The lengths of time (in years) it took a random sample of 32 former smokers to q
ID: 3224815 • Letter: T
Question
The lengths of time (in years) it took a random sample of 32 former smokers to quit smoking permanently are listed. Assume the population standard deviation is 46 years. At alpha = 0.03, is there enough evidence to reject the claim that the mean time it takes smokers to quit smoking permanently is 14 years? Complete parts (a) through (e). 16.1 11.7 13.9 21.2 15.9 17.1 7.6 19.4 10.4 7.8 16.8 14.2 16.7 11.1 17.2 12.2 20.1 7.5 20.7 21.4 21.5 17.8 11.19.5 13.9 11.4 15.5 14.9 13.5 20.2 20.3 13.4 (a) Identify the claim and state the null hypothesis and alternative hypothesis. A. H_0: mu lessthanorequalto 14 (claim) B. H_0: mu > 14 (claim) H_a: mu > 14 H_a: mu lessthanorequalto 14 D. H_0: mu NotEqual 14 (claim) E. H_0: mu = 14 H_a: mu = 14 H_a = NotEqual 14 (b) Identify the standardized test statistic. Use technology. Z = (Round to two decimal places as needed)Explanation / Answer
Using excel function =AVERAGE(16.1,11.7,13.9,21.2,15.9,17.1,7.6,19.4,10.4,7.8,16.8,14.2,16.7,11.1,17.2,12.2,20.1,7.5,20.7,21.4,21.5,17.8,11.1,9.5,13.9,11.4,15.5,14.9,13.5,20.2,20.3,13.4)
We get
Sample mean xbar=15.0625
So, z=(xbar-14)/(/sqrt(n))
=(15.0625-14)/(14/sqrt(32))
=0.43