(d) (4 pt) Select all answers that we can justiably conclude. If you do not have enough information to evaluate whether the statement is true or false, DO NOT select it. I:] If the petrologist convinces 100 of their colleagues to independently take new random samples of 500 rocks, and each colleague generates one approximate 80% condence interval for the true mean rock density in the region, about 95 of the 100 intervals will contain the true mean rock density of the region. I' If the petrologist convinces 100 of their colleagues to independently take new random samples of 500 rocks, and each colleague generates one approximate 90% condence interval for the true mean rock density in the region, about 90 of the 100 intervals will contain the true mean rock density of the region. I If the petrologist's colleague runs the code from part (c), but changes the endpoints of the interval (in the last two lines) to the 0.5th and 99.5th percentile, the resulting interval will be wider than the petrologist's original interval. E] If the interval calculated in part c) is [1.Qg/cm3, 2.8g/cm3], there is a 90% chance that the true mean rock density of the region is in that interval. (e) (4 pt) If the width of a 90% condence interval you calculated was 1 gram/cma, what is an estimate of the stande deviation of the rock densities in the population of rocks from which we drew our sample of size 500? The table below shows percentages of values in a certain range under the normal curve, in addition to those already in the exam reference guide. Normal Distribution about 80% about 90% Percent in Range average :l: 1.3 SDs average :l: 1.65 SDs Show your calculations below. You should NOT simplify arithmetic expressions. Please draw a. box around your nal answer. 8. (24 points) Rock Solid Condence A petrologist (someone who studies rocks) has collected a random sample of 500 sandstone rocks from a desert region. They are interested in knowing the mean density of sandstone rocks in that region. They nd that the mean density of the sample is 2.4 grams/cm3. Instead of giving just one estimate, they wish to provide a range of values. (a) (6 pt) Select one of the options from each parts i-vi to ll in the corresponding blanks in the sentence below. The actual mean density of the sandstone rocks in the region is the population 7(i)7 . Rather than going back to the original population and taking a new sample, the scientist will use the sample they already have. This technique is known as bootstrapping. To use the bootstrap method, the scientist will take many samples _(ii)_ from the _(iii)_ to create one bootstrapped sample. We nd the (iv) of each bootstrap sample. After we've completed this process, we can compute a (v) . We are more condent that our procedure will generate an interval that captures the actual mean density when we use a (vi) interval. i. O parameter Q statistic ii. C) without replacement Q with replacement iii. 0 population 0 original sample iv. 0 middle 95% O mean v. - condence interval 0 p-value vi. Q wider O narrower (b) (4 pt) Select all of the following conditions under which bootstrapping would not be an effective estimation technique. I] The original sample is very big. C] You are trying to estimate the minimum value of a population. h The original sample is very small. |:I The original sample is a random sample from the population. You are trying to estimate the median value of a population. Q The distribution of your population is not roughly bell shaped. (c) (6 pt) The table called rocks has one column, density, containing the density of the 500 sampled rocks in the region. Write code such that left_end and right_end evaluate to the endpoints of a ninety percent (90%) condence interval for the mean density of rocks in the region using 10,000 bootstrapped ruamples. make_arra means = y() for i in np.arange(10000) rocks.sam Ie resampla = p 0 resampla_mean = np.mean(resample) means = np.append(means,resamplemean) ' means left_end = percentile ( 5 ' ) right_end = percentile ( 95 J means )