Question

Cholesterol levels are measured for 15 heart attack patients (two days after their attacks) and 27 other hospital patients who did not have a heart attack. The sample of heart attack patients had a mean cholesterol level of 233.7 and standard deviation 44.9. The sample of other hospital patients had a mean cholesterol level of 206.6 and standard deviation 14.8. The degrees of freedom for the t-distribution, in this case, is df=16.

The doctors leading the study think cholesterol levels will be higher for heart attack patients. Test the claim at the 0.05 level of significance. Use heart attack patients as "Population 1" and non-heart attack patients as "Population 2."

(a) What type of test is this?

(b) What is the test statistic?

(round your answer to three decimal places)

(c) What is the p-value?  

(round your answer to four decimal places)

(d) What is the statistical decision?

This means we  ---Select---   can cannot might always never conclude that the population mean cholesterol level of heart attack patients is higher than the population mean cholesterol level of other hospital patients.

Now create a 95% confidence interval for the difference between population mean cholesterol levels for heart attack patients and other hospital patients.

95% CI =  to

Answer 1

a) This is a t-distribution test for means of two independent samples.

This is a right-tailed test.

b) The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                                 = (233.7 - 206.6)/sqrt((44.9)^2/15 + (14.8)^2/27)

                                 = 2.270

c) P-value = P(T > 2.270)

                 = 1 - P(T < 2.270)

                 = 1 - 0.9813

                 = 0.0187

d) At alpha = 0.05, since the P-value is less than the significance level (0.0187 < 0.05), we should reject the null hypothesis.

We can might always conclude that the population mean cholesterol level of heart attack patients is higher than the population mean cholesterol level of other hospital patients.

At 95% confidence interval the critical value is t_{0.025, 16} = 2.120

The 95% confidence interval for difference in population mean is

() +/- t_{0.025, 16} * sqrt(s1^2/n1 + s2^2/n2)

= (233.7 - 206.6) +/- 2.120 * sqrt((44.9)^2/15 + (14.8)^2/27)

= 27.1 +/- 25.308

= 1.792, 52.408