A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. They randomly selected 850 residents in the community and contacted them by telephone. Of the 850 residents surveyed, 410 supported the property tax levy. Let p represent the proportion of residents in the community that support the property tax levy.
A 90% confidence interval for p is
A. 0.4489 to 0.5159.
B. 0.4542 to 0.5105.
C. 0.4487 to 0.5161.
D. 0.4463 to 0.5185.
also, if you could answer these two i would REALLY appreciate it..
2. In a large midwestern university (the class of entering freshmen is 6000 or more students), an SRS of 100 entering freshmen in 1999 found that 20 finished in the bottom third of their high school class. Admission standards at the university were tightened in 1995. In 2001, an SRS of 100 entering freshmen found that 10 finished in the bottom third of their high school class. Let p1 and p2 be the proportion of all entering freshmen in 1999 and 2001, respectively, who graduated in the bottom third of their high school class.
A 99% confidence interval for p1 – p2 is
A. 0.050 to 0.150.
B. 0.017 to 0.183.
C. 0.002 to 0.198.
D. –0.029 to 0.229.
3. In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study of national smoking trends, two random samples of U.S. adults were selected in different years: The first sample, taken in 1995, involved 1500 adults, of which 576 were smokers. The second sample, taken in 2000, involved 2000 adults, of which 652 were smokers. The samples are to be compared to determine whether the proportion of U.S. adults that smoke declined during the 5-year period between the samples.
Let p1 be the proportion of all U.S. adults that smoked in 1995. Let p2 denote the proportion of all U.S. adults that smoked in 2000.
The hypotheses to test in this problem are
A. H0 : p1 = p2, Ha : p1 < p2.
B. H0 : p1 = p2, Ha : p1 > p2.
C. H0 : p1 = p2, Ha : p1 ≠ p2.
D. H0 : p1 ≠ p2, Ha : p1 = p2.
A 90% confidence interval for p is
A. 0.4489 to 0.5159.
B. 0.4542 to 0.5105.
C. 0.4487 to 0.5161.
D. 0.4463 to 0.5185.
also, if you could answer these two i would REALLY appreciate it..
2. In a large midwestern university (the class of entering freshmen is 6000 or more students), an SRS of 100 entering freshmen in 1999 found that 20 finished in the bottom third of their high school class. Admission standards at the university were tightened in 1995. In 2001, an SRS of 100 entering freshmen found that 10 finished in the bottom third of their high school class. Let p1 and p2 be the proportion of all entering freshmen in 1999 and 2001, respectively, who graduated in the bottom third of their high school class.
A 99% confidence interval for p1 – p2 is
A. 0.050 to 0.150.
B. 0.017 to 0.183.
C. 0.002 to 0.198.
D. –0.029 to 0.229.
3. In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study of national smoking trends, two random samples of U.S. adults were selected in different years: The first sample, taken in 1995, involved 1500 adults, of which 576 were smokers. The second sample, taken in 2000, involved 2000 adults, of which 652 were smokers. The samples are to be compared to determine whether the proportion of U.S. adults that smoke declined during the 5-year period between the samples.
Let p1 be the proportion of all U.S. adults that smoked in 1995. Let p2 denote the proportion of all U.S. adults that smoked in 2000.
The hypotheses to test in this problem are
A. H0 : p1 = p2, Ha : p1 < p2.
B. H0 : p1 = p2, Ha : p1 > p2.
C. H0 : p1 = p2, Ha : p1 ≠ p2.
D. H0 : p1 ≠ p2, Ha : p1 = p2.
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1.CI for probability is p +/- z*sqrt(pq/n)
p is 410/850=.4823
q is 1-p=.5176
z for 90%=1.645
n=850
.4823 +/- 1.645sqrt(.4823*.5176/850)
(.4542,.5105) so its B
2.CI for x-y is (px-py)+zsqrt(sx^2/m+sy^2/n) where sx^2 is sample variance
variance is p(1-p)/n
px=.2 py=.1 px-py=.1
sx^2/m=.0016
sy^2/n=.0009
z=2.5758
.1+/-2.5758sqrt(.0016+.0009)
-.02879,.22879 so its D
3. H0 is p1=p2
HA is p2
p is 410/850=.4823
q is 1-p=.5176
z for 90%=1.645
n=850
.4823 +/- 1.645sqrt(.4823*.5176/850)
(.4542,.5105) so its B
2.CI for x-y is (px-py)+zsqrt(sx^2/m+sy^2/n) where sx^2 is sample variance
variance is p(1-p)/n
px=.2 py=.1 px-py=.1
sx^2/m=.0016
sy^2/n=.0009
z=2.5758
.1+/-2.5758sqrt(.0016+.0009)
-.02879,.22879 so its D
3. H0 is p1=p2
HA is p2
1
keywords: 90%,confidence,the,What,interval,is,for,What is the 90% confidence interval for p