search

# RES 342 Week 2 E-Text Assignments

Homework

### Rating

Showing Page:
1/3
9.54) Faced with rising fax costs, a firm issued a guideline that transmissions of 10 pages or
more should be sent by 2-day mail instead. Exceptions are allowed, but they want the average to
be 10 or below. The firm examined 35 randomly chosen fax transmissions during the next year,
yielding a sample mean of 14.44 with a standard deviation of 4.45 pages.
5a) At the .01 level of significance, is the true mean greater than 10?
(b) Use Excel to find the right-tail p-value.
H0 mu =10
H1: mu>10
Value for one tail-test with a=0.01 and the value for df is 2.3263
Test statistic
T= (14.44) = (14.44-10) / [4.45/( sqrt35)] = 5.903
The p-value = P( Z> Z)
=P (Z> 5.903)
=1.787077e-09
Since the p-value is less than the significance level we reject the null hypothesis and conclude
the alternate hypothesis μ > 10 is true
9.56) A coin was flipped 60 times and came up heads 38 times. (a) At the .10 level of
significance, is the coin biased toward heads? Show your decision rule and calculations. (b)
Calculate a p-value and interpret it.
H0: p= 1/2
Ha: p > 1/2
Let's compute the test statistic:
z = (phat - p)/sqrt(p*(1-p)/N)
Plug in our values:
phat = 38/60
p = 0.5
N = 60 samples
z = (38/60 - 0.5)/sqrt(0.5*(1-0.5)/60)
z = 2.06559

At the 0.10 level, with one tail, the critical value is: 1.2816. Since our test statistic is higher than
that, we reject the null hypothesis, and conclude that the coin IS biased.
We can get the p value of z = 2.06559:
p = 0.019434
Our p value is lower than the 0.10 level, so again, we reject the null hypothesis and conclude that
the coin is biased.
9.62) The Web-based company Oh Baby! Gifts have a goal of processing 95 percent of its orders
on the same day they are received. If 485 out of the next 500 orders are processed on the same
day, would this prove that they are exceeding their goal, using a = .025?
H0 = 95% of orders filled
Ha = greater than 95% of orders filled
The critical value for a one tailed alpha 0.025 test is z = 1.96. We'll reject the null hypothesis if
our z test statistic is greater than that.
Test statistic z:
Phat = 485/500 = 0.97
z = (phat-p)/sqrt (p*(1-p)/N)
z = (0.97-0.95)/sqrt (0.95*0.05/500)
z = 2.0519
This is greater than the z value from the table for a one tailed test, with alpha = 0.025 (z = 1.96),
so we reject the null hypothesis. We can be reasonably sure that they are exceeding their goal.
9.64) An auditor reviewed 25 oral surgery insurance claims from a particular surgical office,
determining that the mean out-of-pocket patient billing above the reimbursed amount was
\$275.66 with a standard deviation of \$78.11. (a) At the 5 percent level of significance, does this
sample prove a violation of the guideline that the average patient should pay no more than \$250
out-of-pocket? State your hypotheses and decision rule. (b) Is this a close decision?
H0: mean = 250
Ha: mean > 250
We have to use a t test since our sample is small. df = 24, one sided, alpha = 0.05.

t = (x-mu)/ (sd/sqrt (N))
t = (275.66-250)/ (78.11/sqrt (25))
t = 1.6426
The critical t value is: 1.711
The t statistic is less than the critical value, so the null hypothesis is accepted and not enough
evidence to conclude there is a violation.
The p value (from the table) is 0.056753, and since this is just over 0.05, it is indeed a close
decision.

User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.
Review
Review

Anonymous
Goes above and beyond expectations!

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4