To study the average price of living spaces in New York City, you collect data on the monthly rent of 30 houses in the city. You find a sample average rent cost of $565.63 with a standard deviation of $697.61. Use this information to construct a 99% confidence interval for the true mean rent cost in this area of New York City. Show all work. You do notneed to interpret the interval. (5 points)
11. It’s a fact that not everyone can roll their tongues (try it!). Many scientists claim that 73% (or 0.73 as a decimal) of people can roll their tongues, but suppose that we have reason to doubt this claim. To investigate, we collect data from 75 people and find that 60 of these people can roll their tongues.
A. What is the parameter of interest? Usecorrect notationin your answer, and describe the parameter in context using a short sentence. (2 points)
H0:
Ha:
C. Calculate the test statistic. Show all supporting work.(4 points)
Here, mean = 565.63, sigma = 697.61, n =30
At 99% confidence interval the z value is,
alpha = 1 - 0.99 = 0.01
alpha/2 = 0.01/2 = 0.0050
zalpha/2 = Z0.005 = 2.576
Margin of error = z *(sigma/sqrt(n))
= 2.576 *(697.61/sqrt(30))
= 328.0937
The 99% confidenc einterval for the mean is,
mean -E < mu < mean +E
565.63 - 328.0937 < mu < 565.63 + 328.0937
237.5363 < mu < 893.7237
11)
a)
Here, parameter of interest is 73% (or 0.73 as a decimal) of people
can roll their tongues
p = 0.73
The true proportion of a population with a certain
trait is denoted by p – this is a parameter
b)
H0 : p = 0.73
HA: p not equals to 0.73
c)
phat = 60/75 = 0.80
n = 75
z = ( phat - p)/sqrt(p*(1-p)/n)
= (0.80 - 0.73)/sqrt(0.73*(1-0.73)/75)
= 1.3655
d)
p value
P(z> 1.3655) = 2 *(1- P(x<1.3655) = 0.1721
p value = 0.1721
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