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# Maths Various Problems

Question # 00013271
Subject: Mathematics
Due on: 05/12/2014
Posted On: 04/23/2014 03:42 PM

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Question

8-2 #28
Original Claim: Women have heights with a standard deviation equal to 5.00cm. The hypothesis test results in a P-value of 0.0055. Assume a significance level of a=0.05 and use the given information for the following.
A. State a conclusion about the null hypothesis. (Reject Ho or fail to reject Ho)
B. Without using technical terms, state a final conclusion that addresses the original claim.

8-3#6
In a Gallup poll of 1003 randomly selected subjects, 373 said that they have guns in their homes. The accompanying Minitab display shows results from a test of the claim that 35% of homes have guns in them. Assume a 0.05 significance level and answer the following
A. Is the test two tailed, left tailed, or right tailed,
B. What is the test statics
C. What is the null hypothesis and what so you conclude
D. what is the final conclusion
E. What is the P-value
(note: in the box is the following information…
Test of P=0.35 vs p not=0.35

Variable X N Sample P 95% CI z-value p-value
Guns(NNN) NNN-NNNN0.371884 (0.341974, 0.401795) 1.45 0.146

8-4 #20
Listed below are the ages (years) of randomly selected race car drivers (based on data reported in USA today) Use a 0.05 significant level to test the claim that the mean age of all race car drivers is greater than 30 years.
32 32 33 33 41 29 38 32 33 23 27 45 52 29 25

9-3#20
Listed below are amounts of strontium-90 (in milibecquerels, or mBq,per gram of calcum) in a simple random sample of baby teeth obtained from Pennsylvania residents and New York residents born after 1979 (based on data from “An Unexpected Rise of Strontium 90 in U.S. Deciduous Teeth in the 1990’s” by Mangano, et al., Science of the Total Environment, Vol 317). Use a 0.05 significance level to test the claim that the mean amount of strontium-90 from Pennsylvania residents is greater than the mean amount from New York residents
Pennsylvania = 155 142 149 130 151 163 151 142 156 133 138 161
New York = 133 140 142 131 134 129 128 140 140 140 137 143

10-2#16
Listed below are altitudes (thousands of feet) and outside air temperatures(degrees Fahrenheit) recorded by the author during Delta Flight 1053 from New Orleans to Atlanta. Is there sufficient evidence to conclude that there is a linear correlation between altitude and outside air temperature? Do the results change if the altitudes are reported in meters and the temperatures are converted to the Celsius scale? Find the P-value or the critical value of r using a= 0.05
Altitude = 3 10 14 22 28 31 33
Temperature = 57 37 24 -5 -30 -41 -54

10-3 #16
At 6327 ft (or 6.327 thousand feet)the author recorded the temperature. Find the best predicted temperature at that altitude. How does the result compare to the actual recorded value of 48 degrees Fahrenheit. (Find the regression equation, letting the first variable be the predictor (x) variable . Find the indicated predicted value by following the prediction procedure summarized in the strategy for predicting values of Y.
Altitude = 3 10 14 22 28 31 33
Temperature = 57 37 24 -5 -30 -41 -54

11-2 #8
Flat Tire and Missed Class
A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn’t have a flat tire, would they be able to identify the same tire that went flat? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author’s claim that the results fit a uniform distribution. What does the result suggest about the ability of the four students to select the same tire when they really didn’t have a flat?
Left Front=11, Right Front=15, Left Rear= 8, Right Rear=6

12-2 #14
Poplar tree weights.
Weights (kg) of poplar trees were obtained from trees planted in a sandy and dry region. The trees were given different treatments identified in the table below. The data are from a study conducted by researchers at Pennsylvania State University and were provided by Minitab, Inc. Use a 0.05 significance level to test the claim that the four treatment categories yield poplar trees with the same mean weight. Is there a treatment that appears to be most effective in the sandy and dry region?

No treatment Fertilizer Irrigation Fertilizer & Irrigation
0.24 0.92 0.96 1.07
1.69 0.07 1.43 1.63
1.23 0.56 1.26 1.39
0.99 1.74 1.57 0.49
1.80 1.13 0.72 0.95

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#### Maths Various Problems

Tutorial # 00012831
Posted On: 04/23/2014 03:44 PM
Posted By:
vikas
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Tutorial Preview …the xxxx age xx all race xxx drivers is xxxxxxx than xx xxxxxxxxxxxx below xxx amounts of xxxxxxxxxxxx (in milibecquerels, xx mBq,per xxxx xx calcum) xx a simple xxxxxx sample of xxxx teeth xxxxxxxx xxxx Pennsylvania xxxxxxxxx and New xxxx residents born xxxxx 1979 xxxxxx xx data xxxx “An Unexpected xxxx of Strontium xx in x x Deciduous xxxxx in the xxxxxxxxxxx by Mangano, xx al x xxxxxxx of xxx Total Environment, xxx 317) Use x 0 xx xxxxxxxxxxxx level xx test the xxxxx that the xxxx amount xx xxxxxxxxxxxx from xxxxxxxxxxxx residents is xxxxxxx than the xxxx amount xxxx xxx York xxxxxxxxxxxxxxxxxxxxx = 155 xxx 149 130 xxx 163 xxx xxx 156 xxx 138 161 xxx York = xxx 140 xxx xxx 134 xxx 128 140 xxx 140 137 xxxxxxxxxxxxxxxxxx hypothesis:Alternative xxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxx p-value xxxxxxxxxxxxx to t x 3 2038, xxxx 11 xxxxxxx xx freedom,for x right-tailed test, xx 0 0042Decision:Since xxx p-value xx xxxx than xxx level of xxxxxxxxxxxxx the decision xx toreject xxx xxxx hypothesis xxxxxxxxxxxxx is sufficient xxxxxxxx at the x 05 xxxxx xx significance xx support theclaim xxxx the mean xxxxxx of xxxxxxxxxxxx xxxx Pennsylvania xxxxxxxxx is greater xxxx the mean xxxxxx from xxx xxxx residents xxxxxxxxxxxxx below are xxxxxxxxx (thousands of xxxxx and xxxxxxx xxx temperatures(degrees xxxxxxxxxxx recorded by xxx author during xxxxx Flight xxxx xxxx New xxxxxxx to Atlanta xx there sufficient xxxxxxxx to xxxxxxxx xxxx there xx a linear xxxxxxxxxxx between altitude xxx outside xxx xxxxxxxxxxxx Do xxx results change xx the altitudes xxx reported xx xxxxxx and xxx temperatures are xxxxxxxxx to the xxxxxxx scale? xxxx xxx P-value xx the critical xxxxx of r xxxxx a= x xxxxxxxxxx = x 10 14 xx 28 31 xxxxxxxxxxxxx = xx xx 24 xx -30 -41 xxxxxx following table xx…
Attachments
Maths_Various_Problems_A++_Solution.docx (501.2 KB)
Preview: hypothesis xxxxxxxxxxxxx is xxxxxxxxxx evidence to xxxxxxxx that there xx a xxxxxx xxxxxxxxxxx between xxxxxxxx and outside xxx temperature No, xxx results xx xxx change xx the altitudes xxx reported in xxxxxx and xxx xxxxxxxxxxxx are xxxxxxxxx to the xxxxxxx scale 10-3 xxxxx 6327 xx xxx 6 xxx thousand feet) xxx author recorded xxx temperature xxxx xxx best xxxxxxxxx temperature at xxxx altitude How xxxx the xxxxxx xxxxxxx to xxx actual recorded xxxxx of 48 xxxxxxx Fahrenheit xxxxx xxx regression xxxxxxxxx letting the xxxxx variable be xxx predictor xxx xxxxxxxx Find xxx indicated predicted xxxxx by following xxx prediction xxxxxxxxx xxxxxxxxxx in xxx strategy for xxxxxxxxxx values of x Altitude x x 10 xx 22 28 xx 33Temperature = xx 37 xx xx -30 xxx -54Using the xxxxx shown in xxx previous xxxxxxxx xxx coefficients xx the linear xxxxxxxxxx equation are:Slope:Intercept:The xxxxx squares xxxxxx xxxxxxxxxx equation xx then:The predicted xxxxxxxxxxx at an xxxxxxxx of xxxx xxxx is:This xxxxx differs from xxx actual temperature xx 48 xxx xx 1 xxxxx or 2 xx Section 11-2: x Flat xxxx xxx Missed xxxxxx A classic xxxxx involves four xxxxxxxxxx students xxx xxxxxx a xxxx and gave xx an excuse x flat xxxx xx the xxxxxx test, the xxxxxxxxxx asked the xxxxxxxx to xxxxxxxx xxx particular xxxx that went xxxx If they xxxxxx didn’t xxxx x flat xxxxx would they xx able to xxxxxxxx the xxxx xxxx that xxxx flat? The xxxxxx asked 41 xxxxx students xx xxxxxxxx the xxxx they would xxxxxx The results xxx listed xx xxx following xxxxx (except for xxx student who xxxxxxxx the xxxxxx xxx a x 05 significance xxxxx to test xxx author’s xxxxx xxxx the xxxxxxx fit a xxxxxxx distribution What xxxx the xxxxxx xxxxxxx about xxx ability of xxx.....
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