One study claimed that 85% of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 108 college students and finds that 86 of them identify themselves as procrastinators. Does this evidence support the professor’s claim that fewer than 85% of college students are procrastinators? Use a 0.01 level of significance.**Step 1 of 3:** State the null and alternative hypotheses for the test. Fill in the blank below.- \(H_0 : p = 0.85\)- \(H_a : p \) _______ 0.85**Answer Options:**- \(>\)- \(\neq\)- \(<\)*There are no graphs or diagrams included in the image.*

Givenclaim is fewer than 85% of college students are procrastinatorssample…

Answered: One study claimed that 85 % of college…

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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Similar questions

A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 297 people over the age of 55, 75 dream in black and white, and among 310 people under the age of 25, 19 dream in black and white. Use a 0.05 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. Complete parts (a) through (c) below. a. Identify the test statistic. z= 6.51 (Round to two decimal places as needed.) Identify the P-value. P-value = 3.770o (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P-value is greater than the significance level of a = 0.05, so fail to reject the null hypothesis. There is insufficient evidence to support the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. b. Test the claim by constructing an appropriate confidence…
In 2005, Health Canada claimed that 16.9% of Canadian adults are obese. A researcher wanted to check whether this percentage has changed in 2009. In 2009, he randomly selected 90 Canadian adults and found 12 of them are obese. Use a 7% significance level. The decision rule is either Reject Ho if Zcalc > 1.48 Reject Ho if Zcalc > 1.81 Reject H0 if Zcalc < -1.81 or Zcalc > 1.81 Reject H0 if Zcalc < -1.48 or Zcalc > 1.48
A company that produces a particular brand of juice knows that about 30% of people select that brand if given a choice of all brands. To determine if a commercial would be an effective advertising tool, the company invites 100 people to view a 30-minute program that includes a commercial for the company’s juice. After the program, the participants are offered a choice of juice brands. In this trial, 36 of the 100 participants chose the company’s juice. When a significance test was performed to determine if the true proportion of people who watch the commercial and select the company’s juice exceeds 30%, the P-value was 0.0952. At the α = 0.05 significance level, what decision should the company make? A. Because the P-value of 0.0952 > α = 0.05, the company has evidence that the true proportion of those who would select the company’s juice is larger than 30%. B. Because the P-value of 0.0952 < 0.3, the company has evidence that the true proportion of those who would select the…
The Bureau of Labor Statistics reports that the official unemployment rate for Black people was 10.4% and 4.7% for White people in February 2015. Select all correct answers for this question. O The samples of white and black people are independent. The explanatory variable is the unemployment rate. The response variable is the unemployment rate. The response variable is race.
Magazine A national publishing house claims that 28% of all weekly magazine readers in SA read their publication. Test this claim at the 1% significance level, if a survey found that 104 out of a random sample of 400 magazine readers said that they read the specific publication.
Suppose that in a random selection of 100100 colored candies, 2323% of them are blue. The candy company claims that the percentage of blue candies is equal to 2525%. Use a 0.100.10 significance level to test that claim.
USA Today ran a report about a UNC poll of 1248 adults from the southern U.S. It was reported that 8% of those surveyed believe that Elvis Presley still lives. The article began with the claim that "almost 1 out of 10" Southerners still thinks Elvis is alive. At the 0.01 significance level, test the claim that the true percentage is 10%.
Suppose that in a random selection of 100 colored candies, 23% of them are blue. The candy company claims that the percentage of blue candies is equal to 26%. Use a 0.05 significance level to test that claim.
You are given two 6-sided fair dice. One is a standard die, with faces numbered 1 through 6. The other die is a non-standard die, it has three faces numbered 0 and three faces numbered 6. You roll these two dice once, record the up faces. Let X be the sum of those two numbers you recorded. What is the probability that 6
A barangay captain claims that at least 25% of the residents in his barangay have household pets. To test this claim a researcher randomly selected a sample of 50 residents and find that 13 of them do have pets. What can you conclude at 0.10 level of significance.
Suppose that in a random selection of 100 colored candies, 20% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%. Use a 0.01 significance level to test that claim.
Suppose that in a random selection of 100 colored candies, 20% of them are blue. The candy company claims that the percentage of blue candies is equal to 22%. Use a 0.01 significance level to test that claim.

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