Question

Question 7

    You work for a hotel chain. You boss wants to know what the mean price for a comparable hotel room in Kansas City to help her determine what she should charge for your chain's rooms. You sample 101 random hotels similar to your own and get the daily cost of a room for each. You compute a mean cost of $96.50. The standard deviation of your sample is $19.50. What is the 95% confidence interval for the mean cost of a hotel room in Kansas City?

    $89.55 to $97.45

      $90.00 to $98.90

        $91.45 to $99.55

    $92.68 to $100.32

1 points

Question 8

    You work for a hotel chain. You boss wants to know what the mean price for a comparable hotel room in Kansas City to help her determine what she should charge for your chain's rooms. You sample 101 random hotels similar to your own and get the daily cost of a room for each. You compute a mean cost of $96.50. The standard deviation of your sample is $19.50. What is the 99% confidence interval for the mean cost of a hotel room in Kansas City?

    $92.88 to $100.00

    $91.49 to $101.51

    $90.00 to $101.99

    $87.50 to $108.50

Question 9

   What is a mathematically based statement describing the mean, standard deviation and shape of a sampling distribution of means. (Hint: look in last week's, Week 8, PowerPoint Presentation)

        Standard Error of the Mean

        Standard Deviation

        Central Limits Theorem

        Null Hypothesis

Question 10

    What is the relationship between sample size and standard error? (Hint: look in last week's, Week 8, PowerPoint Presentation)

        The smaller the sample size, the smaller the standard error.

        The smaller the sample size, the larger the standard error.

        Standard error is the square root of sample size.

       There is no relationship between sample size and standard error.

Answer 1

7)

sample mean 'x̄=	96.50
sample size   n=	101.00
sample std deviation s=	19.50
std error 'sx=s/√n=	1.9403

for 95% CI; and 100 df, value of t=		1.984
margin of error E=t*std error    =		3.850
lower bound=sample mean-E =		92.68
Upper bound=sample mean+E =		100.32
from above 95% confidence interval for population mean =(92.68,100.32)

8)

for 99 % CI value of z=		2.58
margin of error E=z*std error =		4.998
lower bound=sample mean-E=		91.49
Upper bound=sample mean+E=		101.51
from above 99% confidence interval for population mean =(91.49 ,101.51)

9)

  Central Limits Theorem

10)

The smaller the sample size, the larger the standard error.