Question

1. A sample of 36 account balances of a credit company showed an average balance of $1,179 and a standard deviation of $136. You want to determine if the mean of all account balances is significantly greater than $1,150. Use a 0.05 level of significance. Assume the population of account balances is normally distributed.

   Compute the test statistic.

2. A sample of 30 account balances of a credit company showed an average balance of $1,165 and a standard deviation of $125. You want to determine if the mean of all account balances is significantly greater than $1,150. Assume the population of account balances is normally distributed.

   Compute the p-value for this test.

3. A sample of 28 account balances of a credit company was taken to test whether the mean of all account balances is significantly greater than $1,150. Using the sample standard deviation, the test statistic (t) was calculated to be $1.93. We use a 0.05 level of significance. Assume the population of account balances is normally distributed and the population standard deviation is unknown to us.

   We conclude that the mean of all account balances is significantly greater than $1,150. (Enter 1 if the conclusion is correct. Enter 0 if the conclusion is wrong.)

4. The U.S. Bureau of Labor Statistics reports that 11.6% of U.S. workers belong to unions. Suppose a sample of 328 U.S. workers is collected in 2014 to determine whether union efforts to organize have increased union membership. The sample results show that 52 of the workers belonged to unions.

   Compute the test statistics.

Answer 1

1) The test statistic t = ()/(s/)

= (1179 - 1150)/(136/)

= 1.279

2) The test statistic t = ()/(s/)

= (1165 - 1150)/(125/)

= 0.657

DF = 30 - 1 = 29

P-value = P(T > 0.657)

= 1 - P(T < 0.657)

= 1 - 0.7418 = 0.2582

3) DF = 28 - 1 = 27

P-value = P(T > 1.93)

= 1 - P(T < 1.93)

= 1 - 0.9679 = 0.0321

Since the p-value is less than the significance level (0.0321 < 0.05), so we should reject the null hypothesis.

Yes, We can conclude that the mean of all account balances is significantly greater than $1150.

4) = 52/328 = 0.1585

The test statistic z = ( - P)/sqrt(P(1 - P)/n)

= (0.1585 - 0.116)/sqrt(0.116(1 - 0.116)/328)

= 2.40