A newsletter publisher believes that less than 62% of their readers own a laptop. For marketing purposes, a potential advertiser wants to confirm this claim. Afterperforming a test at the 0.10 level of significance, the advertiser decides to reject the null hypothesis.What is the conclusion regarding the publisher's claim?Answer◇ There is sufficient evidence at the 0.10 level of significance that the percentage is less than 62%○ There is not sufficient evidence at the 0.10 level of significance to say that the percentage is less than 62 %.TablesKeypadKeyboard Shortcuts

The objective of the question is to determine the conclusion regarding the publisher's claim that…

Answered: A newsletter publisher believes that…

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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The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that 77% of all fatally injured automobile drivers were intoxicated. A random sample of 53 records of automobile driver fatalities in a certain county showed that 33 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than 77% in Kit Carson County? Use a = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. O Ho: p = 0.77; Hị: p + 0.77 O Ho: p 0.77 O Ho: p = 0.77; H1: p 5 and ng > 5. O The standard normal, since np > 5 and ng > 5. O The Student's t, since np < 5 and ng < 5. O The standard normal, since np < 5 and ng < 5. What is the value of the sample test statistic? (Round your answer to two decimal places.) (c) Find the P-value of the test statistic. (Round your answer to four decimal places.)
A nutrition expert claims that the average American is overweight: To test his claim, a random sample of 25 Americans was selected, and the difference between each person's actual weight and idea weight was calculated. For this data, we have = 16 and 8 = 28.4. Is there sufficient evidence to conclude that the expert's claim is true? Carry out a hypothesis test at a 8% significance level. Hint: Pay attention to the variable being measured: the difference between actual weight and ideal. A positive value indicates that the person is overweight. A. The value of the standardized test statistic (give to 3 decimal places): B. The p-value is (give to 4 decimal places): C. Your decision for the hypothesis test: O A. Reject Ho- OB. Reject Ha C. Do Not Reject Ha- O D. Do Not Reject Ho- E
Historically, the percentage of U.S. residents who support stricter gun control laws has been 55%. A recent Gallup Poll of 2500 people showed 1723 in favor of stricter gun control laws . Assume the poll was given to a random sample of people. QUESTION Test the claim that the proportion of those favoring stricter gun control has changed from 0.55. Perform a hypothesis test, using a significance level of 0.01. What kind of test will you need to perform? A right, one-tailed z-test A left, one-tailed z-test A two-tailed z- test
The Gallup Poll asked 1015 U.S. adults whether they believed that people should pay sales tax on items purchased over the internet. Of these, 477 said they supported such a tax. Does the survey provide convincing evidence that less than 42% of U.S. adults favor an internet sales tax? Use the 0.05 level of significance. 1. STATE THE NULL AND ALTERNATIVE HYPOTHESIS: 2. T STATISTIC/ P VALUE 3.REJECT OR FAIL THE NULL HYPOTHESIS 4. IS THE CLAIM SUPPORTED AT THE 0.05 LEVEL OF SIGNIFICANCE?
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A survey claims that 9 out of 10 doctors recommend brand Wfor their patients who have children. To test this claim against the alternative that the actual proportion of doctors who recommend brand Wis less than 90%, a random sample of 100 doctors results in 85 who indicate that they recommend brand W. 1. What decision should you makeusing a significance level of 5%? 2. Summarize your conclusions in the context of the problem.
With the null hypothesis that a suspect is innocent, an investigation concluded that the suspect should not be convicted at 5% level of significance. The conclusion would be the same if tested at 1% level of significance. Select one: True False
HOw would you figure this out?
You claim that the proportion of students who favored ONLINE ASSESSMENT over FACE-TO-FACE ASSESSMENT is 65% at a 5% significance level. From the analysis, the Z-test is 1.16. The correct conclusion is we have a. sufficient evidence to reject the claim. O b. insufficient evidence to reject the claim. O c. insufficient evidence to support the claim. O sufficient evidence to support the claim.
If the test requires that the results be statistically significant at a level of α = 0.05, fill in the blanks and complete the sentences that explain the test decision and conclusion. The decision is to reject the null hypothesis. There is enough evidence at a significance level of a = 0.05 that the mean weight for U.S. women of age 40-49 who smoke daily is less than 169.976 pounds.
A national organization that conducts research on the cost and quality of health care in the U.S. reported that, in 2012, U.S. families spent an average of $9,590 on health care expenses. Suppose you decide to test whetherthe average in 2015 is greater than the average in 2012. After conducting the appropriate statistical test, you find a P-value of 0.022. If the level of significance is 0.05, which of the following is the best interpretation of the P-value? a. The P-value of 0.022 indicates that there is a 2.2% chance that the 2015 average is greater than the average amount spent in 2012.b. The P-value of 0.022 provides weak evidence that the average in 2015 average is greater than the average amount spent in 2012.c. The P-value of 0.022 provides strong evidence that the 2015 average is greater than the average amount spent in 2012.d. The P-value of 0.022 indicates that there is a very low probability that the 2015 average is different than average amount spent in 2012.
The US Department of Energy reported that 45% of homes were heated by natural gas. A random sample of 348 homes in Oregon found that 156 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 1% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) Ho: P Test Statistic = =✓ ✓ OF H₁: P Based on the hypotheses, compute the following: p-value = 45 0° 45 OB vo

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