1) In a study of the cost per case of treating patients admitted with a disease

Question

1) In a study of the cost per case of treating patients admitted with a disease of the digestive system, the following data wrere obtained: $5,000, $5,500, $9,000, $8,000, $10,500, $11,000,$490 Use these data to construct a 98% confidence interval for true means cost per case. 2) Suppose that we selected 12 outstanding accounts and measured the time was required to convert each into cash after the first billing. The results in days are as follows: 120, 50, 40, 180, 230, 250, 30, 60, 70, 180, 270, 190 use these data to construct a 98% confidence interval for the true mean time required to transform accounts into cash after the first billing. Please show work.. I am a little confused. Thanks

Explanation / Answer

Q.1 For the given data

xbar = $ 7070 and standerd error (s) = $3693

SO we have to calculate the 98% confidence interval. Here we don't know population variance so we will use t- value.

Here t - value will be calculated from the t- table for degree of freedom dF = 7-1 = 6 and confidence interval 98%. t - value = +/- 3.1426

so Here mean of the population = mean = $ 7070

so confidence interval = +/- 3.1426 *s/n = 7070 +- 3.1426* 3693/7

= (2683.49, 11456.51)

2) For the given data

xbar =139.17 days and standerd error (s) = 87.75 days

SO we have to calculate the 98% confidence interval. Here we don't know population variance so we will use t- value.

Here t - value will be calculated from the t- table for degree of freedom dF = 12-1 = 11 and confidence interval 98%. t - value = +/- 2.718

so Here mean of the population = mean = 139.17 days

so confidence interval = +/- 2.718 *s/n = 139.17 +- 2.718* 87.75/12

= (70.32, 208.02)

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.