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Hypothesis Testing and Statistical Calculations

1. HO: µ = 70
H1: µ > 70
? = 20, n = 100, xbar = 80, ? = .01
a) calculate the value of the test statistic
b) set up the rejection region.
c) determine the p-value
d) interpret the results
attach file please

2. Draw the operating characteristic curve for n = 10, 50, and 100 for the following test:
Ho: ? = 400
Hi: ? > 400
? = .05, ? = 50
Attach file please

3. Determine the sample size necessary to estimate a population proportion to within .03 with 90% confidence assuming you have no knowledge of the approximate value of the sample proportion.

4. Some traffic experts believe that the major cause of highway collisions is the differing speeds of cars. That is, when some cars are driven slowly while others are driven at speeds well in excess of the speed limit, cars tend to congregate in bunches, increasing the probability of accidents. Thus, the greater the variation in speeds, the greater will be the number of collisions that occur. Suppose that one expert believes that when the variance exceeds 18 mph, the number of accidents will be unacceptably high. A random sample of the speeds of 245 cars on a highway with one of the highest accidents rates in the country is taken. Can we conclude at the 10% significance level that the variance in speeds exceeds 18 mph? Data is in file below.

5. How much time do executives spend each day reading and sending e-mail? A survey was conducted to obtain this information. The response (in minutes) in file below. Can we infer from these data that the mean amount of time spent reading and sending e-mail differs from 60 minutes each day? Data is in file below.

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1.
HO: µ = 70
H1: µ > 70
? = 20, n = 100, xbar = 80, ? = .01
a) calculate the value of the test statistic
b) set up the rejection region.
c) determine the p-value
d) interpret the results
attach file please
a)t=(80-70)/(20/sqrt(100))=5.
b)since this is one tailed t test, the critical value is TINV(0.02,99)=2.36. The rejection region is when t>2.36. (TINV is a function in Excel).
c)P value=TDIST(5,99,1)= 1.2407E-06
d) since P value is less than 0.01, we could reject the null hypothesis. Based on the test, we could conclude that µ>70.

2. Draw the operating characteristic curve for n = 10, 50, and 100 for the following test:
Ho: ? = 400
Hi: ? > 400
? = .05, ? = 50
Attach file please
when n=10, critical value is 1.83, (x-400)/(50/sqrt(10))=1.83. x=428.93.

When test value t>1.83, we could reject Ho and accept H1 ( shadowed area is the rejection ...

Solution Summary

The expert examines the hypothesis testing and statistical calculations.

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