8-2 #28Original 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#6In 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 followingA. Is the test two tailed, left tailed, or right tailed,B. What is the test staticsC. What is the null hypothesis and what so you concludeD. what is the final conclusionE. 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-valueGuns(NNN) NNN-NNNN0.371884 (0.341974, 0.401795) 1.45 0.146
8-4 #20Listed 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#20Listed 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 residentsPennsylvania = 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#16Listed 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= 0.05Altitude = 3 10 14 22 28 31 33Temperature = 57 37 24 -5 -30 -41 -54
10-3 #16At 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 33Temperature = 57 37 24 -5 -30 -41 -54
11-2 #8Flat Tire and Missed ClassA 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 & Irrigation0.24 0.92 0.96 1.071.69 0.07 1.43 1.631.23 0.56 1.26 1.390.99 1.74 1.57 0.491.80 1.13 0.72 0.95
Tags problems various maths test tire claim level significance results flat trees students following mean poplar york variable treatment selected identify data pennsylvania right residents hypothesis state temperature recorded author value does reported select
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