A credit bureau analysis of undergraduate students\' credit records found that t

Question

A credit bureau analysis of undergraduate students' credit records found that the average number of credit cards In an undergraduate's wallet was 4.06. It was also reported that in a random sample of 27 undergraduates, the sample mean number of credit cards that the students said they earned was 3.2, The sample standard deviation was not reported, but for purposes of this exercise, suppose that it was 1.1, Is there convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of 4.06? (Use a = 0.05. Round your test statistic to one decimal place and your p-value to three decimal places.) t = -1.4 p-value = State your conclusion. Do not reject H_0. we do not have convincing evidence that the mean number of credit cards carried by undergraduates is less than the credit bureau's figure of 4.06. Reject H_0. We have convincing evidence that the mean number of credit cards earned by undergraduates is less than the credit bureau's figure of 4.06. Reject H_0. We do not have convincing evidence that the mean number of credit cards carried by undergraduates is less than the credit bureau's figure of 4.06. Do not reject H_0. We have convincing evidence that the mean number of credit cards carried by undergraduates is less than the credit bureau's figure of 4.06. You may need to use the appropriate table in Appendix A to answer this question.

Explanation / Answer

Solution :-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: <= 4.06
Alternative hypothesis: > 4.06

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n) = 1.1 / sqrt(27) = 0.2116950987
DF = n - 1 = 27 - 1 = 26
t = (x - ) / SE = (3.2 - 4.06)/0.2116950987 = -4.062

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Here is the logic of the analysis: Given the alternative hypothesis ( > 4.06), we want to know whether the observed sample mean is large enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of -4.062. We use the t Distribution Calculator to find P(t < -4.062) = 0.000199.

Interpret results. Since the P-value (0.000199) is smaller than the significance level (0.05), we can reject the null hypothesis.

Conclusion. Reject H0, We have convincing evidence that the mean number number of credit cards carried by undergraduated is less than the credit bureau's figure of 4.06.

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