Question

1. Suppose we are interested in analyzing how much people spend on travel during the summer. We know that on average, people spend $1005 on travel with a population standard deviation of $1740. Suppose we take a sample of 30 people and we find that the average spent from that sample is $1850. We are interested in seeing if the amount spent on travel is increasing.
   1. What is the standard error?
   2. What is the margin of error at 90% confidence?
   3. Using my sample of 30, what would be the 90% confidence interval for the population mean?
   4. If I wanted to control my margin of error and set it to 500 at 90% confidence, what sample size would I need to take instead of the 30?
   5. What are the null and alternative hypotheses?
   6. What is the critical value at 90% confidence?
   7. Calculate the test statistic (using the sample of 30 and NOT the answer from part d).
   8. Find the p-value.
   1. What conclusion would be made here at the 90% confidence level?

Answer 1

Answer:

=1005, =1740, n=30, =1850

a)

Standard error= 317.679

b)

Margin of error = Zc * standard error

Where zc is the critical value for c=90%

Margin of error= 1.645* 317.679

Margin of error= 522.582

C)

Formula for confidence interval is

margin of error

1850 522.582

Thus we get confidence interval as

( 1327.418 , 2372.582)

d)

Formula is

n=32.771

Sample size required is= 33

e)

Ho: 1005

Ha: > 1005

f)

Critical value = 1.28

g)

Test statistics= 2.66

h)

Calculate p-value for right tailed test we get

P-value= 0.0039

I)

Since p-value < ( =0.10)

Hence Reject null hypothesis.