According to a study conducted by a statistical organization, the proportion of people who are satisfied with the way things are going in their lives is 0.78. Suppose thata random sample of 100 people is obtained. Complete parts (a) through (e) below.responaing to the question truthfully.O C. The sample proportion p is a random variable because the value of p represents a random person included in the sample. The variability is due to the fact thatdifferent people feel differently regarding their satisfaction.The sample proportion p is a random variable because the value of p represents a random person included in the sample. The variability is due to the fact thatpeople may not be responding to the question truthfully.(c) Describe the sampling distribution of p, the proportion of people who are satisfied with the way things are going in their life. Be sure to verify the modelrequirements.Since the sample size is no more than 5% of the population size and np(1 - p) = 2 10, the distribution of p is approximately normal with µa = 0.780 andp(Round to three decimal places as needed.)

From the given data fill the blanks given

Answered: (c) Describe the sampling distribution…

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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#13).
Researchers conducted a study to determine whether magnets are effective in treating back pain. Pain was measured using the visual analog scale, and the results shown below are among the results obtained in the study. Higher scores correspond to greater pain levels. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) to (c) below. Reduction in Pain Level After Magnet Treatment (u): n=22, x=0.58, s=0.89 Reduction in Pain Level After Sham Treatment (H2): n=22, x=0.51, s = 1.29 OA. Ho: H₁₂ H₁₁₂ H₁₁
A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effectiveness periods of the tranquilizer for each patient (in hours) were as follows: 2.7, 2.8, 3.0, 2.3, 2.3, 2.2, 2.8, 2.1and 2.4. If the error bound is 0.2444, then the level of confidence interval of the mean is: None of the other options 0.99 0.98 0.95
You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. A sample of 28 night students and 29 day students and their GPA's are recorded below. Test the claim using a 10% level of significance. Assume the population variances are unequal and that GPAS are normally distributed. Give answers to 4 decimal places. GPA-Night GPA-Day 3.46 3.18 2.77 2.84 3.44 3.22 3.16 3.22 3.37 3.46 2.94 3.33 2.74 2.98 3.1 3.03 2.8 3.1 3.2 3.23 3.43 3.22 3.03 3.1 3.42 3.23 2.88 3.27 2.65 2.93 3.53 3.16 3.32 3.24 2.94 3.25 2.21 3.17 2.81 2,83 2.89 3.35 2.32 3.28 2.25 3.07 3.4 3.24 2.84 3.43 3.13 3.05 2.25 2.73 2.62 2.83 What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem. Ho: Select an answer Select an answer H: Select an answer Select an answer Based on the hypotheses, find the following: Test Statistic- p-value- The correct decision is to Select an answer e that…
Are phone calls equally likely to occur any day of the week? The day of the week for each of 420 randomly selected phone calls was observed. The results are displayed in the table below. Use an a = 0.05 significance level. a. Complete the rest of the table by filling in the expected frequencies: Frequencies of Phone Calls for Each Day of the Week Outcome Frequency Expected Frequency Sunday 61 Monday 70 Tuesday 69 Wednesday 50 Thursday 40 Friday 64 Saturday 66 b. What is the correct statistical test to use? Select an answer ✓ c. What are the null and alternative hypotheses? Ho: The distribution of phone calls is not uniform over the days of the week. Phone calls and days of the week are dependent. Phone calls and days of the week are independent. O The distribution of phone calls is uniform over the days of the week.
J2.
The undergraduate grade point averages (UGPA) of students taking an admissions test in a recent year can be approximated by a normal distribution, as shown in the figure. (a) What is the minimum UGPA that would still place a student in the top 15% of UGPAS? (b) Between what two values does the middle 50% of the UGPAS lie? H=3.26 G=0.16 3.26 2.52 Grade point average (a) The minimum UĞPA that would still place a student in the top 15% of UGPAS is (Round to two decimal places as needed.) (b) The middle 50% of UGPAS lies between on the low end and on the high end. (Round to two decimal places as needed.)
You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. A sample of 21 night students and 22 day students and their GPA's are recorded below. Test the claim using a 10% level of significance. Assume the population variances are unequal and that GPAs are normally distributed. Give answers to 4 decimal places. GPA-Night GPA-Day 3.32 3.1 3.22 3.26 3.55 3.25 3.03 3.51 3.22 3.21 3.12 3.05 2.86 3.38 3.32 3.08 3.16 2.88 3.13 3.43 3.15 3.17 3.13 3.26 2.97 2.46 3.3 3.28 3.5 3.19 3.37 3.46 3.88 2.86 3.61 2.89 3.51 2.91 3.11 2.32 3.06 3.16 2.03 What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem.H0: H1: Based on the hypotheses, find the following:Test Statistic=p-value=The correct decision is to .The correct summary would be:…
A sample of size 5 was collected from an unknown population 15 15 16 16 17 1. Use the table of expected z-scores (rounded to the nearest tenth) below to construct the normality plot with expected z-scores on the x-axis and the observations on the y-axis: 6 -1.3 -0.6 -0.2 0.2 0.6 1.3 5 -1.2 -0.5 0 0.5 1.2 -1 19 18 normal O not normal Clear All Draw: 17- 16- 14 T 15- رها 1 2 7 -1.4 -0.8 -0.4 0 0.4 0.8 1.4 8 -1.4 -0.9 -0.5 -0.2 0.2 0.5 0.9 1.4 9 -1.5 -0.9 -0.6 -0.3 0 0.3 0.6 0.9 1.5 2. Based on whether the pattern above is linear or not, in your opinion, was the sample drawn from a normal population or not?
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…
See below. I really need help with this.
Chapter 5, Section 2, Exercise 043 Hearing Loss in TeenagersA recent study1 found that, of the 1771 participants aged 12 to 19 in the National Health and Nutrition Examination Survey, 19.5% had some hearing loss (defined as a loss of 15 decibels in at least one ear). This is a dramatic increase from a decade ago. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error for the proportion is SE=0.009.Find a 95% confidence interval for the proportion of teenagers with some hearing loss.Round your answers to two decimal places.The 95% confidence interval is to .

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