Question

A billing company that collects bills for​ doctors' offices in the area is concerned that the percentage of bills being paid by medical insurance has risen.​ Historically, that percentage has been 30%. An examination of 7924 recent, randomly selected bills reveals that 31​% of these bills are being paid by medical insurance. Is there evidence of a change in the percent of bills being paid by medical​ insurance? (Consider a​ P-value of around 1​% to represent reasonable​ evidence.)

Question: What is the test statistic ?

_______________________________ (round to two decimal places if needed.)

Question: What is the P-Value ?

_______________________________ (round to three decimal places if needed.)

Answer 1

SOLUTION:

From given data,

A billing company that collects bills for​ doctors' offices in the area is concerned that the percentage of bills being paid by medical insurance has risen.​ Historically, that percentage has been 30%. An examination of 7924 recent, randomly selected bills reveals that 31​% of these bills are being paid by medical insurance. Is there evidence of a change in the percent of bills being paid by medical​ insurance?

= 30% = 30/100 = 0.30

Sample size = n = 7924

= 31% = 31/100 = 0.31

Test hypothesis:

Null hypothesis: : = 0.30

Alternative hypothesis: : 0.30

Question: What is the test statistic ?

Test statistic = Z = ( - ) / (sqrt( ((1-) ) / n ))

Z = (0.31 - 0.30 ) / (sqrt( (0.30(1-0.30) ) / 7924 ))

Z = 0.01 / (sqrt( (0.30(1-0.30) ) / 7924 )

Z = 1.94 (round to two decimal places if needed.)

Question: What is the P-Value ?

The test is two tailed , the p-value of the test statistic is:

P-Value = 2 * P(N(0,1) > Z)

P-Value = 2 * P(N(0,1) > 1.94)

P-Value = 2 * [1 - P(N(0,1) < 1.94) ]

P-Value = 2 * [1 - 0.97381]

P-Value = 2 * 0.02619

P-Value = 0.052 (round to three decimal places if needed.)