## Questions and Answers

I gotten more confused in solving these problems. And these are the only ones I can't solve. Please help.

1. A certain family restaurant X is studying the length of time customers spend in their establishment. If the times are normally distributed, with a mean of 18 minutes and a standard deviation of 5.5 minutes:

a. What percent of the customers spend less than 30 minutes in the given restaurant?

B. What is the probability that of a customer spending exactly 30 min in the restaurant?

C. How would you estimate the probability of a customer spending 30 minutes in the restaurant.

2. Experienced over a long period of time has shown that, on average, a mother stayed 2.0 days in Hospital Y after childbirth. With a standard deviation of 0.5 days, hospital administrators and other groups decided to make a joint effort to reduce the average time a new mother spends in the hospital. Following their campaign, a sample file of 100 mothers revealed that the new mean length of stay is 1.8 days. Are mothers staying in the hospital less time, or could the difference between 2.0 and 1.8 be due to sampling error?

A. State the null and alternative hypothesis

b. At 0.05 level, state the decision rule

c. State the appropriate decision of the situation

3. Patients entering the hospital have complained that it takes 30min to fill out forms required for admittance. As a result of their complaints, the forms and procedure have been revised. A recent analysis of a random sample of 40 incoming patients reveals that the mean time to fill out the forms now is 28.5 min, with a standard deviation of 5 min. Is there sufficient evidence at the 0.02 level of signing=ficance to show that the new system is an improvement?

**Custer SD**

1. A) X ~ normal(18,5.5) You want. P(X<30). We don't know what the crap that is. But if get that into a STANDARD normal (Z) with mean 0 and sd 1, then we could look it up on a chart. Subtract the mean and divide by the sd on both sides. You get: P( (X – 18)/5.5) < (30-18)/5.5 ) = P( Z < 2.181)

See how the left side turns into a Z. Now, just use your chart.

B) There is zero probability that a customer would stay 30.00000000…..min The normal dist. Is a continuous dist. Any single point has no "weight" to it. However, P( cust stays between 30.000001 and 30.000002) is not zero.

C) P ( 29.5 < X < 30.5)? Since its not exact, I guess we are rounding. Solve like above. Subtract the mean, divide by sd to everything in sight. You get a Z in the middle.

2. A) the null is that there is mean1 = mean2 (no significant difference in stay times). Alternative is the opposite (mean1 not equal mean2).

B)

One-Sample Z

Test of mu = 2 vs not = 2

The assumed standard deviation = 0.5

N Mean SE Mean 95% CI Z P

100 1.80000 0.05000 (1.70200, 1.89800) -4.00 0.000

c) reject null, there is evidence that the stay time is less

3. We dont have the population variance. So, we use the T distribution.

One-Sample T

Test of mu = 30 vs not = 30

N Mean StDev SE Mean 95% CI T P

40 28.5000 5.0000 0.7906 (26.9009, 30.0991) -1.90 0.065

notice the p value. It is greater than .02, so we do not reject.