To test the belief that sons are taller than their fathers, a student randomly selects 13 fathers who have adult male children. She records the height of both the father and son in inches and obtains the following data. Are sons taller than theirfathers? Use the a = 0.025 level of significance. Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.Click the icon to view the table of data....Which conditions must be met by the sample for this test? Select all that apply.D A. The sampling method results in an independent sample.- XO B. The sample size must be large.c. The sampling method results in a dependent sample.Table of height dataYD. The sample size is no more than 5% of the population size.Height ofFather, X,Height of OSon, YLE. The differences are normally distributed or the sample size is large.71.176.2Let d, = X, - Y,. Write the hypotheses for the test.68.872.367.970.3Họ:72.474.170.469.471.5H:71.570.171,471.473.172.566.765.568.666.968.766.167.263.767.662.5PrintDoneHeln me solve thic

There are two dependent samples which are height of father and height of son. We have to test…

Answered: To test the belief that sons are taller…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Assume the samples are random and independent, the populations are nomally distributed, and the population variances are equal. The table available below shows the prices (in dollars) for a sample of automobile batteries. The prices are classified according to battery type. At a = 0.10, is there enough evidence conclude that at least one mean battery price is different from the others? Complete parts (a) through (e) below. E Click the icon to view the battery cost data. (a) Let u1. P2. H3 represent the mean prices for the group size 35, 65, and 24/24F respectively. Identify the claim and state Ho and H. H Cost of batteries by type The claim is the V hypothesis. Group size 35 Group size 65 Group size 24/24F 101 111 121 124 D 146 173 182 278 124 140 141 89 (b) Find the critical value, Fo, and identify the rejection region. 90 79 84 The rejection region is F Fo, where Fo = (Round to two decimal places as needed.) (c) Find the test statistic F. Print Done F= (Round to two decimal places as…
The football coach randomly selected fifteen players and timed how long each player took to perform a certain drill. The times (in minutes) were: 3.5, 3.3, 3.2, 4.7, 4.4, 3.5, 3.9, 4.5, 4.8, 5.2, 5.4, 4.0, 4.3, 4.1, 5.1 (a) Show that this data approximates a normal distribution. (b) Determine a 90% confidence interval for the mean time for all players.
Calculate the z-score representing the bottom 10% of an approximately normal distribution. (Use a table or technology. Round your answer to two decimal places.)
Beef Consumption. According to Food Consumption, Prices, and Expenditures, published by the U.S. Department of Agriculture, the mean consumption of beef per person in 2011 was 57.5 lb. A sample of 40 people taken this year yielded the data, in pounds, on last year’s beef consumption given on the WeissStats site. Use the technology of your choice to do the following. a. Obtain a normal probability plot, a boxplot, a histogram, and a stem-and-leaf diagram of the data on beef consumptions. b. Decide, at the 5% significance level, whether last year’s mean beef consumption is less than the 2011 mean of 57.5 lb. Apply the onemean t-test. c. The sample data contain four potential outliers: 0, 0, 0, and 13. Remove those four observations, repeat the hypothesis test in part (b), and compare your result with that obtained in part (b). d. Assuming that the four potential outliers are not recording errors, comment on the advisability of removing them from the sample data before performing the…
In a normal distribution, what z-score cuts off the top 65% of the distribution?a. z = 0.7422b. z = 0.2578c. z =0.39d. z = -0.39
A study found that there was 48% chance that the driver fails a breathalyzer test when it is administered on a Friday night. The test is administered to 80 drivers. What distribution that can be used to model the above situation. How many drivers do we expect to fail the test
Number of Number of patients lesions Cholestyramine group Placebo group 64GAONTO 3 5 7 8 >10 Total 5 4 6 5 7 8 4649 00 6 7 2 60 2 4 6 697524 8 57 You have a choice between Poisson and negative binomial distributions, as possible models for the number of lesions in the cholestyramine group. Using the x²-test check which distribution better fits the data (e.g., results in smaller chi-square statistic). For the negative binomial use r=4 and p = 0.44, and for the Poisson use λ = 4.6.
Thirty percent of the students in a class of 100 are planning to go to graduate school. The Variance of this binomial distribution is: A. 16 B. 21 C. 25 D. 36 Cat
The accompanying histogram shows the lengths of hospital stays (in days) for all female patients admitted to a certain hospital during one year with a primary diagnosis of acute myocardial infarction (heart attack). Complete parts (a) through (c) below. a) From the histogram, determine whether the mean or median is larger. Explain. A. Because the distribution is nearly symmetric, the mean is expected to be larger than the median. B. Because the distribution is nearly symmetric, the mean is expected to be about the same as the median. C. Because the distribution is skewed to the high end, the mean is expected to be larger than the median. D. Because the distribution is skewed to the high end, the mean is expected to be smaller than the median. E. Because the distribution is skewed to the low end, the mean is expected to be larger than the median. F. Because the distribution is skewed to the low end, the mean is expected to be…

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