Statistics Assignment on Confidence Intervals and Z-Scores

This statistics assignment solution explores several key statistical concepts. The first question involves calculating the percentage of punts within a specific yardage range, determining a z-score, and finding the value corresponding to a percentile. The second question focuses on calculating a 95% confidence interval for a sample mean, given the population variance and sample size. The third question analyzes a binomial distribution, calculating a 90% confidence interval for the proportion of high school students taking at least one college course and comparing it to a given population percentage. The assignment demonstrates the application of statistical methods in analyzing data and drawing inferences.

STATISTICS
STUDENT NAME/ID
[Pick the date]

Question 1
Average = 45 yards
Standard deviation = 4 yards
(a) % of punts travel a distance between 50 yards and 55 yards
Hence, there are 9.94% punts that travel a distance between 50 yards and 55 yards.
(b) Let the punt is in 42th percentile then the ball travelled
1

(c) The z score of a punt that travel 53 yards
z score=?
x=53 yards
z= x−μ
σ
z= 53−45
4 = 8
4
z=2
2

Question 2
Sample size n = 196
Variance of population σ 2=8.41
Mean of sample x=10
(a) Standard deviation of the population
σ = √ Variance of population= √ σ2= √ 8.41=2.9
(b) Standard deviation of the mean of sample
σ x= Standard deviation of the population
√ Sample ¿ ¿= 2.9
√n = 2.9
√196 =0.207 ¿ ¿
(c) The given test would be two tailed considering that confidence interval is used which tends
to consider the lower and upper end of values in the form of an interval.
(d) The z value for 95%confidence interval = 1.96
Here, z= 1.96
0.207 =9.462
(e) The k value
Here, z= k
σ x
k =z∗σ x=1.96∗0.207=0.406
(f) 95% confidence interval for the sample mean of the population
(xbar – k) < μ < (xbar + k)
Lower limit of 95% confidence interval = x−k=10−0.406=9.594
Upper limit of 95% confidence interval = x +k =10+0.406 = 10.406
3

Hence, 95% confidence interval [9.594 10.406]
(g) Mean = 4
Standard deviation σ x=0.207
xbar = 10
Question 3
Sample size = 900 students
Number of students took at least 1 college course = 625
% of high school students in US take some college = 65%
4

Binomial distribution
x=n ^p
σ = √ n ^p ^q
(a) The value of n
Sample size or total number of trials (n) = 900
(b) The value of x
Number of students took at least 1 college course x= 625
(c) The value of p
Sample proportion (probability of success) p= 625/900 = 0.694
(d) The value of q
Probability of failure q=1− p=1−0.694=0.306
(e) The value of p (based on population)
p=0.65
(f) The value of σ
σ = √n ^p ^q= √900∗0.694∗0.306=13.82
(g) The value of σ p
σ p= σ
n = 13.82
900 =0.015356
(h) 90% confidence interval
z= k
σ p
5

In present case, population variance is not given and hence, two tailed z score would be taken
into consideration.
(i) The z value for 90% confidence interval
Here, level of significance = 0.10
The z score =1.645(from z table)
(j) The k value
z= k
σ p
k =z∗σ p=1.645∗0.015356=0.0253
(k) 90% confidence interval
( ^p – k) <ptrue< ( ^p + k)
Lower limit of 90% confidence interval ¿ 0.694−0.0253=0.6692
Upper limit of 90% confidence interval ¿ 0.694+ 0.0253=0.7197
90% confidence interval = [0.6692 0.7197]
(l) It can be seen from the above that 0.65 does not fall within the 90% confidence interval and
hence, it cannot be said that with 90% confidence that 65% of high school students have
taken at least 1 college class.
6