M7 Problem Set

M7: Problem Set Due No due date Points 5 Questions 12 Time Limit None Attempt History Attempt Time Score LATEST Attempt 1 7,058 minutes 5 out of 5 Score for this quiz: 9 out of 5 Submitted Sep 4 at 12:41pm This attempt took 7,058 minutes. Question 1 0/0 pts A lettuce grower claims that the mean weight of her heads of lettuce is 2 pounds. You believe that the mean weight of the lettuce is less than two pounds. Please state the null and alternative hypothesis. Your Answer: Here claim is that mean is less than two pounds As null hypothesis is always having equality sign so null hypothesis is H() U= 2 And so claim is the alternative hypothesis that is | H; : u < 2 Ho: u=2 pounds. H4: u<2 pounds.

Question 2 0/0 pts According to Time magazine, in April 2014 the average gas price in the U.S. was $ 3.70. A financial analyst believes that the average gas price is different today. Please state the null and alternative hypothesis. Your Answer: Claim: financial analyst believes that the average gas price is different today. it mean s the average price can be greater than $3.70 or less than $3.70 \Hy : p = 3.70| \Hy : p# 3.70] Ho: u=9% 3.70. Hq: p#$ 3.70. Question 3 0/0 pts According to the Wall Street Journal, in 2011 the average (or mean) cable bill was $ 128 per month. You believe that the mean cable bill is higher today. Please state the null and alternative hypothesis. Your Answer: \Ho : p =128 ‘Hl U > 128‘

a) Type | error: Reject the null hypothesis that the mean weight is 2 pounds when the mean weight is at least 2 pounds b) Type Il error: Do not reject the null hypothesis that the mean weight is 2 pounds when the mean weight is less than 2 pounds Question 5 0/0 pts According to Time magazine, in April 2014 the average gas price in the U.S. was $ 3.70. A financial analyst believes that the average gas price is different today. Please state the null and alternative hypothesis. a) Explain what it would mean to make a Type | error. b) Explain what it would mean to make a Type Il error. Your Answer: \Hy : p = 3.70} \Hy @ p # 3.70| a. A type | error occurs if we reject the null hypothesis when the null hypothesis is actually true. Concluding that the average gas price was not $ 3.70. when in reality the claim is correct. b. Atype Il error occurs if we do not reject the null hypothesis when the null hypothesis is false. Supporting the claim of that the average gas price was $ 3.70 when in reality it is not.

a) Type | error: Reject the null hypothesis that the mean gas price is $ 3.70 when the mean gas price is $ 3.70. b) Type Il error: Do not reject the null hypothesis that the mean gas price is $ 3.70 when the mean gas price is different from $ 3.70. Question 6 0/0pts According to the wall street journal, in 2011 the average (or mean) cable bill was $ 128 per month. You believe that the mean cable bill is higher today. Please state the null and alternative hypothesis. a) Explain what it would mean to make a Type | error. b) Explain what it would mean to make a Type Il error. Your Answer: \Hy : p = 128 ‘HO U > 128‘ a. A type | error occurs if we reject the null hypothesis when the null hypothesis is actually true. Concluding that the average cable bill was not $128 per month. when in reality the claim is correct. b. A type Il error occurs if we do not reject the null hypothesis when the null hypothesis is false. Supporting the claim of that the average cable bill was $128 per month when in reality it is not.

It is important to note that this is a right-tailed test. Therefore, we want to find a z that satisfies: P(Z>z)=.05. It is easier to find the equivalent z that satisfies P(Z<z)=1-.05=.95. Corresponding to a probability of .95, we find on the standard normal chart that, approximately, z=1.64. Question 8 0/0 pts Suppose that we have a problem for which the null and alternative hypothesis are given by: Ho: u = 890. Hq:u # 890. Is this a right-tailed test, left-tailed test, or two-tailed test. Find the z value based on a level of significance of a=.09. Your Answer: Two tailed test since ‘Hl - HO‘ IP(Z<2)=1-.09=.91 z=1.34 and z=-1.34

It is important to note that this is a two-tailed test. Therefore, we want to find a z that satisfies: P(Z<-2)=.09/2=.045 and P(Z>z)=.09/2=.045. When we find the corresponding z on the standard normal table we find z=-1.70 and z=1.70. Question 9 0/0 pts Suppose that we have a problem for which the null and alternative hypothesis are given by: Ho: M = 88.5 ounces. Hq:M < 88.5 ounces. Is this a right-tailed test, left-tailed test, or two-tailed test. Find the z value based on a level of significance of a=.10. Your Answer: Left tail since ‘Hl < HO‘ \P(Z < z) = .10| z=-1.28

Ho: M = 50.5 hours. H¢:u < 50.5 hours. This is a left-tailed test, so we must find a z that satisfies P(Z<z)=.02. In the standard normal table, we find z..qg» =-2.05. For a left-tailed test, we will reject the null hypothesis if the z-score is less than -2.05. We find the z-score: o 12.7 5008 Og=—= —= 2 T vn 40 _¥-u_458-505_ 2= s~ 2008 ' Notice that since the z-score is less than -2.05, we reject the null hypothesis. Question 11 0/0pts A social worker in a certain city claimed that only 25% of the children between 19 and 35 months had not had all of their vaccines. You believe the percentage is higher. In order to test the social worker’s claim, you contact 145 families that have a child between 19 and 35 months and find that 44 of the children had not had all of their vaccines. Can the social worker’s claim be supported to a level of significance of a = .02, test the hypothesis. Your Answer: p=.25, n=145, x=44, a=.02 \Hy : p=.25] ‘Hl U > 25‘ Sample proportion ‘13 == = 1% = .30

30—.20 1.39 p-p _ \/W 25(1—.25) n 145 1.39<20.53 Fail to reject Test static|z = Suppose p represents the probability that the child has not had of his or her vaccines Hoi P = 25. Hq: p > .25. Since this is a right-tailed test, we must find the z that satisfies P(Z>z)=.02. In the standard normal table, we find that z..q2 =2.05. This is a right-tailed test, we reject the null hypothesis if the z-score is greater than 2.05. Recall that for proportions, the mean and standard deviation are found by: u=np = 145(.25) = 36.25 os = np(1 — p) =/145(.25)(1 - .25) =5.21 If we find the z-score: _F-u_44-3625 Z — SCore = e = 571 — 1.9, Since this is a right-tailed test, and the z-score is less than 2.05, we do not reject the null hypothesis. Question 12 515 pts As a reminder, the questions in this review quiz are a requirement of the course and the best way to prepare for the module exam. Did you complete all questions in their entirety and show your work?? Your Answer: yes