Question

In: Statistics and Probability

A sample of 55 account balances of a credit card company showed a sample mean balance...

A sample of 55 account balances of a credit card company showed a sample mean balance of $1,050. Assume that the population standard deviation is $175. At a 5% level of significance, test to determine if the population mean account balance is significantly different from $1,000. What is your conclusion?

Solutions

Expert Solution

Solution :

= 1000

= 1050

= 175

n = 55

This is the two tailed test .

The null and alternative hypothesis is ,

H0 :   = 1000

Ha :     1000

Test statistic = z

= ( - ) / / n

= (1050 - 1000) /175 / 55

= 2.119

P(z >2.119 ) = 1 - P(z <2.119 ) = 0.0341

P-value = 0.0341

= 0.05  

p = 0.0341 < 0.05, it is concluded that the null hypothesis is rejected.

There is enough evidence to claim that the population mean μ is different than 1000, at the 0.05 significance level.


Related Solutions

1. A sample of 30 account balances of a credit company showed an average balance of...
1. A sample of 30 account balances of a credit company showed an average balance of $1,187 and a standard deviation of $129. 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 60 account balances of a credit company showed an average balance of $1,165 and a standard deviation of...
A sample of 100 account balances of a credit company showed an average balance of $3,200...
A sample of 100 account balances of a credit company showed an average balance of $3,200 with a standard deviation of $316. Formulate the hypotheses that can be used to determine whether the mean of all account balances is significantly different from $3000. Conduct a full hypothesis test using the p-value approach. Let α = .05. #2 During the recent primary elections, the democratic presidential candidate showed the following pre-election voter support in Alabama and Mississippi. State Voters Surveyed Voters...
A sample of 64 account balances from a credit company showed an average daily balance of...
A sample of 64 account balances from a credit company showed an average daily balance of $1,040. The standard deviation of the population is known to be $200. We are interested in determining if the mean of all account balances (i.e., population mean) is significantly different from $1,000.What is the probability of making a of the Type II error if the mean is 945?
55. (same story as previous question) A sample of 64 account balances from a credit company...
55. (same story as previous question) A sample of 64 account balances from a credit company showed an average daily balance of $1145. The standard deviation of the population is known to be $200. We are interested in determining if the mean of all account balances (i.e., population mean) is significantly different from $1100. (In other words, we are testing the Null Hypothesis that the mean account balance for the population is $1100) Which of the following is true The...
In June 2012, the average Amercian credit card account balance was US$3,127. Assuming that these balances...
In June 2012, the average Amercian credit card account balance was US$3,127. Assuming that these balances are approximately normally distributed with a mean of US$3,127 and a standard deviation of US$1,500: a) P(that an account balance is less than $2,500). (Use diagram to show.) b) one quarter of credit card accounts carry at least what balance? (Use diagram to Show)
In a sample of credit card holders the mean monthly value of credit card purchases was...
In a sample of credit card holders the mean monthly value of credit card purchases was $ 298 and the sample variance was 53 ($ squared). Assume that the population distribution is normal. Answer the following, rounding your answers to two decimal places where appropriate. A.) Suppose the sample results were obtained from a random sample of 13 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders. B.)...
credit card payments: The outstanding balance on Bill’s credit card account is $3200. The bank issuing...
credit card payments: The outstanding balance on Bill’s credit card account is $3200. The bank issuing the credit card is charging 9.3%/year compounded monthly. If Bill de cides to pay off this balance in equal monthly installments at the end of each month for the next 18 months, how much will be his monthly payment? What is the effective rate of interest the bank is charging Bill?
Credit card balances are normally​ distributed, with a mean of​ $2870 and a standard deviation of​...
Credit card balances are normally​ distributed, with a mean of​ $2870 and a standard deviation of​ $900. You randomly selected 25 credit card holders. What is the probability that their mean credit card balance is less than​ $2500? Put your work in the SHOW YOUR WORK. nothing
A credit card company claims that the mean credit card debt for individuals is greater than...
A credit card company claims that the mean credit card debt for individuals is greater than $ 5 comma 300. You want to test this claim. You find that a random sample of 34 cardholders has a mean credit card balance of $ 5 comma 554 and a standard deviation of $ 650. At alpha equals 0.10?, can you support the? claim? Complete parts? (a) through? (e) below. Assume the population is normally distributed
Example 1: A credit card company claims that the mean credit card debt for individuals is...
Example 1: A credit card company claims that the mean credit card debt for individuals is greater than $5,300. You want to test this claim. You find that a random sample of 27 cardholders has a mean credit card balance of $5,560 and a standard deviation of $575. At α = 0.05​, can you support the​ claim? Assume the population is normally distributed. Write out the hypothesis statements below and identify the parameter of interest. Ho: _________________________         Ha: _________________________         Which...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT