Answer:
yes
Step-by-step explanation:
lets use a significance level of = 0.1
<u> Determine if the sequence indicates randomness </u>
First step :
H0 : pattern is random
H1 : pattern not random
n1 ( number of true answers ) = 10
n2 ( number of false answers ) = 10
also number of runs for T = 5
number of runs for F = 5
Total number of runs = 5+ 5 = 10
Given that critical value at 0.05 = 23
we will reject the null hypothesis ( i.e the sequence departs from randomness )
Answer:
As per dot plots we see the distribution of prices is close but majority of prices are concentrated in different zones. So MAD would be more similar by the look.
<u>Let's verify</u>
<h3>Neighborhood 1</h3>
<u>Data</u>
- 55, 55, 60, 60, 70, 80, 80, 80, 90, 120
<u>Mean</u>
- (55*2+ 60*2+ 70+ 80*3 + 90+ 120)/10 = 75
<u>MAD</u>
- (20*2+15*2+5+5*3+15+45)/10 = 15
<h3>Neighborhood 2</h3>
<u>Data</u>
- 100, 110, 110, 110, 120, 120, 120, 140, 150, 160
<u>Mean</u>
- (100 + 110*3+ 120*3+ 140 + 150+ 160)/10 = 124
<u>MAD</u>
- (24+14*3+4*3+16*3+16+26+36)/10 = 20.4
As we see the means are too different (75 vs 124) than MADs (15 vs 20.4).
Answer:
6
Step-by-step explanation:
3/4 / 6 = 3/4 x 1/6, which equals 3/24
and prodigy is a good game :D
Answer:
The 95% confidence interval for the percent of all black adults who would welcome a white person into their families is (0.8222, 0.8978).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
For this problem, we have that:
323 blacks, 86% of blacks said that they would welcome a white person into their families. This means that 
95% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the percent of all black adults who would welcome a white person into their families is (0.8222, 0.8978).
