According to a lending institution, students graduating from college have an ave

Question

According to a lending institution, students graduating from college have an average credit card debt of $4, 300. A random sample of 40 graduating seniors was selected, and their average credit card debt found to be $4, 615. Assume the standard deviation for student credit card debt is $1, 200. Using alpha = 0.10, complete parts a and b below. a. Does this sample provide enough evidence to challenge the findings by the lending institution? Determine the null and alternative hypotheses. The z-test statistic is (Round to two decimal places as needed.) The critical z-score(s) is(are) (Round to two decimal places as needed. Use a comma to separate answers as needed.) Because the test statistic the null hypothesis b. Determine the p-value for this test. The p-value is (Round to three decimal places as needed.)

Explanation / Answer

Here we have to test the hypothesis that,

H0 : mu = 4300          Vs               H1 : mu not= 4300

where mu is the population mean.

Alpha = significance level = 0.1

Given that,

Xbar = sample mean = 4615

n = sample size = 40

sigma= standard deviation = 1200

Now here population standard deviation is known so we use one sample z-test.

The test statistic is,

Z = (Xbar - mu) / (sigma/sqrt(n))

= (4615-4300) / (1200/sqrt(40))

= 1.66

Now we have to find critical value.

Critical value we can find using EXCEL.

syntax :

=NORMSINV(probability)

where probability = 1 - alpha/2

Critical values are -1.64 and 1.64 since the test is two sided.

P-value also we can find using EXCEL.

syntax :

=2*(1 - NORMSDIST(z))

where z is test statistic

P-value = 0.097

P-value < alpha (0.10)

Reject H0 at 0.1 significance level.

Conclusion : There is suffiicent evidence to say that population mean is differ from $4,300.

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