Use the given confidence interval limits to find the point e. timate p and the margin of error E. 0.12
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The given confidence interval limits to find the point e timate p and the margin of error E.
0.12 < p < 0.28
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- Suppose a certain species bird has an average weight of x = 3.2 grams. Based on previous studies we can assume that the weights of these birds have a normal distribution with sigma = 0.34 grams. For a small group of 18 birds , find a 98% confidence interval for the average weights of these birds. Write answer in interval notation with two decimal places.Find the critical t-value that corresponds to 90% confidence and n = 15. A. 1.761 OB. 2.624 O C. 1.345 O D. 2.145 18Decide whether to reject or fail to reject the null hypothesis. Z (right - tailed test), a = 0.05 Zcomputed = 1.42
- Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value tα/2, (b) find the critical value zα/2, or (c) state that neither the normal distribution nor the t distribution applies. The confidence level is 95%, σ is not known, and the histogram of 61 player salaries (in thousands of dollars) of football players on a team is as shown. 040008000120001600020000010203040Salary (thousands of dollars)Frequency A histogram has a horizontal axis labeled "Salary (thousands of dollars)" from below 0 to above 20000 in increments of 2000 and a vertical axis labeled "Frequency" from 0 to 40 in increments of 10. The histogram contains vertical bars of width 2000, where one vertical bar is centered over each of the horizontal axis tick marks. The heights of the vertical bars are as follows, where the salary is listed first and the height is listed second: 0, 34; 2000, 16; 4000,…Compute the confidence interval for the following problem. This time we will be using estimation of mean, standard deviation unknown. Write your answer in inequality notation, in the p +- e notation and graph the interval. x bar (mean)= -2.34 Sx= 0.42 n= 65 C-int: 90%A statistics class is estimating the mean height of all female students at their college. They collect a random sample of 42 female students and measure their heights. The mean of the sample is 65.3 inches and the sample standard deviation is 5.4 inches. Find the 90% confidence interval for the mean height of all female students in their school? Assume that the distribution of individual female heights at this school is approximately normal. 1. The t value is 1.3025 2. The standard error has a value of -- 3. The margin of error has a value of 4. Find the 90% confidence interval for the mean height of all female students in their school
- Assuming the population of interest is approximately normallydistributed, construct an 80% confidence interval estimate for the population mean given the values below. x overbarequals15.9 sequals4.2 nequals24 The 80% confidence interval for the population mean is from --- to ---.Find the critical value zc necessary to form a confidence interval at the level of confidence shown below. c=0.92 zc=Construct the indicated confidence interval for the population mean μ using the t-distribution. Assume the population is normally distributed. c=0.99, x=13.1, s=0.51, n=12 , (Round to one decimal place as needed.)
- Construct the indicated confidence interval for the population mean µ using the t-distribution. Assume the population is normally distributed. c = 0.99, x= 14.4, s=4.0, n= 10 (Round to one decimal place as needed.)Use the given confidence interval limits to find the point estimate ?̂ and the margin of error ?. 0.69<?<0.710 ?̂ = ?=A warehouse owner wants to know the average weight of books at his facility. He weighs 22 books and finds that the sample has a mean weight of 9.39 pounds and standard deviation of 2.43 pounds. Construct a 99% confidence interval for the population mean of book weights. Assume the population is normally distributed.Choose the correct confidence interval and critical value used in the calculation. A. (8.06,10.72) (Zc=2.576) B. (7.94,10.84)(tc=2.8073) C. (7.92,10.86) (tc=2.8314) D. (8.18,10.6) (Zc=2.326) E. (7.93,10.85)(tc=2.8187)