A college administrator claims that 84% of college students purchase their books from the campus bookstore. You this is inaccurate and form a random sample of 66 students at that college and find that 61 of them purchased their books from the bookstore. Test the administrator's claim using a level of significance of 5%.

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A college administrator claims that 84% of college students purchase their books from the campus bookstore. You this is inaccurate and form a random sample of 66 students at that college and find that 61 of them purchased their books from the bookstore. Test the administrator's claim using a level of significance of 5%.

  1. What type of test will be used in this problem?

 



  • What evidence justifies the use of this test? Check all that apply
    • The sample size is larger than 30
    • np>5

and nq>5

    •  
    • The population standard deviation is known
    • The original population is approximately normal
    • The population standard deviation is not known
    • There are two different samples being compared
    • The sample standard deviation is not known


  • Enter the null hypothesis for this test.
    H0


  • Enter the alternative hypothesis for this test.
    H1:


  • Is the original claim located in the null or alternative hypothesis?


  • What is the test statistic for the given statistics?


  • What is the p-value for this test?


  • What is the decision based on the given statistics?


  • What is the correct interpretation of this decision?

    Using a % level of significance, there

sufficient evidence to the claim that 84% of college students purchase their books from the campus bookstore.

 

Expert Solution
Step 1

The type of test used is one-proportion test because the claim is to test whether 84% of college students purchase their books from the campus bookstore or not.

Thus, the type of test used is One-proportion test.

From the given information, the population proportion is 0.84, and sample size is 66.

The evidence justifies the use of this test is np>5 and nq>5.

Checking conditions:

np=66×0.84=55.44>5

np=66×1-0.84=66×0.16=10.56>5

Thus, the conditions are satisfied.

The hypothesis is,

Null hypothesis:

H0:p=0.84

Alternative hypothesis:

H1:p0.84

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