1. The Carolina Tobacco company advertised that its best-selling nonlfitered cigarettes
contain at most 40 mg of nicotine, but Consumer Advocate magazine ran tests of
10 randomly selected cigarettes and found the amounts (in mg) shown below. It’s a
serious matter to charge that the company advertising is wrong, so the magazine editor
chooses a significance level of 0.01 in testing her belief that the mean nicotine content
is greater than 40 mg. Perform this test to determine if the mean is greater than 40
mg.
47.3 39.3 40.3 38.3 46.3 43.3 42.3 49.3 46.3 40.3
H0 :
Ha :
Test Statistic (value):
p−value:
Results:
2. Analysis of the last digits of sample data values sometimes reveals whether the data
have been accurately measured and reported. When single digits 0 through 9 are ran-
domly selected with replacement, the mean should be 4.50 and the standard deviation
should be 2.87. Reported data such as weights for heights are often rounded so that
the last digits include disproportionately more 0′s and 5′s. The last digits in the re-
ported lengths (in feet) of the 73 home runs hit by Barry Bonds in 2001 are reported
as follows: Mean, 1.753, standard deviation, 2.65. At the 0.05 significance level, does
it appear that the distances were accurately measured?
H0 :
Ha :
Test Statistic (value):
p−value:
Results:
3. In clinical tests of the drug Lipitor, 863 patients were treated with 10 mg doses, and
19 of those patients experienced
u symptoms (based on data from Parke-Davis). Use
a 0.01 significance level to test the claim that the percentage of treated patients with
u symptoms is greater than the 1.9% rate for patients not given treatments. Does it
appear that
u symptoms are an adverse reaction to the treatment?
H0 :
Ha :
Test Statistic (value):
p−value:
Results:
4. The percentage of physicians who are women is 27.9%. In a survey of physicians
employed by a large university health system, 45 of 120 randomly selected physicians
were women. Is there suffcient evidence at the 0.05 level of signifcance to conclude that
the proportion of women physicians at the university health system exceeds 27.9%?
H0 :
Ha :
Test Statistic (value):
p−value:
Results:
Chosen Answer:
1. ANSWER: A level of significance = 0.01 Conclusion: H0 is true. Not enough data to say greater than 40.
Why????
SINGLE SAMPLE TEST, ONE-TAILED, 6 – Step Procedure for t Distributions, “one-tailed test”
Step 1: Determine the hypothesis to be tested.
Lower-Tail
H0: μ ≥ μ0 H1: μ < μ0
or
Upper-Tail
H0: μ ≤ μ0 H1: μ > μ0
hypothesis test (lower or upper) = upper
Step 2: Determine a planning value for α [level of significance] =0.01
Step 3: From the sample data determine x-bar, s and n; then compute Standardized Test Statistic: t = (x-bar – μ0)/(s/SQRT(n))
x-bar: Estimate of the Population Mean (statistical mean of the sample) = 43.3
n: number of individuals in the sample = 10
s: sample standard deviation = 3.801
μ0: Population Mean = 40
significant digits =3
Standardized Test Statistic t = ( 43.3 – 40 )/( 3.801 / SQRT( 10 )) = 2.745
Step 4: Use Students t distribution, ‘lookup’ the area to the left of t (if lower-tail test) or to the right of t (if upper-tail test) using Students t distribution Table or Excel TDIST(x, n-1 degrees_freedom, 1 tail) =TDIST( 2.745 , 9 , 1 )
Step 5: Area in Step 4 is equal to P value [based on n -1 = 9 df (degrees of freedom)] = 0.011
Table look-up value shows area under the 9 df curve to the right of t = 2.745 is (approx) probability = 0.011
Step 6: For P ≥ α, fail to reject H0; and for P < α, reject H0 with 99% confidence.
Conclusion: H0 is true
Note: level of significance [α] is the maximum level of risk an experimenter is willing to take in making a “reject H0″ or “conclude H1″ conclusion (i.e. it is the maximum risk in making a Type I error).