A certain drug is used to treat asthma. In a clinical trial ofthe drug, 30 of 281 treated subjects experiencedheadaches (based on data from the manufacturer). Theaccompanying calculator display shows results from a testof the claim that less than 10% of treated subjectsexperienced headaches. Use the normal distribution as anapproximation to the binomial distribution and assume a0.05 significance level to complete parts (a) through (e)below.a. Is the test two-tailed, left-tailed, or right-tailed?Right tailed testTwo-tailed testLeft-tailed testb. What is the test statistic?Z=(Round to two decimal places as needed.)c. What is the P-value?1-PropZTestprop <0.1z = 0.377814991p=0.6472159817p=0.1067615658^n = 281P-value =(Round to four decimal places as needed.)d. What is the null hypothesis, and what do you conclude about it?Identify the null hypothesis.O A. Ho: p<0.1B. Ho: p=0.1C. Ho: p>0.1D. Ho: p = 0.1 A certain drug is used to treat asthma. In a clinical trial ofthe drug, 30 of 281 treated subjects experiencedheadaches (based on data from the manufacturer). Theaccompanying calculator display shows results from a testof the claim that less than 10% of treated subjectsexperienced headaches. Use the normal distribution as anapproximation to the binomial distribution and assume a0.05 significance level to complete parts (a) through (e)below.C. Ho:p> 0.1D. Ho: p = 0.11-PropZTestprop <0.1z = 0.377814991p=0.6472159817p=0.1067615658n = 281Decide whether to reject the null hypothesis. Choose the correct answer below.A. Fail to reject the null hypothesis because the P-value is less than or equal tothe significance level, a.B. Reject the null hypothesis because the P-value is less than or equal to thesignificance level, a.C. Reject the null hypothesis because the P-value is greater than the significancelevel, a.D. Fail to reject the null hypothesis because the P-value is greater than thesignificance level, a.e. What is the final conclusion?A. There is not sufficient evidence to support the claim that less than 10% oftreated subjects experienced headaches.B. There is not sufficient evidence to warrant rejection of the claim that less than10% of treated subjects experienced headaches.C. There is sufficient evidence to support the claim that less than 10% of treatedsubjects experienced headaches.D. There is sufficient evidence to warrant rejection of the claim that less than10% of treated subjects experienced headaches.

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Answered: A certain drug is used to treat asthma.…

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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A 0.1 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 than 0.5. Assume that sample data consists of 78 girls in 169 births, so the sample statistic of 613 results in a z score that is 1 standard deviation below 0. Complete parts (a) through (h) below.Click here to view page 1 of the Normal table. LOADING... Click here to view page 2 of the Normal table. LOADING...Question content area bottomPart 1a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below.A.H0: p≠0.5H1: p<0.5B.H0: p=0.5H1: p≠0.5C.H0: p=0.5H1: p<0.5D.H0: p=0.5H1: p>0.5Part 2b. What is the value of α?α=enter your response here (Type an integer or a decimal.)Part 3c. What is the sampling distribution of the sample statistic? Normal distribution Student (t) distribution χ2Part 4d. Is the test two-tailed, left-tailed, or right-tailed? Right-tailed…
. 3. A researcher is interested in knowing if preterm infants with late metabolic acidosis and preterm infants without the condition differ with respect to urine levels of a certain chemical. The mean levels, standard deviations, and sample sizes from the two samples are given below. Assuming that the populations have equal variances: a. Fill in the missing pieces of the Stata output. ttesti 38 8.5 3.5 32 6.85 2.6 X y Two-sample t test with A variances combined diff Obs Ho: diff=0 38 32 Sample With condition 70 diff mean(x) n 38 Without condition 32 Ha: diff |t|) = F [95% Conf. Intervall 7.349579 5.9126 6.98079 .1552544 t = degrees of freedom = 9.650421 7.7874 8.510638 3.144746 E b. What is the conclusion of your test? c. Use this output to construct a 99% confidence interval for the difference between the two means. 68 Ha: diff0 Pr(Tt) = 0.0155
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7
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14.
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