In: Statistics and Probability
Word Problems with a Sample Data Set:
16 18 12 17 20 18 15 14
*Round all final answers to 2 decimal places if necessary
a)
sample mean x= | 16.250 |
sample size n= | 8 |
sample std deviation s= | 2.550 |
std error sx=s/√n= | 0.9014 |
for 90% CI; and 7 df, critical t= | 1.8950 | |
margin of error E=t*std error = | 1.7081 | |
lower bound=sample mean-E = | 14.5419 | |
Upper bound=sample mean+E= | 17.9581 | |
from above 90% confidence interval for population mean =(14.54,17.96) |
c)
here n = | 8 | ||
s2= | 6.500 | ||
Critical value of chi square distribution for n-1=7 df and 95 % CI | |||
Lower critical value χ2L= | 1.690 | ||
Upper critical valueχ2U= | 16.013 |
for Confidence interval of standard deviation: | |||||
Lower bound =√((n-1)s2/χ2U)=(8-1)*6.5/16.013= |
1.686 | ||||
Upper bound =√((n-1)s2/χ2L)=(8-1)*6.5/2.841 = | 5.189 | ||||
from above 95% confidence interval for population standard deviation =(1.686<σ<5.189) |
c)
null hypothesis: Ho: μ | = | 18 | |
Alternate Hypothesis: Ha:μ | < | 18 | |
0.05 level with left tailed test and n-1= 7 df, critical t= | -1.895 | ||
Decision rule : reject Ho if test statistic t<-1.895 | |||
population mean μ= | 18 | ||
sample mean x= | 16.250 | ||
sample size n= | 8 | ||
sample std deviation s= | 2.550 | ||
std error sx=s/√n= | 0.9014 | ||
test stat t='(x-μ)*√n/s= | -1.941 |
since test statistic falls in rejection region we reject null hypothesis | |||
we have sufficient evidence to conclude that their students average less than 18 credit hours per semester |