All answers must include: 1. the rationale of using a test, 2. the hypotheses 3.

Question

All answers must include: 1. the rationale of using a test, 2. the hypotheses 3. the critical-value, 4. the test-statistics 5. the p-value, 6. the test result, 7. the distribution chart, and 8. the conclusion rooms. Constrresults for the number of bacteria per cubic foot for both types of indicate that the data are approximately nor confidence interval and determine whether carpeted rooms have more rooms at the = 0.07 level of significance. Normal probability plots A researcher wanted to ca determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table s a 93% interval 67.3 13.6 13.8 15.4 12.2 11. 4 7.4 8.1 8.6 8.9 mal and boxplots indicate that there are no outliers. Carpeted Uncarpeted 11.7 13.5 One of the more popular statistics reported in the media is the president's job approval rating. The approval rating is reported as the proportion of the population who approve of the job that the sitting president is doing and is typically based on a random sample of registeredv poll showed the approval rating to be 45%. A second poll based on 2500 randomly selected voters showed that 1150 approved of the job the president was doing. Do the results of second p is significantly lower than the original level? Assume the 0.03 level of Write your answer on the backside of the paper It has long been stated that the mean temperature of humans is 98.6°F, researchers currently involved in the subject thought that the me different from 98.6°F. They for 3 days, obtaining 275 measurements with is 983 and s = 11 oll indicate that the proportion of voters who approve of the job the president is doing the significance. However, two an temperature of humans is measured the temperatures of 61 healthy adults 1 to 4 times daily °F. Use a hypothesis test to interval of the = 0.01 level of significance. Construct a 98% confidence judge their claim at human body temperature. Write your answer on the backside of the paper

Explanation / Answer

Last question::

we are given that

xbar = 98.3
mu = 98.6
n = 275
s = 11

so we calculate the z stat as

Z = (xbar - mu)/(s/sqrt(n))

= (98.3-98.6)/(11/sqrt(275)) = -0.4522

so we check the p value from the z table as
0.3255

as the p value is not less than 0.01 , hence we fail to reject the null hypothesis

we know that the confidence interval is given as

mean +-z* s/sqrt(n)
here z = 2.33 for 98% CI from the Z table

98.3 +- 2.33**11/sqrt(275)

96.75 lower bound
99.84 upper bound

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