{ "cells": [ { "cell_type": "markdown", "metadata": { "deletable": false, "editable": false, "nbgrader": { "checksum": "c08d9aae7522da83f957f4eceea5bf55", "grade": false, "grade_id": "cell-bb1f11f852c7e361", "locked": true, "schema_version": 1, "solution": false } }, "source": [ "## Stat 140-03/05 Fall 2018\n", "# Homework 6R (Ch 17)" ] }, { "cell_type": "code", "execution_count": 23, "metadata": {}, "outputs": [], "source": [ "library(dplyr)\n", "library(ggplot2)\n", "library(mosaic)\n", "library(repr)\n", "options(repr.plot.height = 3)" ] }, { "cell_type": "markdown", "metadata": { "deletable": false, "editable": false, "nbgrader": { "checksum": "b935dc40babfe413f655f6232b8bce99", "grade": false, "grade_id": "cell-05d4443899ffa522", "locked": true, "schema_version": 1, "solution": false } }, "source": [ "## 1. The sampling distribution of a mean.\n", "The code below randomly generates some data with a bimodal distribution. You do not need to know how it works. Assume for this exercise that this data is your entire population." ] }, { "cell_type": "code", "execution_count": 24, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

value
100
23
0
0
0
32

\n" ], "text/latex": [ "\\begin{tabular}{r|l}\n", " value\\\\\n", "\\hline\n", "\t 100\\\\\n", "\t 23\\\\\n", "\t 0\\\\\n", "\t 0\\\\\n", "\t 0\\\\\n", "\t 32\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "value | \n", "|---|---|---|---|---|---|\n", "| 100 | \n", "| 23 | \n", "| 0 | \n", "| 0 | \n", "| 0 | \n", "| 32 | \n", "\n", "\n" ], "text/plain": [ " value\n", "1 100 \n", "2 23 \n", "3 0 \n", "4 0 \n", "5 0 \n", "6 32 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "BiPop <- data.frame(value=round(100*rbeta(10000,0.2,0.2)))\n", "head(BiPop)" ] }, { "cell_type": "markdown", "metadata": { "deletable": false, "editable": false, "nbgrader": { "checksum": "543688bbb292bc8de88a17ae116a9a0c", "grade": false, "grade_id": "cell-6e69de2ad146b4fd", "locked": true, "schema_version": 1, "solution": false } }, "source": [ "### 1a) Plot a histogram of the values in `BiPop` with 101 bins, and calculate the mean and standard deviation for this population." ] }, { "cell_type": "code", "execution_count": 25, "metadata": {}, "outputs": [ { "data": {}, "metadata": {}, "output_type": "display_data" }, { "data": { "image/png": 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"text/plain": [ "plot without title" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "ggplot(data = BiPop, aes(x = value)) + geom_histogram(bins = 101)" ] }, { "cell_type": "code", "execution_count": 27, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

value	Mean	SD
100	49.9093	42.28913
23	49.9093	42.28913
0	49.9093	42.28913
0	49.9093	42.28913
0	49.9093	42.28913
32	49.9093	42.28913

\n" ], "text/latex": [ "\\begin{tabular}{r|lll}\n", " value & Mean & SD\\\\\n", "\\hline\n", "\t 100 & 49.9093 & 42.28913\\\\\n", "\t 23 & 49.9093 & 42.28913\\\\\n", "\t 0 & 49.9093 & 42.28913\\\\\n", "\t 0 & 49.9093 & 42.28913\\\\\n", "\t 0 & 49.9093 & 42.28913\\\\\n", "\t 32 & 49.9093 & 42.28913\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "value | Mean | SD | \n", "|---|---|---|---|---|---|\n", "| 100 | 49.9093 | 42.28913 | \n", "| 23 | 49.9093 | 42.28913 | \n", "| 0 | 49.9093 | 42.28913 | \n", "| 0 | 49.9093 | 42.28913 | \n", "| 0 | 49.9093 | 42.28913 | \n", "| 32 | 49.9093 | 42.28913 | \n", "\n", "\n" ], "text/plain": [ " value Mean SD \n", "1 100 49.9093 42.28913\n", "2 23 49.9093 42.28913\n", "3 0 49.9093 42.28913\n", "4 0 49.9093 42.28913\n", "5 0 49.9093 42.28913\n", "6 32 49.9093 42.28913" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "MeanSD <- BiPop %>% mutate(Mean = mean(value), SD = sd(value))\n", "head(MeanSD)" ] }, { "cell_type": "markdown", "metadata": { "deletable": false, "editable": false, "nbgrader": { "checksum": "d3be95fb8d107d5c545640d009d9075d", "grade": false, "grade_id": "cell-3b2ae9187e9b7b06", "locked": true, "schema_version": 1, "solution": false } }, "source": [ "### 1b) Take a sample of 10 values from `BiPop` and calculate the mean." ] }, { "cell_type": "code", "execution_count": 29, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\n", "

sample_mean_value
49.3

\n" ], "text/latex": [ "\\begin{tabular}{r|l}\n", " sample\\_mean\\_value\\\\\n", "\\hline\n", "\t 49.3\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_mean_value | \n", "|---|\n", "| 49.3 | \n", "\n", "\n" ], "text/plain": [ " sample_mean_value\n", "1 49.3 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 10\n", "\n", "sample_results <- {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_mean_value = mean(value))\n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "markdown", "metadata": { "deletable": false, "editable": false, "nbgrader": { "checksum": "8fb4158ef042bcbdec8f9d1677e99597", "grade": false, "grade_id": "cell-ee604390be7728df", "locked": true, "schema_version": 1, "solution": false } }, "source": [ "### 1c) Take 1000 samples of size n=5 and calculate the mean of each. Compute the mean and standard deviation of this sampling distribution." ] }, { "cell_type": "code", "execution_count": 32, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

sample_mean_value
41.0
46.8
20.2
47.2
70.4
32.6

\n" ], "text/latex": [ "\\begin{tabular}{r|l}\n", " sample\\_mean\\_value\\\\\n", "\\hline\n", "\t 41.0\\\\\n", "\t 46.8\\\\\n", "\t 20.2\\\\\n", "\t 47.2\\\\\n", "\t 70.4\\\\\n", "\t 32.6\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_mean_value | \n", "|---|---|---|---|---|---|\n", "| 41.0 | \n", "| 46.8 | \n", "| 20.2 | \n", "| 47.2 | \n", "| 70.4 | \n", "| 32.6 | \n", "\n", "\n" ], "text/plain": [ " sample_mean_value\n", "1 41.0 \n", "2 46.8 \n", "3 20.2 \n", "4 47.2 \n", "5 70.4 \n", "6 32.6 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 5\n", "\n", "simulation_results <- do(1000) * {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_mean_value = mean(value))\n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "markdown", "metadata": { "deletable": false, "editable": false, "nbgrader": { "checksum": "ec33e8bc1624ed739ff7049f84517c13", "grade": false, "grade_id": "cell-96c8291a720434c7", "locked": true, "schema_version": 1, "solution": false } }, "source": [ "### 1d) Repeat this for n = 1, 2, 10, 40, 4000." ] }, { "cell_type": "code", "execution_count": 33, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

sample_mean_value
31
100
76
100
99
0

\n" ], "text/latex": [ "\\begin{tabular}{r|l}\n", " sample\\_mean\\_value\\\\\n", "\\hline\n", "\t 31\\\\\n", "\t 100\\\\\n", "\t 76\\\\\n", "\t 100\\\\\n", "\t 99\\\\\n", "\t 0\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_mean_value | \n", "|---|---|---|---|---|---|\n", "| 31 | \n", "| 100 | \n", "| 76 | \n", "| 100 | \n", "| 99 | \n", "| 0 | \n", "\n", "\n" ], "text/plain": [ " sample_mean_value\n", "1 31 \n", "2 100 \n", "3 76 \n", "4 100 \n", "5 99 \n", "6 0 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 1\n", "\n", "simulation_results <- do(1000) * {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_mean_value = mean(value))\n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "code", "execution_count": 35, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

sample_mean_value
98.5
47.0
96.0
90.0
85.0
99.5

\n" ], "text/latex": [ "\\begin{tabular}{r|l}\n", " sample\\_mean\\_value\\\\\n", "\\hline\n", "\t 98.5\\\\\n", "\t 47.0\\\\\n", "\t 96.0\\\\\n", "\t 90.0\\\\\n", "\t 85.0\\\\\n", "\t 99.5\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_mean_value | \n", "|---|---|---|---|---|---|\n", "| 98.5 | \n", "| 47.0 | \n", "| 96.0 | \n", "| 90.0 | \n", "| 85.0 | \n", "| 99.5 | \n", "\n", "\n" ], "text/plain": [ " sample_mean_value\n", "1 98.5 \n", "2 47.0 \n", "3 96.0 \n", "4 90.0 \n", "5 85.0 \n", "6 99.5 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 2\n", "\n", "simulation_results <- do(1000) * {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_mean_value = mean(value))\n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "code", "execution_count": 36, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

sample_mean_value
57.4
39.8
61.8
73.5
57.5
42.1

\n" ], "text/latex": [ "\\begin{tabular}{r|l}\n", " sample\\_mean\\_value\\\\\n", "\\hline\n", "\t 57.4\\\\\n", "\t 39.8\\\\\n", "\t 61.8\\\\\n", "\t 73.5\\\\\n", "\t 57.5\\\\\n", "\t 42.1\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_mean_value | \n", "|---|---|---|---|---|---|\n", "| 57.4 | \n", "| 39.8 | \n", "| 61.8 | \n", "| 73.5 | \n", "| 57.5 | \n", "| 42.1 | \n", "\n", "\n" ], "text/plain": [ " sample_mean_value\n", "1 57.4 \n", "2 39.8 \n", "3 61.8 \n", "4 73.5 \n", "5 57.5 \n", "6 42.1 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 10\n", "\n", "simulation_results <- do(1000) * {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_mean_value = mean(value))\n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "code", "execution_count": 37, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

sample_mean_value
52.025
41.775
52.825
45.725
47.150
61.950

\n" ], "text/latex": [ "\\begin{tabular}{r|l}\n", " sample\\_mean\\_value\\\\\n", "\\hline\n", "\t 52.025\\\\\n", "\t 41.775\\\\\n", "\t 52.825\\\\\n", "\t 45.725\\\\\n", "\t 47.150\\\\\n", "\t 61.950\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_mean_value | \n", "|---|---|---|---|---|---|\n", "| 52.025 | \n", "| 41.775 | \n", "| 52.825 | \n", "| 45.725 | \n", "| 47.150 | \n", "| 61.950 | \n", "\n", "\n" ], "text/plain": [ " sample_mean_value\n", "1 52.025 \n", "2 41.775 \n", "3 52.825 \n", "4 45.725 \n", "5 47.150 \n", "6 61.950 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 40\n", "\n", "simulation_results <- do(1000) * {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_mean_value = mean(value))\n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "code", "execution_count": 39, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

sample_mean_value
50.06875
50.07725
50.14000
49.82650
49.73400
50.21750

\n" ], "text/latex": [ "\\begin{tabular}{r|l}\n", " sample\\_mean\\_value\\\\\n", "\\hline\n", "\t 50.06875\\\\\n", "\t 50.07725\\\\\n", "\t 50.14000\\\\\n", "\t 49.82650\\\\\n", "\t 49.73400\\\\\n", "\t 50.21750\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_mean_value | \n", "|---|---|---|---|---|---|\n", "| 50.06875 | \n", "| 50.07725 | \n", "| 50.14000 | \n", "| 49.82650 | \n", "| 49.73400 | \n", "| 50.21750 | \n", "\n", "\n" ], "text/plain": [ " sample_mean_value\n", "1 50.06875 \n", "2 50.07725 \n", "3 50.14000 \n", "4 49.82650 \n", "5 49.73400 \n", "6 50.21750 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 4000\n", "\n", "simulation_results <- do(1000) * {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_mean_value = mean(value))\n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "markdown", "metadata": { "deletable": false, "editable": false, "nbgrader": { "checksum": "6a16fad4a0881dcef34d251f33d386a6", "grade": false, "grade_id": "cell-c6c790b314536427", "locked": true, "schema_version": 1, "solution": false } }, "source": [ "### 1e) What do you notice about the means and standard deviations of your sampling distributions of means?" ] }, { "cell_type": "markdown", "metadata": { "deletable": false, "nbgrader": { "checksum": "e53a3ebbc89cacaf4b16e5afd1dc8661", "grade": true, "grade_id": "cell-3b5229fa28fe0183", "locked": false, "points": 6, "schema_version": 1, "solution": true } }, "source": [ "As n gets bigger, the means get closer to 50." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 2. The sampling distribution of other statistics.\n", "What's so special about means?" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### 2a) Choose ONE of the statistics below. Then for the `BiPop` values, simulate the sampling distribution of your statistic for different values of $n$. Find the mean and standard deviation for each $n$.\n", "* minimum\n", "* first quartile\n", "* median\n", "* IQR\n", "* standard deviation\n" ] }, { "cell_type": "code", "execution_count": 52, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

sample_min_value	sample_sd_value	sample_mean_value	.row	.index
100	NA	100	1	1
0	NA	0	1	2
3	NA	3	1	3
0	NA	0	1	4
3	NA	3	1	5
3	NA	3	1	6

\n" ], "text/latex": [ "\\begin{tabular}{r|lllll}\n", " sample\\_min\\_value & sample\\_sd\\_value & sample\\_mean\\_value & .row & .index\\\\\n", "\\hline\n", "\t 100 & NA & 100 & 1 & 1 \\\\\n", "\t 0 & NA & 0 & 1 & 2 \\\\\n", "\t 3 & NA & 3 & 1 & 3 \\\\\n", "\t 0 & NA & 0 & 1 & 4 \\\\\n", "\t 3 & NA & 3 & 1 & 5 \\\\\n", "\t 3 & NA & 3 & 1 & 6 \\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_min_value | sample_sd_value | sample_mean_value | .row | .index | \n", "|---|---|---|---|---|---|\n", "| 100 | NA | 100 | 1 | 1 | \n", "| 0 | NA | 0 | 1 | 2 | \n", "| 3 | NA | 3 | 1 | 3 | \n", "| 0 | NA | 0 | 1 | 4 | \n", "| 3 | NA | 3 | 1 | 5 | \n", "| 3 | NA | 3 | 1 | 6 | \n", "\n", "\n" ], "text/plain": [ " sample_min_value sample_sd_value sample_mean_value .row .index\n", "1 100 NA 100 1 1 \n", "2 0 NA 0 1 2 \n", "3 3 NA 3 1 3 \n", "4 0 NA 0 1 4 \n", "5 3 NA 3 1 5 \n", "6 3 NA 3 1 6 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 1\n", "\n", "simulation_results <- do(1000) * {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_min_value = min(value), sample_sd_value = sd(value), sample_mean_value = mean(value)) \n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "code", "execution_count": 53, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

sample_min_value	sample_sd_value	sample_mean_value	.row	.index
14	60.81118	57.0	1	1
3	68.58936	51.5	1	2
0	12.72792	9.0	1	3
2	67.88225	50.0	1	4
100	0.00000	100.0	1	5
100	0.00000	100.0	1	6

\n" ], "text/latex": [ "\\begin{tabular}{r|lllll}\n", " sample\\_min\\_value & sample\\_sd\\_value & sample\\_mean\\_value & .row & .index\\\\\n", "\\hline\n", "\t 14 & 60.81118 & 57.0 & 1 & 1 \\\\\n", "\t 3 & 68.58936 & 51.5 & 1 & 2 \\\\\n", "\t 0 & 12.72792 & 9.0 & 1 & 3 \\\\\n", "\t 2 & 67.88225 & 50.0 & 1 & 4 \\\\\n", "\t 100 & 0.00000 & 100.0 & 1 & 5 \\\\\n", "\t 100 & 0.00000 & 100.0 & 1 & 6 \\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_min_value | sample_sd_value | sample_mean_value | .row | .index | \n", "|---|---|---|---|---|---|\n", "| 14 | 60.81118 | 57.0 | 1 | 1 | \n", "| 3 | 68.58936 | 51.5 | 1 | 2 | \n", "| 0 | 12.72792 | 9.0 | 1 | 3 | \n", "| 2 | 67.88225 | 50.0 | 1 | 4 | \n", "| 100 | 0.00000 | 100.0 | 1 | 5 | \n", "| 100 | 0.00000 | 100.0 | 1 | 6 | \n", "\n", "\n" ], "text/plain": [ " sample_min_value sample_sd_value sample_mean_value .row .index\n", "1 14 60.81118 57.0 1 1 \n", "2 3 68.58936 51.5 1 2 \n", "3 0 12.72792 9.0 1 3 \n", "4 2 67.88225 50.0 1 4 \n", "5 100 0.00000 100.0 1 5 \n", "6 100 0.00000 100.0 1 6 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 2\n", "\n", "simulation_results <- do(1000) * {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_min_value = min(value), sample_sd_value = sd(value), sample_mean_value = mean(value)) \n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "code", "execution_count": 54, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

sample_min_value	sample_sd_value	sample_mean_value	.row	.index
0	49.58024	55.2	1	1
2	42.07374	43.8	1	2
0	49.76244	62.6	1	3
5	35.40198	48.6	1	4
0	48.15288	34.8	1	5
0	51.88256	56.6	1	6

sample_sd_value	sample_mean_value	.row	.index
47.48392	43.5	1	1
42.91659	70.5	1	2
40.80918	50.5	1	3
43.00917	40.7	1	4
43.30050	31.4	1	5
34.84952	69.4	1	6

sample_sd_value	sample_mean_value	.row	.index
42.92567	54.550	1	1
41.01306	52.075	1	2
45.81254	49.675	1	3
43.35116	52.100	1	4
40.10752	53.450	1	5
44.51095	52.425	1	6

\n" ], "text/latex": [ "\\begin{tabular}{r|lllll}\n", " sample\\_min\\_value & sample\\_sd\\_value & sample\\_mean\\_value & .row & .index\\\\\n", "\\hline\n", "\t 0 & 42.92567 & 54.550 & 1 & 1 \\\\\n", "\t 0 & 41.01306 & 52.075 & 1 & 2 \\\\\n", "\t 0 & 45.81254 & 49.675 & 1 & 3 \\\\\n", "\t 0 & 43.35116 & 52.100 & 1 & 4 \\\\\n", "\t 0 & 40.10752 & 53.450 & 1 & 5 \\\\\n", "\t 0 & 44.51095 & 52.425 & 1 & 6 \\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_min_value | sample_sd_value | sample_mean_value | .row | .index | \n", "|---|---|---|---|---|---|\n", "| 0 | 42.92567 | 54.550 | 1 | 1 | \n", "| 0 | 41.01306 | 52.075 | 1 | 2 | \n", "| 0 | 45.81254 | 49.675 | 1 | 3 | \n", "| 0 | 43.35116 | 52.100 | 1 | 4 | \n", "| 0 | 40.10752 | 53.450 | 1 | 5 | \n", "| 0 | 44.51095 | 52.425 | 1 | 6 | \n", "\n", "\n" ], "text/plain": [ " sample_min_value sample_sd_value sample_mean_value .row .index\n", "1 0 42.92567 54.550 1 1 \n", "2 0 41.01306 52.075 1 2 \n", "3 0 45.81254 49.675 1 3 \n", "4 0 43.35116 52.100 1 4 \n", "5 0 40.10752 53.450 1 5 \n", "6 0 44.51095 52.425 1 6 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 40\n", "\n", "simulation_results <- do(1000) * {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_min_value = min(value), sample_sd_value = sd(value), sample_mean_value = mean(value)) \n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "code", "execution_count": 57, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "

sample_sd_value	sample_mean_value	.row	.index
42.39141	49.48700	1	1
42.18403	50.48300	1	2
42.39267	49.29975	1	3
42.35040	49.64325	1	4
42.21866	48.95575	1	5
42.46241	49.56650	1	6

\n" ], "text/latex": [ "\\begin{tabular}{r|lllll}\n", " sample\\_min\\_value & sample\\_sd\\_value & sample\\_mean\\_value & .row & .index\\\\\n", "\\hline\n", "\t 0 & 42.39141 & 49.48700 & 1 & 1 \\\\\n", "\t 0 & 42.18403 & 50.48300 & 1 & 2 \\\\\n", "\t 0 & 42.39267 & 49.29975 & 1 & 3 \\\\\n", "\t 0 & 42.35040 & 49.64325 & 1 & 4 \\\\\n", "\t 0 & 42.21866 & 48.95575 & 1 & 5 \\\\\n", "\t 0 & 42.46241 & 49.56650 & 1 & 6 \\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "sample_min_value | sample_sd_value | sample_mean_value | .row | .index | \n", "|---|---|---|---|---|---|\n", "| 0 | 42.39141 | 49.48700 | 1 | 1 | \n", "| 0 | 42.18403 | 50.48300 | 1 | 2 | \n", "| 0 | 42.39267 | 49.29975 | 1 | 3 | \n", "| 0 | 42.35040 | 49.64325 | 1 | 4 | \n", "| 0 | 42.21866 | 48.95575 | 1 | 5 | \n", "| 0 | 42.46241 | 49.56650 | 1 | 6 | \n", "\n", "\n" ], "text/plain": [ " sample_min_value sample_sd_value sample_mean_value .row .index\n", "1 0 42.39141 49.48700 1 1 \n", "2 0 42.18403 50.48300 1 2 \n", "3 0 42.39267 49.29975 1 3 \n", "4 0 42.35040 49.64325 1 4 \n", "5 0 42.21866 48.95575 1 5 \n", "6 0 42.46241 49.56650 1 6 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "n <- 4000\n", "\n", "simulation_results <- do(1000) * {\n", " sample_n(BiPop, size = n) %>%\n", " summarize(sample_min_value = min(value), sample_sd_value = sd(value), sample_mean_value = mean(value)) \n", "}\n", "\n", "head(simulation_results)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### 2b) Does the sampling distribution of your statistic become more normal as $n$ increases? How do the mean and standard deviation change?" ] }, { "cell_type": "markdown", "metadata": { "deletable": false, "nbgrader": { "checksum": "b8992c3612b352d406ada236143ec948", "grade": true, "grade_id": "cell-dc93eeef63050cd9", "locked": false, "points": 4, "schema_version": 1, "solution": true } }, "source": [ "The minimum appears to more frequently be 0 as n increases. The " ] } ], "metadata": { "kernelspec": { "display_name": "R", "language": "R", "name": "ir" }, "language_info": { "codemirror_mode": "r", "file_extension": ".r", "mimetype": "text/x-r-source", "name": "R", "pygments_lexer": "r", "version": "3.4.4" } }, "nbformat": 4, "nbformat_minor": 2 }

Homework 6R.ipynb