A bank wonders whether omitting the annual credit card fee for customers who charge ar least $3000 in a year will increase the amount charged on its cards. The bank makes this offer to an SRS of 400 of its customers. It then compares how much these customers charge this year with the amount they charged last year. The mean increase in the sample $246 and the standard deviation is $112. Is there significant evidence at the 1% level that the mean amount charged increases under the no-fee offer? State Ho and Ha and carry out a significance test. Use significance level 0.01
Z = (mean1 - mean2)/standard error (SE) of difference between means = 246/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Compare to .01.
Ho: The mean amount charged does not increase under the no-fee offer (μ = 0)
Ha: The mean amount charged increases under the no-fee offer (μ > 0)
To carry out the significance test, we need to calculate the test statistic and compare it to the critical value.
Step 1: Calculate the test statistic:
Z = (sample mean - hypothesized mean) / (standard deviation / √sample size)
= (246 - 0) / (112 / √400)
= (246 - 0) / (112 / 20)
= 246 / 5.6
≈ 43.93
Step 2: Find the critical value.
Since the significance level is 0.01, the critical value for a one-tailed test with a 0.01 level of significance is 2.33 (from the z-table or calculator).
Step 3: Compare the test statistic to the critical value.
Since the test statistic (43.93) is greater than the critical value (2.33), we reject the null hypothesis.
Conclusion: There is significant evidence at the 1% level that the mean amount charged increases under the no-fee offer.
To conduct a significance test, we first need to establish the null hypothesis (Ho) and the alternative hypothesis (Ha).
Ho: The mean amount charged does not increase under the no-fee offer.
Ha: The mean amount charged increases under the no-fee offer.
Next, we need to calculate the test statistic. In this case, we will use a t-test since we have a sample mean and want to test if it is significantly different from a population mean.
To calculate the test statistic, we will use the formula:
t = (sample mean - population mean) / (sample standard deviation / √n)
Given:
Sample mean (x̄) = $246
Population mean (μ) = 0 (since we are testing if there is an increase)
Sample standard deviation (s) = $112
Sample size (n) = 400
Now, let's plug in the values:
t = ($246 - 0) / ($112 / √400)
t = $246 / ($112 / 20)
t = $246 / $5.6
t ≈ 43.93
To determine if the test statistic is significant at the 1% level, we need to compare it with the critical value for a t-distribution with (n-1) degrees of freedom. In this case, we have 399 degrees of freedom (400-1).
Looking up the critical value at a 0.01 significance level and 399 degrees of freedom in a t-distribution table, we find it to be approximately 2.617.
Since the test statistic (43.93) is much larger than the critical value (2.617), we reject the null hypothesis Ho.
Therefore, the evidence suggests that there is significant evidence at the 1% level that the mean amount charged increases under the no-fee offer.