Question

Suppose that BC financial aid allots a textbook stipend by claiming that the average textbook at BC bookstore costs $71.84. You want to test this claim.

The null and alternative hypothesis in symbols would be:

• H0:μ≤71.84
  H1:μ>71.84
• H0:μ=71.84
  H1:μ≠71.84
• H0:p=71.84
  H1:p≠71.84
• H0:p≥71.84
  H1:p<71.84
• H0:μ≥71.84
  H1:μ<71.84
• H0:p≤71.84
  H1:p>71.84

The null hypothesis in words would be:

• The average of price of all textbooks from the store is greater than $71.84.
• The proportion of all textbooks from the store that are less than 71.84 is equal to 50%
• The average of price of all textbooks from the store is less than $71.84.
• The average price of all textbooks from the store is $71.84
• The average price of textbooks in a sample is $71.84

Based on a sample of 200 textbooks at the store, you find an average of 79.04 and a standard deviation of 26.8.

The point estimate is:  (to 3 decimals)

The 99 % confidence interval (use z*) is:  to  (to 3 decimals)

Based on this we:

• Reject the null hypothesis
• Fail to reject the null hypothesis

Please explain how to solve this problem by using a TI-84 calculator. Thank you!!!

Answer 1

The null and alternative hypothesis is

• H0:μ=71.84
  H1:μ≠71.84

The null hypothesis in words would be:

• The average price of all textbooks from the store is $71.84

Sample size = n = 200

Sample mean = = 79.04

Standard deviation = s = 26.8

The point estimate is: Sample mean = = 79.04

We have to construct a 99% confidence interval.

By using the TI-84 calculator we have to find a confidence interval.

Here population standard deviation is unknown so we have to use t-interval.

Click on STAT --------> TESTS --------> TInterval ----------> Stats -------->

: 79.04

Sx: 26.8

n: 200

C-Level: 0.99

Calculate

We get confidence interval ( 74.111 , 83.969)

The value 71.84 is not belongs to the above interval so we reject null hypothesis.

• Reject the null hypothesis