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Statistics Problems

Question # 00009117
Subject: Statistics
Due on: 02/25/2014
Posted On: 02/25/2014 10:01 AM

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-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

Tags problems statistics test temperature claim pvalue level following residents altitude tailed pennsylvania conclusion listed recorded hypothesis variable value york results mean significance author conclude selected reported guns predicted sample randomly original 0

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Statistics Problems Solution with Working

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Posted On: 02/25/2014 10:03 AM
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Tutorial Preview …heights xx different xxxx 5 00 xx 8-3#6In a xxxxxx poll xx xxxx randomly xxxxxxxx subjects, 373 xxxx that they xxxx guns xx xxxxx homes xxx accompanying Minitab xxxxxxx shows results xxxx a xxxx xx the xxxxx that 35% xx homes have xxxx in xxxx xxxxxx a x 05 significance xxxxx and answer xxx following x xx the xxxx two tailed,…
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Preview: use xxx given xxxxxxxxxxx for the xxxxxxxxx A State x conclusion xxxxx xxx null xxxxxxxxxx (Reject Ho xx fail to xxxxxx Ho)Since xxx xxxxxxx is xxxxxx than the xxxxx of significance, xxx decision xxxxx xx to xxxx to reject xx B Without xxxxx technical xxxxxx xxxxx a xxxxx conclusion that xxxxxxxxx the original xxxxx There xx xxxxxxxxxxxx evidence xx conclude that xxx standard deviation xx women’s xxxxxxx xx different xxxx 5 00 xx 8-3#6In a xxxxxx poll xx xxxx randomly xxxxxxxx subjects, 373 xxxx that they xxxx guns xx xxxxx homes xxx accompanying Minitab xxxxxxx shows results xxxx a xxxx xx the xxxxx that 35% xx homes have xxxx in xxxx xxxxxx a x 05 significance xxxxx and answer xxx followingA xx xxx test xxx tailed, left xxxxxxx or right xxxxxxxxxx test xx xxxxxxxxxx B xxxx is the xxxx statisticThe value xx the xxxx xxxxxxxxx is x 45 C xxxx is the xxxx hypothesis xxx xxxx do xxx concludeThe null xxxxxxxxxx is Ho x p x x 35 xxx critical values xxx a z-test xx = x xx are xx 96 and xx 96 Since xxx test xxxxxxxxx xx between xxxxx two values, xxx decision is xx fail xx xxxxxx the xxxx hypothesis D xxxx is.....
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