a. Question 3: For each of the following situations determine which hypothesis t

Question

a. Question 3: For each of the following situations determine which hypothesis test procedure is the most appropriate. The significant level is 0.05 You DO have to perform the test, following 5 steps procedure. The hypothesis test procedures you have a choose from are One sample Z-test for proportions T wo sample Z-test for proportions One sample t-test for means Two sample t-test for means Paried t-test b. A committee at the College Board has been asked to study the SAT math scores for students in Pennsylvania and Ohio. A sample of 45 students from Pennsylvania had an average score of 580, whereas a sample of 38 students had an average score of 530. The sample standard deviations for Pennsylvania and Ohio are 105 and 114 respectively. Does the study suggest that the SAT math score for students in Pennsylvania and Ohio differ? c. A field biologist examined the sex ratio at birth of the lesser snow geese. A random sample of nests containing four eggs was taken. For each egg resulting in a live gosling, the laying order and the gender of the gosling were recorded Of the 27 successfully hatched first eggs, 17 were male. Is there significant evidence that the proportion of male goslings is not 50% d. A manufacturer claims that a new design for a portable phone has increased the phone's range to 150 feet, allowing many customers to use the phone throughout their homes and yards. An independent testing laboratory found that a random sample of 44 of these phones worked over an average distance of 142 feet, with a standard deviation of 12 feet. Is there evidence that the manufacturer's claim is false? e. In 2001, one county reported that, among 3132 white women who had babies, 94 were multiple births. There were also 20 multiple births to 606 black women Does this indicate any racial difference in the likelihood of multiple births? Question 4: For Question 3b. The 95% confidence interval for the proportion is (0.4475, .8118) Based on the confidence interval given, is there significant evidence that the proportion of male goslings is not 50%? WHY? If you had instead created a 99% confidence interval, how would it compare to the one given here? Question 5: Does the diet program help participants lose weight? A new diet program claims that participants will lose more than 15 pounds after completion of the program. Fifteen participants were randomly selected to take part in the diet. Each participant was weighed at the beginning of the diet and then weighed again 12 weeks later. The data of the before and after weights for the 15 participants who completed the program are included in a data file on Canvas (weight.txt). Does the data provide sufficient evidence to support the theory that the diet programs leads to a weight loss of more than 15 pounds? Perform a level 0.05 hypothesis test to test this claim. (You can use Minitab to perform the hypothesis test. However, make sure that you still write out the assumptions of the test, write out your conclusion to the hypothesis test, and attach your Minitab output.) Question 6: Construct and interpret a 90% confidence interval for the situation described in Question 3a

Explanation / Answer

3. (a,b). Hypotheses:

H0:mu1-mu2=0 (there is no difference in SAT math score for students in Pennsylvania and Ohio)

H1:mu1-mu2=/=0 (there is difference in SAT math score for students in Pennsylvania and Ohio)

where, group1 denote SAT math score for Pennsylvania and 2 denote SAT math score for Ohio.

Assumptions: Independence group assumption: randomizing the experiment gives independent groups. Independence assumption: this is an experiment, so there is no need for the subjects to be randomly selected, one needs to check whether they were assigned randomly to treatment groups. Randomization condition:the experiment was randomized, as subjects were randomly assigned to treatment groups. Nearly normal condition:though raw data is not available, it is assumed that the SAT math score for the two states are nearly normal.

The assumptions are reasonable and conditions are met, use Students' t model to perform a two-sample t test for means [3(a)].

Test statistic: t=(x1bar-x2bar)/sqrt[s1^2/n1+s2^2/n2], where, xbar is sample mean, s is sample standard deviation, and n is sample size.

=(580530)/sqrt[105^2/45+114^2/38]

=2.06

Using technology, pvalue at t(76) is 0.042.

Decision rule:Reject null hypothesis if p value is less than alpha=0.05. Here, p value is les than 0.05, therefore, reject null hypothesis.

Conclusion: there is sufficient sample evidence to conclude that there is significant difference in SAT math score for students in Pennsylvania and Ohio.

3(a,c) Hypotheses: H0:p=0.5 (proportion of male gosling is 50%)

H1:p=/=0.5 (proportion of male gosling is not 50%)

Assumptions: Model:random sampling; level of measurement:nominal, sampling distribution:normal. Both np=27*0.5=13.5 and n(1-p)=27*0.5=13.5 are atleast 5. Therefore, use one-sample Z test for proportions [3(a)].

Test statistic: z=(phat-p)/sqrt[p(1-p)/n], where, phat is sample proportion, p is population proportion, n is sample size.

=(17/27-0.5)/sqrt[0.5(1-0.5)/27]

=1.35

pvalue is 0.178.

Decsiion rule: reject null hypothesi if p value is less than alpha=0.05. Here, p value is not less than 0.05, therefore, fail to reject null hypothesis.

Conclusion: There is insufficient sample evidence to conclude that proportion of male gosling is not 50%.

3(a,d) Hypothesis: H0: mu=150 (the average distance for new design phone is 150 feet)

H1:mu=/=150 (the average distance for new design phone is not 150 feet)

Assumption: Randomization condition: the phones were sampled in a randomized way, so the distance of working is mutually exclusive. Nearly normal condition: it is assumed that the workingdistance is nearly normal.

The conditions are met, use student's t model with (n-1)=43 degrees of freedom and compute one-sample t test. for means [3(a)].

test statistic: t=(xbar-mu)/(s/sqrt n), where, xbar is sample mean, mu is population mean, s is sample standard deviation and n is sample size.

=(142-150)/(12/sqrt 44)

=-4.42

p value is 0.0000.

Decision rule:reject null hypothesi if p value is less than alpha=0.05. Here, p value is less than 0.05, therefore, reject null hypothesis.

Conclusion: there is sufficient sample evidence to conclude that the average distance for new design phone is not 150 feet.

3(a,e) Hypotheses: H0: p1-p2=0 (there is no difference in likelihood of multiple births for white and black women)

H1:p1-p2=/=0 (there is difference in likelihood of multiple births for white and black women)

Assumptions: Model: independnet random samples; level of measurement: nominal; sampling distribution: normal; for two samples, x1=94, n1-x1=3132-94=3038, x2=20, n2-x2=586 are all greater than 10. Thus, use two sample z test for proportions [3(a)].

Test ststaitic: pooled sample proportion, phatp=(x1+x2)/(n1+n2)=(94+20)/(3132+606)=0.03

Z=(p1hat-p2hat)/[sqrt{phatp(1-phatp)}sqrt(1/n1+1/n2)]

=(94/3132-20/606)/[sqrt{0.03(1-0.03)}sqrt(1/3132+1/606)]

=-0.39

p value:0.695

Decison rule: reject null hypothesis if p value is less that 0.05. Here, p value is not less than 0.05, therefore, fail to reject null hypothesis.

Conclusion: there is insufficient sample evidence to suggest that there is difference in likelihood of multiple births for white and black women.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.