And then the next one I will you 420 times 12. And the nice thing about this calculator is that it does have the ability to recall information. So I could have made this a little easier on myself and I could have stored some of this information and the calculator. All right. And I multiply this one by a 100 And it looks like it's probably gonna be 56 But I'll write 55 .59 for right now I'm going to do the same thing for end three. So 516 times six I'm going to press second, enter second, enter Cursor over 5 16 Times six and delete that extra value. And when I multiply this by 100. looks like I'm gonna get 34, so I probably am not going to have to round On that one. So let's see if N1 is 10 An end to is 56 and N three is 34. Does that add to 100? It sure does. So that's the first part of the question. So for the next part of the question I'm asked to find the mean of the population given that the mean of each sample is as given £82.150 pounds and £90. Now this is a weighted mean. So I have to go back and I have to use the values that I found out in the previous part of the problem because that would be my sample size. So in my weighted mean, I will take 10 times 82 plus 56 times 115 Plus 34 times 90. So I'm multiplying each of these values together, that gives me 10,320 and then remember, it was supposed to be out of 100. So when that is divided by 100, 103.2 becomes mu, or the mean for my population.

So in this particular problem were asked to do several things. So first of all, we need to identify Alpha, the significance level is 1% 0.01. And we know that the null hypothesis is that P is 0.301. And then the alternate hypothesis would be that P is less than now to figure out what kind of distribution it is. It's going to be standard, normal. And a quick check Is N. Times P. Greater than five. And indeed it is. So to 15 time zero point 301 is indeed greater than five because it equals 64.7 about And then his end times Q Greater than five. So if P is 301, Then 1 -301 will give us Q. So Q will be 0.699. So when I multiply those together To 15 times 0.699, I get approximately 150. So yes, that is indeed greater than five. Now I need to find the test statistic which in this case will be P hat and they're asking us to find the Z value. So I'm going to do this all at once with my calculator. Probably got it worked out but let me walk through it with you. So stat tests. This is a one proportion Z test. So number five. Now the probability of success is .301. We're told that our is 46. So in the calculator that's the x the total is population is 215. Were testing if it's actually a less than cursor down to calculate and the information they need. It's all right there. So I have this on the other screen. So P hat Is .21 depending on rounding here 214 and Z is -2.78. And in this particular situation I'm also given the p. value Which is this one. I know you got all these peas, you got the little P. For probability and you've got row and you've got your P. Value 27 So that's part C. Were asked for to find the P value. So if we're if we're actually shading this then we make our normal curve And here's negative 2.78. And I'm shading to the left. So that's if we had to shade it. Now I need to check this. So is this P value less than greater than or equal to my significance level? And I can see that my P value is less than or equal to. So that means I need to reject the no and then how do I write that out? How do I explain that? I would say something like At the 1% level of significance. The sample data indicate that the population proportion in the revenue is less Than 0.301. So this next part is asking for your opinion. And if you're doing a multiple choice question, the answer might be a little bit different than what I worded here. But This indicates the fact that P is in fact if he is in fact less than 0.31, Then that indicates that there are not enough numbers that start with one. So, yes. So what does that mean as a stockholder? Well, as a stockholder, that could mean that the value of your stock is inflated for the FBI. That might be a red flag to investigate. Because, according to Bedford's law, there should be a certain amount of Values that start with one. So for a stockholder it could mean that your stock is not worth as much as you think it is. And then for FBI this could be an indication that there's fraud. Now. Finally, just because we reject the null hypothesis doesn't mean that we have proved anything. So we did not prove H. Of zero, which was the fact that the probability should be this. Mhm. All we did was take some sample data and because the sample data let us to reject the null, then there could be too few numbers with leading digits of one. So you need to investigate more. So it's not an indication that this is actually false, but it's an indication that more investigation needs to be done. Maybe another sample or maybe a larger sample.


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