A survey of 1 comma 567 randomly selected adults showed that 570 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 38% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (d. Sample proportion: 0.363752 Test statistic, z: negative 1.3251 Critical z: plus or minus1.9600 P-Value: 0.1852 a. Is the test two-tailed, left-tailed, or right-tailed? Left-tailed test Two-tailed test Right tailed test b. What is the P-value? P-valueequals 0.1852 (Round to four decimal places as needed.) c. What is the null hypothesis and what do you conclude about it? Identify the null hypothesis. A. Upper H 0 : p greater than 0.38 B. Upper H 0 : p less than 0.38 C. Upper H 0 : p equals 0.38 D. Upper H 0 : p not equals 0.38
A survey of 1 comma 567 randomly selected adults showed that 570 of them have heard...
A survey of 1,680 randomly selected adults showed that 549 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 37% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts a through e Sample proportion: 0.326786 Test statistic Critical z: P-Value z:-3.6687 ± 2.5758 0.0002 a. Is the test...
A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The technology display below ret from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the bin distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p+0.92 Z-Value P-value Sample p 95% CI N Sample X 0.000 4.01 (0.930869,0.956847) 1 1967 2,084 0.943858 a....
A poll of 2,142 randomly selected adults showed that 92% of them own cell phones. The technology display below results from a test of the claim that 91% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (e). Test of p=0.91 vs p≠0.91 Sample X N Sample p 95% CI Z-Value P-Value 1 1970 2,142 0.919701 (0.908193,0.931210) 1.57 0.117 a. Is the...
a poll of 2094 randomly selected adults showed that 94% of
them own cell phones
A poll of 2,094 randomly selected adults showed that 94% of them own cell phones. The technology display below results from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p*0.92 Samplex N...
A certain drug is used to treat asthma. In a clinical trial of the drug, 2727 of 287287 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 1212% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.010.01 significance level to complete parts (a) through (e) below. 1-PropZTest propless than<0.120.12 zequals=negative 1.351450141−1.351450141 pequals=0.08827564060.0882756406 ModifyingAbove...
A 0.01 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is less less than 0.5. Assume that sample data consists of 45 girls in 100 births, so the sample statistic of StartFraction 9 Over 20 EndFraction 9 20 results in a z score that is 1 standard deviation below below 0. Complete parts (a) through (h) below. Click here to view page...
In a study of 807807 randomly selected medical malpractice lawsuits, it was found that 487487 of them were dropped or dismissed. Use a 0.010.01 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed. Which of the following is the hypothesis test to be conducted? A. Upper H 0 : p not equals 0.5H0: p≠0.5 Upper H 1 : p equals 0.5H1: p=0.5 B. Upper H 0 : p greater than 0.5H0: p>0.5 Upper H...
A genetic experiment involving peas yielded one sample of offspring consisting of 435 green peas and 133 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 27 % of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution. What are...
The test statistic of zequals=2.952.95 is obtained when testing the claim that pnot equals≠0.4970.497. a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed. b. Find the P-value. c. Using a significance level of alphaαequals=0.010.01, should we reject Upper H 0H0 or should we fail to reject Upper H 0H0? Click here to view page 1 of the standard normal distribution table. LOADING... Click here to view page 2 of the standard normal distribution table. LOADING... a. This is...
1. Claim Fewer than 97 % of adults have a cell phone. In a reputable poll of 1038 adults, 89 % said that they have a cell phone. Find the value of the test statistic.2. The test statistic of z=1.38 is obtained when testing the claim that p>0.2a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.b. Find the P-value by the calculator or by the table.c. Using a significance level of α=0.05 should we reject H₀ or should...