Question

In: Statistics and Probability

A) As part of a larger project to study the behavior of stressed-skin panels, a structural...

A)

As part of a larger project to study the behavior of stressed-skin panels, a structural component being used extensively in North America, an article reported on various mechanical properties of Scotch pine lumber specimens. Data on the modulus of elasticity (MPa) obtained 1 minute after loading in a certain configuration and 4 weeks after loading for the same lumber specimens is presented here.

Observation       1 min 4 weeks Difference
1             10,490 9,160   1330
2             16,620 13,250   3370
3             17,100 14,720   2380
4             15,480 12,750   2730
5             12,960 10,120   2840
6             17,260 14,570   2690
7             13,400 11,220   2180
8             13,900 11,100   2800
9             13,630 11,420   2210
10             13,240 10,930   2310
11             14,370 12,110   2260
12             11,700 8,620   3080
13             15,470 12,590   2880
14             17,840 15,090   2750
15             14,070 10,550   3520
16             14,760 12,230   2530

Calculate an upper confidence bound for the true average difference between 1-minute modulus and 4-week modulus; first check the plausibility of any necessary assumptions. (Use α = 0.05. Round your answer to two decimal places.)
MPa=_______

B)

It is thought that the front cover and the nature of the first question on mail surveys influence the response rate. An article tested this theory by experimenting with different cover designs. One cover was plain; the other used a picture of a skydiver. The researchers speculated that the return rate would be lower for the plain cover.

Cover     Number Sent     Number Returned
Plain 207 102
Skydiver 214 109

Does this data support the researchers' hypothesis? Test the relevant hypotheses using α = 0.10 by first calculating a P-value.

Compute the test statistic value and find the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z =
P-value =

C)

Do teachers find their work rewarding and satisfying? An article reports the results of a survey of 396 elementary school teachers and 261 high school teachers. Of the elementary school teachers, 224 said they were very satisfied with their jobs, whereas 125 of the high school teachers were very satisfied with their work. Estimate the difference between the proportion of all elementary school teachers who are satisfied and all high school teachers who are satisfied by calculating a 95% CI. (Use pelementaryphigh school. Round your answers to four decimal places.)

( , )

Solutions

Expert Solution

Ans A ) using minitab>stat>basic stat>Paired t

we have

Paired T-Test and CI: 1 min, 4 weeks

Paired T for 1 min - 4 weeks

N Mean StDev SE Mean
1 min 16 14518 2040 510
4 weeks 16 11902 1897 474
Difference 16 2616 522 130


95% upper bound for mean difference: 2844.91
T-Test of mean difference = 0 (vs < 0): T-Value = 20.06 P-Value = 1.000

an upper confidence bound for the true average difference between 1-minute modulus and 4-week modulus is 2844.91

Ans B ) using excel>addin>phstat>two sample test

we have

Z Test for Differences in Two Proportions
Data
Hypothesized Difference 0
Level of Significance 0.1
Group 1
Number of Items of Interest 102
Sample Size 207
Group 2
Number of Items of Interest 109
Sample Size 214
Intermediate Calculations
Group 1 Proportion 0.492753623
Group 2 Proportion 0.509345794
Difference in Two Proportions -0.01659217
Average Proportion 0.5012
Z Test Statistic -0.3404
Lower-Tail Test
Lower Critical Value -1.2816
p-Value 0.3668
Do not reject the null hypothesis
z =-0.34
P-value =0.3668

C) using excel>addin>phstat >2 sample test

\we have

Z Test for Differences in Two Proportions
Data Confidence Interval Estimate
Hypothesized Difference 0 of the Difference Between Two Proportions
Level of Significance 0.05
Group 1 Data
Number of Items of Interest 224 Confidence Level 95%
Sample Size 396
Group 2 Intermediate Calculations
Number of Items of Interest 125 Z Value -1.9600
Sample Size 261 Std. Error of the Diff. between two Proportions 0.0397
Interval Half Width 0.0778
Intermediate Calculations
Group 1 Proportion 0.565656566 Confidence Interval
Group 2 Proportion 0.478927203 Interval Lower Limit 0.0089
Difference in Two Proportions 0.086729363 Interval Upper Limit 0.1646
Average Proportion 0.5312
Z Test Statistic 2.1799
Two-Tail Test
Lower Critical Value -1.9600
Upper Critical Value 1.9600
p-Value 0.0293

95% confidence interval is ( 0.0089 ,0.1646)


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