Question

number one has been answered! please help with the rest

2- A randon sample of College students were asked how moch money they spent on books and supplies this senester. Beze are the sumary statistics: a-10 $249, s-$30.construct a 90% confidence interval tor the nean coats for all college students. Assune that the anount spent on books and supplies by college students is nonally distributed. 3- A researchez wants to estinate the mean beights of 8- year- old gizis. It is coc that the population standard deviation is 2.9 inches. Bow large the sample should be in order to be 95% confident that the estimate is within 0.3 inches of the popalation ean? -& consner group ciains that the mean annual consunption person in the 0s is at most 10.3 lbs. A randon sanple of 100 people in the DS has a nean ancoal cheddar cheese consnption of 9.9 1bs. Assune the population standard deviation is 2.1 lbs. Use a 0.05 significant level to test the clain ehat the popalation ean of cheddar cheese consunption per person in the Us is at most 10.3 lbs of cheddar cheese by a 5- consider the 90% confidence interval for the mean of student evaluation ratings of coarses at a certain sniversity (3.67, 4.17 al Calcalate the point estinate of b) Calcalate tbe nargis of ezzor 6-Asedical researcher says that less than 25% of us are smokers. In a randon s-le of 200 CS adults 19.3% say that they are smokers. a) state the bypothesis Testing problen and specity the claim and the type of cest b) 14st tbe assnption and calculate the test statistic. 7- True or False: a) α is the probability of Type-1 Error b) For a given standard error lower conidence levels produce wider confidence c) Oze way to get a nazzover confidence interval is to increase the sample d) The equal sig is alvays zeserved for the null hypothesis. intervals size. e) The z score corresponding to 99% confidence level is 1.96

Answer 1

2:

Here we have x =249, s-30, n=10 Since populaiton SD is unknown so t critical value will be used. Level of significance: α-0.1 Critical value: .833 Degree of freedom df -n-1-9 The requried confidence interval is: 249 (30*1.833/sqrt (10)) Nm -249 t 17.389 (231.611,266.389) Excel function for critical value: TINV(0.1,9) Hence,the required cofidence intervalis (231,611266.389).

3:

Here we have ME-03,ơ-29, CL-0.95 Level of significance: -0.05 Critical value of z is 21-a12-1.96 Required sample size is: -2 ((1.96*2.9)/0.3)A2 ME Hence, the required sample size is 359. Excel function for critical value: NORMSINV(1-(0.05/2))