Question

1.In a survey, 25 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $42 and standard deviation of $6. Construct a confidence interval at a 98% confidence level.

Give your answers to one decimal place.

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2. CNNBC recently reported that the mean annual cost of auto insurance is 980 dollars. Assume the standard deviation is 269 dollars. You take a simple random sample of 71 auto insurance policies.

Find the probability that a single randomly selected value is less than 962 dollars.
P(X < 962) =

Find the probability that a sample of size n=71n=71 is randomly selected with a mean less than 962 dollars.
P(X̅ < 962) =

Enter your answers as numbers accurate to 4 decimal places.

3. The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.63 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.2 inches and a standard deviation of 2.54 inches.

What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent.

Answer 1

TRADITIONAL METHOD
given that,
sample mean, x =42
standard deviation, s =6
sample size, n =25
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 6/ sqrt ( 25) )
= 1.2
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 2.492
margin of error = 2.492 * 1.2
= 2.99
III.
CI = x ± margin of error
confidence interval = [ 42 ± 2.99 ]
= [ 39.01 , 44.99 ]
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DIRECT METHOD
given that,
sample mean, x =42
standard deviation, s =6
sample size, n =25
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 2.492
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 42 ± t a/2 ( 6/ Sqrt ( 25) ]
= [ 42-(2.492 * 1.2) , 42+(2.492 * 1.2) ]
= [ 39.01 , 44.99 ]
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interpretations:
1) we are 98% sure that the interval [ 39.01 , 44.99 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean
Answer:
98% sure that the interval [ 39.01 , 44.99 ] =(39.0,44.9)$

2.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 980
standard Deviation ( sd )= 269/ Sqrt ( 71 ) =31.9244
sample size (n) = 71
the probability that a single randomly selected value is less than 962 dollars
P(X < 962) = (962-980)/269/ Sqrt ( 71 )
= -18/31.9244= -0.5638
= P ( Z <-0.5638) From Standard NOrmal Table
= 0.2864

3.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 64.2
standard Deviation ( sd )= 2.54
percentage of women are TALLER than 5 feet 11 inches(71 inches)

P(X > 71) = (71-64.2)/2.54
= 6.8/2.54 = 2.677
= P ( Z >2.677) From Standard Normal Table
= 0.004
=0.4%