Question

A bank found that in recent​ years, the average monthly charge on its credit card was $1,350. With an improving​ economy, they suspect that this amount has increased. A sample of 42 customers resulted in an average monthly charge of ​$1,375.94 with a standard deviation of ​$183.78. Do these data provide statistical evidence that the average monthly charge has​ increased? Formulate the appropriate hypothesis test and draw a conclusion. Use a level of significance of 0.05.

Is there sufficient evidence at the 0.05 level of significance that the average monthly charge has​ increased?

Determine the null​ hypothesis, H0​, and the alternative​ hypothesis,H1.

H0​:

H1​:

​(Type whole​ numbers.)

Compute the test statistic__

​(Round to two decimal places as​ needed.)

Using a level of significance of​ 0.05, what is the critical​ value?___

​(Round to two decimal places as​ needed.)

Find the​ p-value for the test.___

​(Round to three decimal places as​ needed.)

What is your​ conclusion?

The​ p-value is ___ the chosen value of α​, so____the null hypothesis. There is_____ evidence to conclude that mean is greater than 1,350.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 1350
Alternative Hypothesis, Ha: μ > 1350

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (1375.94 - 1350)/(183.78/sqrt(42))
t = 0.915

Rejection Region
This is right tailed test, for α = 0.05 and df = 41
Critical value of t is 1.68
Hence reject H0 if t > 1.68

P-value Approach
P-value = 0.183

As P-value >= 0.05, fail to reject null hypothesis.

The​ p-value is greater that the chosen value of α​, so fail to reject the null hypothesis. There is not sufficient evidence to conclude that mean is greater than 1,350.