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Emily Miller Math 1040 Project Part 4 Part A: A confidence interval is an estimate of a population parameter, something you observe, or from a sample of the population parameter. In other words the purpose of a confidence interval will show a close or true value of the parameter. For a 99% confidence interval estimate for the true population, we gather the appropriate data and compute the interval. All class yellows = n = 1867 All class yellow candies = x= 387 π₯ =387/1867=0.207=πΜ π 99% confidence interval = 1-0.99=0.01 ZΞ±/2=0.01/2=0.005 look up in the body of z score table = 2.575 βπΜ (1-πΜ )/n = β. 207 β .793/1867 = 0.009 E=ZπΌ/2βπΜ πΜ/n-2.575*.009 = 0.23 Confidence interval = πΜ ±E = (.184, .23) It is a true population parameter, because the 99% confidence level is between .184, and .23. My single bag had 23 yellow candies out of 63 total candies 23 = 0.365 My bag would not be a good estimate or unusual because my proportion was not within the confidence interval. 63 To construct 95% confidence interval, gather required data: n= 31 df= 30 π₯Μ = 60.2 s= 3.127 E= 1.147 π 3.127 tΞ±/2*βπ= 2.042* β31 = E =1.147 60.2 ± 1.147 = (59.053, 61.347) I can say with 95% confidence that the population mean of candies per a bag of skittles will be 59.053 to 61.347 of candies per bag, which is a good representation of the sample population. My single bag contained 63 candies that that was not in the confidence interval because it was not between 59.053, and 61.347 98% confidence interval for the estimate of the standard deviation is computed with the following information S=3.127 N=31 β β β (πβ1)π 2 <Ο<β 2 π₯π 30β3.1272 50.892 30β3.1272 14.954 (πβ1)π 2 π₯πΏ2 = 2.355 = 4.492 2.355 < Ο < 4.492 We can say with a 98% confidence that the mean of candies per bag are within 2.355 and 4.492 of the population standard deviation. Conditions for true proportion: 1- Must be SRS 2- Conditions for binomial distribution must be satisfied: fixed # of trials, independent trials, 2 possible outcomes, probability of success is constant for each trial. 3- Sample size large enough, at least 5 successes and 5 failures Conditions for population mean 1- Must be SRS 2- Population is normally distributed or n>30, with our class Conditions for estimating a population standard deviation: 1- Must be SRS 2- Population must have normally distributed values (regardless of sample sizes) The entire class used a convenience sample to select the bags of skittles for this project so the requeriment for that was not met. The other conditions howeverwere met, we had all conditions required to get the ture proportion. The conditions for estimating a population standard deviation I donβt think were meet because the population is not normally distributed according to the graphs from the part 3 b of the project the histogram is skewed to the left. Part B: The meaning and purpose of a hypothesis test uses data from a study to show you if something is likely or unlikely to occur. In other words it shows the outcome of a study. For a .05 level to test the claim that 20% of all skittles candies are red we need the following information to compute. p Μ=.184 p=.2 q=.8 n=1867 a=.05 H0: p=.20 H1: pβ .20 πΜβπ .184β.2 Z= ππ = 0.2β0.8 = -1.73= test statistic β π β 1867 P-value = 0.0418 Significance level (a) = .05/2=.025 Critical values are -1.96 and 1.96 -1.96>critical region>1.96 We would fail to reject the null hypothesis because the (p-value) 0.0418>.a (.025) and the test statistic -1.73 is not less than -1.96 so it is not in the critical region of being less than1.96 Therefore, there is insufficient evidence to reject the claim. To use a .01 significance level to test the claim that the mean number of candies per bag is 55 we would need the following information to compute. π₯Μ = 60.2 ΞΌ = 55 s = 3.127 n = 31 a=.01 H0: ΞΌ = 55 H1: ΞΌ β 55 π₯Μ βπ t= π βπ t= 60.2β55 3.127 β31 = 9.259= test statistic P-value < .01 Significance level =.01 Critical values= -2.750 and 2.750 -2.750>critical region>2.750 Our rejection region is ±2.750 and our T.S. is 9.259 which is greater that the critical value rejection which means it is in the rejection region. So we would reject the H0 because there is sufficient evidence to reject the claim. Conditions for testing a claim about p: 1- Must be SRS 2- Binomial distribution 3- npβ₯ 5 and nqβ₯ 5 and both must be satisfied. Conditions for testing a claim about π with Ο unkown: 1- Must be SRS 2- Either or both, normally distributed and n>30 Conditions were for both tests because they were both SRS, and np > 5 and nq > 5, conditions for binomial distribution are met, and the sample is greater than 30. n>5