A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 65 specimens and counts the number of seeds in each. Use her sample results (mean = 77.3, standard deviation = 5.5) to find the 80% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i. e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place). 80% C. I. =
Answer should be obtained without any preliminary rounding.

Confidence interval = [76.4, 78.2]    to one decimal place

Step-by-step explanation:

Given:

N = size of sample = 65

m = sample mean = 77.3

s = sample standard deviation = 5.5

alpha = significance level = (100%-80%) = 20%

percentile = 100(1-alpha/2) 90%  ( assume two-sided interval)

Solution:

Confidence limits

= m +/- t(1-alpha/2, N-1) * s / sqrt(N)    t=t-distribution

= 77.3 +/- t(0.9,64) * 5.5 / sqrt(65)

= 77.3 +/- 1.29492 * 5.5 / sqrt(65)            [t-table from R]

= [76.42, 78.18]

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.

steve should create a type of business report called a business plan. the correct option among the options that are already given in the question is the last or the fourth option. this business report created by steve will actually influence the investors about investing their money in steve's business. this business report will have all the details about the business, the amount of investment required and the amount of expected profit.