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Laura Walsh

Laura W.

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The domestic supply and demand curves for hula beans are as follows: \[ \begin{aligned} \text {Supply:} & P=50+Q \\ \text {Demand:} & P=200-2 Q \end{aligned} \] where $P$ is the price in cents per pound and $Q$ is the quantity in millions of pounds. The U.S. is a small producer in the world hula bean market, where the current price (which will not be affected by anything we do) is 60 cents per pound. Congress is considering a tariff of 40 cents per pound. Find the domestic price of hula beans that will result if the tariff is imposed. Also compute the dollar gain or loss to domestic consumers, domestic producers, and government revenue from the tariff

Microeconomics

The domestic supply and demand curves for hula beans are as follows: \[ \begin{aligned} \text {Supply:} & P=50+Q \\ \text {Demand:} & P=200-2 Q \end{aligned} \] where $P$ is the price in cents per pound and $Q$ is the quantity in millions of pounds. The U.S. is a small producer in the world hula bean market, where the current price (which will not be affected by anything we do) is 60 cents per pound. Congress is considering a tariff of 40 cents per pound. Find the domestic price of hula beans that will result if the tariff is imposed. Also compute the dollar gain or loss to domestic consumers, domestic producers, and government revenue from the tariff.

Microeconomics

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4. IA survey is conducted to learn about bome broadband esage in Iriah Hoeseholds. A sample of 10 households with home broadband had an average monthly usage of \( 190 \mathrm{~GB} \) and varied with a standard deviation of \( 28 \mathrm{~GB} \). Which of the following is a \( 95 \% \) confidence imarval fos the true mean monthly broadband usage for the population of all lrish Houcholds if we ; assume the variable, monthly broadband usage, has a normal distribution for the population? (a) \( 190 \pm 14.61 \). (b) \( 190 \pm 16.23 \). (c) \( 190 \pm 17.35 \). (d) \( 190 \pm 20.03 \) (e) \( 190 \pm 22.84 \) (f) \( 190 \pm 54.88 \) (g) \( 190 \pm 63.34 \). 5. Which one of the following woald result in an interval estimate that is eore preccise, ic.

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(a). The president of a large chain of fast-food restaurants is interested in estimating the true average net profit among all of their branches over the last year. The net profit (in million euros) from a random sample of 35 branches were recorded and a summary of the responses is shown below: (b). Another sample of 25 branches were selected for which the last year's drive-through sales were recorded. A summary of the drive-throngh sales for the 25 branches is as follow: \( \begin{array}{rrrr} & \text { a Minimum Maximum Mean Std.Deviation } \\ 125 & 2 & 9.85 .38 & 2.12\end{array} \) (i) Provide an interval estimate of the population mean, \( \mu \), calculated at the \( 90 \% \) Confidence (i) The president believes the last year's true average drive-throtgh sales among all branches should be more than \( 4.5 \) million euros. Is there evidence from this sample that the level, and interpret the result with reference to this application. true average average drive-through sales is greater than \( 4.5 \) million euros, i.e, test the (ii) The president believes the true average net profit should not be any different than 1 . hypotheses \( H_{0}: \mu=4.5 \) versus \( H_{a}: \mu>4.5 \). Carry out the test at \( \alpha=0.1 \). Include the ii) The president believes the true average net profit shotuld not be any different than 1 calculation of the test statistic, the rejection region for the test statistic and a conclusion. [5 marks]. branches is a value of 1 million euro. Use the confidence interval calculated in part (i) to justify your answer. hypothesis test carried out in part (i) to be valid. Do the features of the distribution hypothesis test carried out in part (i) to be vatid. Bo the features of the distribution respect to the validity of the hypothesis test calculated in part (i)? Give a reason for your answer. (iii) If an interval estimate of \( \mu \) is desired that is more precise, i.e. narrower, than that calculated in part (i) should this be calculated at a higher or lower confidence level? (iii) Provide an interval estimate of the population mean, calculating at the \( 90 \% \) Confidence level, and interpret the result with reference to this application. 6 [1 marks] [5 marks]

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Qudsiya Anis verified

Numerade educator

A researcher conducted a large sample two-sided test of the null hypothesis that H0 : µ = 100. She reports a p.value of 0.054. Which one of the following is correct? (a) The 95% confidence interval for µ would contain 100. (b) The null hypothesis is not rejected at – = 0.10. (c) The null hypothesis is rejected at – = 0.05. (d) The 99% confidence interval for µ would not contain 100

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Qudsiya Anis verified

Numerade educator

A developer of a dating app wishes to estimate the proportion of the adult Irish population that are single. A survey showed that out of a random sample of 50 individuals from the population, 21 of them stated they were single, i.e. a sample proportion of 0.42. Which of the following is a 95% confidence interval for the true proportion of the adult Irish population that are single? (a) 21 ± 0.0698 (b) 0.42 ± 0.1368 (c) 21 ± 0.1368 (d) 0.42 ± 0.1152 (e) 21 ± 0.1152 (f) 0.42 ± 0.1801 (g) 21 ± 0.1164 (h) 0.42 ± 0.1164

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Rashmi Sinha verified

Numerade educator

Suppose that a sample of 50 boxes of chocolates are weighed and the sample mean weight is 495.9321. The output below, shows the corresponding tolerance interval. Which one of the following interpretations of this tolerance interval is correct? library(tolerance) normtol.int(x=choc.dat, alpha = 0.05, P = 0.99, side = 2) alpha P x.bar 2-sided.lower 2-sided.upper 1 0.05 0.99 495.9321 429.775 562.0893 (a) We are 95% confident that the true average weight of all boxes of chocolates in the population of interest is likely to be between 429.775 and 562.0893. (b) We are 99% confident that the true average weight of all boxes of chocolates in the population of interest is likely to be between 429.775 and 562.0893. (c) We are 95% confident that the true weight of 99% of the boxes of chocolates is likely to be between 429.775 and 562.0893. (d) We are 99% confident that the true weight of 95% of the boxes of chocolates is likely to be between 429.775 and 562.0893.

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Qudsiya Anis verified

Numerade educator

A sample of 15 popcorn servings selected at random from a cinema food outlet had an average salt content of 2.4 grams. The standard deviation of this sample was 1 gram. Which of the following is a 99% confidence interval for the true average salt content of all popcorn servings if we assume the variable salt content has a normal distribution for the population of all servings? (a) 2.4 ± 0.5061 (b) 2.4 ± 0.5538 (c) 2.4 ± 0.6662 (d) 2.4 ± 0.7687

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2. A random variable \( X \) is being measured for a representative sample of individuals from a propulation. If values of \( X \) for individuals in a population are known to vary by a standard deviation of \( \sigma=15 \), what is the minhmum sample size required to ensure a \( 95 \% \) confidence interval for the true mean \( \mu \) has a width of no more than 10 ? (a) 9 (b) 10 (c) \( 34.57 \) (d) 35 (e) \( 138.30 \) (f) 139 [4 marks]

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Sanchit Jain verified

Numerade educator

Question 2 Complete both parts (a) and (b). (a). The president of a large chain of fast-food restaurants is interested in estimating the true average net profit among all of their branches over the last year. The net profit (in million euros) from a random sample of 35 branches were recorded and a summary of the responses is shown below: n Minimum Maximum Mean Std. Deviation 1 35 0.12 2.8 1.01 0.58 Net profit (million euros) (i) Provide an interval estimate of the population mean, ?, calculated at the 90% Confidence level, and interpret the result with reference to this application. [5 marks] (ii) The president believes the true average net profit should not be any different than 1 million euro. Would you agree or disagree that the true average net profit for all the branches is a value of 1 million euro. Use the confidence interval calculated in part (i) to justify your answer. [2 marks] (iii) If an interval estimate of ? is desired that is more precise, i.e. narrower, than that calculated in part (i) should this be calculated at a higher or lower confidence level? [1 marks] (b). Another sample of 25 branches were selected for which the last year's drive-through sales were recorded. A summary of the drive-through sales for the 25 branches is as follows: n Minimum Maximum Mean Std. Deviation 1 25 2 9.8 5.38 2.12 Drive-Through Sales (million euros) (i) The president believes the last year's true average drive-through sales among all branches should be more than 4.5 million euros. Is there evidence from this sample that the true average drive-through sales is greater than 4.5 million euros, i.e. test the hypotheses H0 : ? = 4.5 versus Ha : ? > 4.5. Carry out the test at ? = 0.1. Include the calculation of the test statistic, the rejection region for the test statistic and a conclusion. [5 marks] (ii) Provide details of any assumptions that are required to be made in order for the hypothesis test carried out in part (i) to be valid. Do the features of the distribution of the sample of drive-through sales indicated by the boxplot above worry you with respect to the validity of the hypothesis test calculated in part (i)? Give a reason for your answer. [2 marks] (iii) Provide an interval estimate of the population mean, calculating at the 90% Confidence level, and interpret the result with reference to this application. [5 marks] Total: 20 marks

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Question 3 Question 4 Complete all parts (a), (b) and (c). Complete part (a). (a). A study in a Galway fast-food branch showed that out of a sample of 200 'single-item' orders (a). In order to explore whether there is any difference, comparing drive-through sales and counter received on a particular day, 125 of them were drive-through orders. sales, in the average amount of sales (mensured in million euros) in a Galway fast food branch random samples of each type of sales were the output obtained is provided below. Carry out a hypothesis test to determine if this gives evidence that more than \( 60 \% \) of the all single-item orders received in the Galway branch in that particular day are drive-through orders, i.e test the hypothesis \( H_{\mathbf{a}}: p>0.6 \), at significance level \( \alpha=0.05 \). (b). For the 125 drive-through orders the customer's choice of food was observed and categorised as one of four categories of 'Pizza', 'Beef Burger', 'Chicken Burger' and 'Chicken Fillet', giving the following summary counts: \begin{tabular}{|l|l|} \hline Orders & Observed Frequency \\ \hline Pizza & 5 \\ \hline Beef Burger & 5 \\ \hline Chicken Burger & 15 \\ \hline Chicken Fillet & 25 \\ \hline Total & 125 \\ \hline \end{tabular} Use the sample observations to test whether the population of all drive-through single-item orders in the Galway branch have an equal versus unequal distribution of proportions across these types of foods at significance level \( \alpha=0 \).1. Include statements of the null and alternative hypotheses being tested, in terms of \( p_{1} ; p_{2} ; p_{3} ; \) \( p_{4} \), the population proportions for each of the four categories. Include calculation of the value of the appropriate test statistic, the rejection region and a conclusion. (c). Part (b) showed that out of the 125 drive-through orders 50 were Beef Burger. Of the 75 Di F value \( \mathrm{Pr}(>\mathrm{F}) \) orders received at the counter, 24 of these were Beef BurgerCalculate and interpret a \( 95 \% \) Confidence Interval for \( p_{1}-p_{2} \), where poef Burger orders of all counter orders. Does this indicate that proportions? Give a reason for your answer. A\# t. test() output when assuming the population standard deviations are equal. t. test(Sales - Sale. Type, alternative = "two. a1ded", mu = 0 , pared = FALSE, var . equal = TRUE, data \( = \) Sales. dat) Two Sample t-test data: Sales by Sale.Type \( t=1.8678, d f=23, p \)-value \( =0.07459 \) alternative hypothesis: true difference in means between two groups ls not equal to 0 95 percent confidence interval: sample estimates: mean in group Counter Sales mean in group Drive Through Sales \( 6.393333 \)

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Question 2

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