Exercise 3. You believe that a particular r.v. is binomial distributed. Empirically you find that the mean value of the variable is 10 and its variance is 10/3. Calculate the parameters in the binomial distribution that match these data.
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- Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.97.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 2424 samples is 8.18.1 ppm with a variance of 0.250.25. Does the data support the claim at the 0.10.1 level? Assume the population distribution is approximately normal. Step 2 of 5: Find the value of the test statistic. Round your answer to three decimal places.A professor claims that exam scores for non-accounting majors are more variable than for accounting majors. Random samples of 29 non-accounting majors and 29 accounting majors are taken from a final exam. Test at the 1% level the null hypothesis that the two population variances are equal against the alternative that the true variance is higher for the non-accounting majors. Assume the populations are normally distributed. Non-Accounting Accounting 2 A. H₂:0²=0² H₁:0² > 0² Ho: H₁:0² +0² The test statistic is F = = 29 nx= ny = 29 2 S S 2 the alternative hypothesis is represented by H₁. Choose the correct answer below. B. Ho: 0²20² H₁:0² <0 = 1611.016 = 446.238 ○D. H₂:0² = 0² H₁:0² <0² (Round to three decimal places as needed.)If the variances for independent random variables X and Y is Var(X)=0.94 and Var(Y)=4.22 then Var (2Y-0.5X+1) equals: a. 18.1150 b. 17.1150 c. 8.9700 d. 17.6450
- A company has found that its profits Y are related to its sales X by the formula Y = 0.2X + 100. If next year’s sales have a Normal distribution with a mean of 7,500 and a variance of 160,000. What is the distribution of next year's profit?(a) Compute a 95% CI for u (b) Compute a 99% CI for µ. (c) Which of the two intervals has bigger length, and why?A magazine claims that the mean amount spent by a customer at Burger Stop is greater than the mean amount spent by a customer at Fry World. The results for samples of customer transactions for the two fast food restaurants are shoun below. At g= 0.01, can you support the magazine's claim? Assume the population variances are equal. Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) through (e) below Burger Stop X = $8.65 Fry World X2 = $8.02 S, = $0.77 S2 = $0.75 n1 = 12 n2 = 10 (a) Identify the claim and state H, and H. Which is the correct claim below? O A. "The mean amount spent by a customer at Burger Stop is greater than the mean amount spent by a customer at Fry World." O B. "The mean amount spent by a customer at Burger Stop is equal to the mean amount spent by a customer at Fry World." O C. "The mean amount spent by a customer at Fry World is greater than the mean amount spent by a customer at Burger Stop." O D. The…
- can you help me understand these questions?A pet association claims that the mean annual cost of food for dogs and cats are the same. The results for samples for the two types of pets are shown below. At a = 0.01, can you reject the pet association's claim? Assume the population variances are not equal. Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) through (e) below. Dogs Cats X = $261 X2 = $226 S, = $39 ৪ = $35 n, = 15 n2 = 9 (a) Identify the claim and state Ho and H. Which is the correct claim below? O A. "The mean annual costs of food for dogs and cats are not equal." OB. "The mean annual costs of food for dogs and cats are equal." OC. The mean annual cost of food for cats is greater than the cost for dogs." O D. "The mean annual cost of food for dogs is greater than the cost for cats." What are Ho and H,? The null hypothesis, Ho, is V The alternative hypothesis, H, is 2 H SH2Let x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately ?2 = 47.1. However, a random sample of 18 colleges and universities in Kansas showed that x has a sample variance s2 = 86.1. Use a 5% level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a 95% confidence interval for the population variance. (a) What is the level of significance?State the null and alternate hypotheses. Ho: ?2 = 47.1; H1: ?2 < 47.1Ho: ?2 = 47.1; H1: ?2 > 47.1 Ho: ?2 = 47.1; H1: ?2 ≠ 47.1Ho: ?2 < 47.1; H1: ?2 = 47.1 (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)What are the degrees of freedom?What assumptions are you making about the original distribution? We assume a uniform population distribution.We…
- A professor claims that exam scores for non-accounting majors are more variable than for accounting majors. Random samples of 29 non-accounting majors and 29 accounting majors are taken from a final exam. Test at the 1% level the null hypothesis that the two population variances are equal against the alternative that the true variance is higher for the non-accounting majors. Assume the populations are normally distributed. Non-Accounting Accounting H₁:0² = 0² H₁:0² > 0² C. H₂:0² = 0² H₁:0² #0² nx = 29 ny = 29 S The critical value is X 2 represented by Ho and the alternative hypothesis is represented by H₁. Choose the correct answer below. OB. Ho: 0²20² H₁:0² <0² = 1611.016 = 446.238 2 OD. H₂:0² = 0² o H₁:0² <0² The test statistic is F = 3.610. (Round to three decimal places as needed.) (Round to two decimal places as needed.)Let x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately ?2 = 47.1. However, a random sample of 18 colleges and universities in Kansas showed that x has a sample variance s2 = 86.1. Use a 5% level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a 95% confidence interval for the population variance. ased on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Since the P-value > ?, we fail to reject the null hypothesis.Since the P-value > ?, we reject the null hypothesis. Since the P-value ≤ ?, we reject the null hypothesis.Since the P-value ≤ ?, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, there is insufficient…Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 6.3 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 870 samples is 6.4 ppm. Assume the variance is known to be 1.00. Does the data support the claim at the 0.1 level? Step 2 of 5 : Enter the value of the z test statistic. Round your answer to two decimal places.
