In: Statistics and Probability
2. Accounting procedures allow a business to evaluate their inventory costs based on two methods: LIFO (last in first out) or FIFO (first in first out). A manufacturer evaluated its finished goods inventory (in $000s) for five products with the LIFO and FIFO methods. To analyze the difference, they computed FIFO − LIFO for each product. We would like to determine if the LIFO method results in a lower cost of inventory than the FIFO method.
Product | FIFO (F) | LIFO (L) |
1 | 225 | 221 |
2 | 119 | 100 |
3 | 100 | 113 |
4 | 212 | 200 |
5 | 248 | 245 |
What is the alternate hypothesis?
Multiple Choice
H1: µd ≤ 0
H1: µd ≠ 0
H1: µd = 0
H1: µd > 0
3. A recent study focused on the number of times men and women send a Twitter message in a day. The information is summarized here.
Sample Size | Sample Mean | Population Standard Deviation | |
Men | 25 | 23 | 5 |
Women | 30 | 28 | 10 |
At the 0.01 significance level, is there a difference in the mean number of times men and women send a Twitter message in a day? Based on the p-value, what is your conclusion?
Multiple Choice
Fail to reject the null hypothesis.
Reject the null hypothesis and conclude the means are different.
Reject the null hypothesis and conclude the means are the same.
Fail to reject the null hypothesis and conclude the means are different.
6. Accounting procedures allow a business to evaluate its inventory costs based on two methods: LIFO (last in first out) or FIFO (first in first out). A manufacturer evaluated its finished goods inventory (in $000s) for five products with the LIFO and FIFO methods. To analyze the difference, they computed FIFO − LIFO for each product. We would like to determine if the LIFO method results in a lower cost of inventory than the FIFO method.
Product | FIFO (F) | LIFO (L) |
1 | 225 | 221 |
2 | 119 | 100 |
3 | 100 | 113 |
4 | 212 | 200 |
5 | 248 | 245 |
What is the value of calculated t?
Multiple Choice
+0.933
+0.47
−2.028
±2.776
7. Use the following table to determine whether or not there is a significant difference between the average hourly wages at two manufacturing companies.
Manufacture 1 | Manufacture 2 |
n1 = 81 | n2 = 64 |
x1=$15.80x1=$15.80 | x2=$15.00x2=$15.00 |
σ1 = $3.00 | σ2 = $2.25 |
The p-value is ______.
rev: 12_06_2019_QC_CS-192965
Multiple Choice
2.58
0.036
0.4664
0.0672
9. A recent study focused on the number of times men and women send a Twitter message in a day. The information is summarized here.
Sample Size | Sample Mean | Population Standard Deviation | |
Men | 25 | 23 | 5 |
Women | 30 | 28 | 10 |
At the 0.01 significance level, we ask if there is a difference in the mean number of times men and women send a Twitter message in a day. What is the value of the test statistic for this hypothesis test?
Multiple Choice
2.40
2.672
2.58
2.668
11. A company is researching the effectiveness of a new website design to decrease the time to access a website. Five website users were randomly selected, and their times (in seconds) to access the website with the old and new designs were recorded. To compare the times, they computed new website design time − old website design time. The results follow.
User | Old Website Design | New Website Design |
A | 30 | 25 |
B | 45 | 30 |
C | 25 | 20 |
D | 32 | 30 |
E | 28 | 27 |
For a 0.01 significance level, what is the critical value?
Multiple Choice
−1.895
−3.747
−2.447
−2.256
2)
H1: µd > 0
3)
std error σx1-x2=√(σ21/n1+σ22/n2) = | 2.082 | ||
test stat z =(x1-x2)/σx1-x2 = | -2.40 | ||
p value : = | 0.0164 |
since p value >0.01 ;
Fail to reject the null hypothesis.
6)
mean dbar= | d̅ = | 5.000 |
degree of freedom =n-1 = | 4.000 | |
Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) = | 11.979 | |
std error=Se=SD/√n= | 5.3572 | |
test statistic = | (d̅-μd)/Se = | +0.933 |
7)
std error σ1-2=√(σ21/n1+σ22/n2) = | 0.436 | ||
test stat z =(x1-x2-Δo)/σ1-2 = | 1.83 | ||
p value : = | 0.0672 |
9)
std error σx1-x2=√(σ21/n1+σ22/n2) = | 2.082 | ||
test stat z =(x1-x2-Δo)/σx1-x2 = | -2.40 |
11)
for 0.01 level with left tailed test and n-1= 4 df, critical value of t= | -3.747 |