Find the P-value for the indicated hypothesis test. sample of 79 children select
ID: 3319526 • Letter: F
Question
Find the P-value for the indicated hypothesis test. sample of 79 children selected randomly from one town, t is found that 10 of them1 In a suffer from asthma. Find the P-value for a test of the claim that the proportion of all children in the town who suffer from asthma is less than 11%. Find the confidence interval specified random sample of 100 full-grown lobsters had a mean weight of 23 ounces. Assume that = 3.0 ounces. Construct a 98% confidence interval for the population mean . Perform a hypothesis test for the population mean. GDA test of sobriety involves measuring the subject's motor skills. Twenty randomly select ed sober subjects take the test and produce a mean score of 41.0 with a standard deviation of 3.7. At the 0.01 level of significance, test the claim that the true mean score for all sober subjects is equal to 35.0Explanation / Answer
Question 1:
below are the null and alternate hypothesis
H0: p = 0.11
H1: p < 0.11
SE = sqrt(0.11*0.89/79) = 0.0352
pcap = 10/79 = 0.1266
test statistics, t = (0.1266 - 0.11)/0.0352 = 0.4716
p-value = 0.3186
Question 2:
Question 3:
below are the null and alternate hypothesis
H0: mu = 35
H1: p not equals to 35
test statistics, t = (41 - 35)/(3.7/sqrt(20)) = 7.252
p-value = 0.00000
Reject null hypothesis
There are sufficient evidence to conclude that true mean score is not equals to 35
CI for 98% n 100 mean 23 z-value of 98% CI 2.3263 std. dev. 3 SE = std.dev./sqrt(n) 0.30000 ME = z*SE 0.69790 Lower Limit = Mean - ME 22.30210 Upper Limit = Mean + ME 23.69790 98% CI (22.3021 , 23.6979 )Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.