Question

7. Use the following collection of 30 test scores: 100, 65, 67, 61, 62, 70, 75, 73, 88, 77, 83, 79, 95, 79, 80, 102, 86, 87, 87, 91, 87, 89, 92, 90, 99, 87, 72, 93, 79,52 a. Construct a frequency distribution table using 6 classes: include classes, frequencies, class marks, class boundaries, relative frequencies and relative percentages. b. Find the mean to the nearest tenth and sample standard deviation to the nearest tenth. c. Use the results from part b to determine the z-score associated with a data value of 79. 8. In a recent poll on education 64% of all college students stay up all night studying for final exams. A professor at a local college says this is too low and wants to test this claim. He surveys 5,000 college students and finds that 3250 have stayed up all night studying for final exams. Which statement can we conclude. Use a = 5% a. State the null and alternative hypothesis b. Determine and draw the hypothesis testing model (label everything) c. Determine the critical value(s) and test statistic d. Determine the conclusion

Answer 1

Ans :

a)

Classes	Frequency	Class Maker	Class Boundaries	relative frequency	relative percentage
50 - 59	1	54.5	49.5 - 59.5	1/30=0.03	0.03*100=3 5
60 - 69	4	64.5	59.5 - 69.5	4/30=0.13	13 %
70 - 79	8	74.5	69.5 - 79.5	8/30=0.27	27 5
80 -89	9	84.5	79.5 - 89.5	9/30 =0.3	30 %
90 - 99	6	94.5	89.5 - 99.5	6/30 =0.2	20 %
100 - 109	2	104.4	99.5 - 109.5	2/30 = 0.07	7 %
Total	30			1	100%

Class Maker = ( Lower limit + Upper limit )/2

B) To find mean and standard deviation

mean for grouped data is given by

Σ.fm Cf

X

Standard deviation for grouped data is given by :

$\large \sigma =\sqrt{\frac{\sum (f*m^2)-(n*mean^2)}{n-1}}$

Classes	Frequency	Class Maker (m)	f*m	f*m^2
50 - 59	1	54.5	54.5	2970.25
60 - 69	4	64.5	258	16641
70 - 79	8	74.5	596	44402
80 -89	9	84.5	760.5	64262.25
90 - 99	6	94.5	567	53581.5
100 - 109	2	104.4	208.8	21798.72
Total	30		2444.8	203655.7

mean = $\small \bar{X}$ = 2444.8 / 30

mean = $\small \bar{X}$ = 81.4933

Standard deviation :

$\large \sigma =\sqrt{\frac{203655.7-(30*81.4933^2)}{30-1}}$

$\large \sigma$ = 12.3469

c) z - score associate with data values of 79

$\large z-score = \frac{X-\mu }{\sigma }$

z=(79-81.4933)/12.3469

z =-0.20

Q. 8) Ans :

P = Proportion of certain type of item in population

n = sample size

X= number of items of certain type in sample of size n

p= X/n = proportion of certain type of item in sample

P0 = specified value P

from the given problem

sample size = 5000

X = 3250

then P= 3250/5000 = 0.65

p=0.65

p₀ = 0.64

a) Null and alternative hypothesis :

Null hypothesis

H₀ : P = P0

alternative hypothesis :

H₁: P < P0

To test :

H₀ : P = P0 versus  H₁: P < P0

X : Students stay up all night studying

X follows binomial distribution with parameter n = 5000 and p = 0.65

hence we can write

X $\large \rightarrow$ B(n,p)

E(x) =n*p = 5000*0.65

E(x) =3250

E(p) = E(X/n) = 1/n

var(X) = npq = 5000*0.65*(0.35)

var(x) = 11375.7

var(p) = 1/n^2 * var(x)

Using central limit theorem , we get limiting distribution of

$\large z = \frac{(p - E(p))}{\sqrt{var(p)}}$

here we test the population proportion equal to specified value (P0) :

Test Statistics :

$\large z = \frac{(p - P_0)}{\sqrt{pq/n}}$

$\large z= \frac{0.65-0.64}{\sqrt{\frac{0.64*0.36}{5000}}}$

z = 1.5

Critical region is Z <z$\small \alpha$ then we reject H0

$\small \alpha$ = 0.05

c) critical value = 1.64

d) Since z = 1.5 <  critical value = 1.64 Then we reject null hypothesis at 0.05 significance level

we conclude that the there is no sufficient condition to conclude that proportion of Students stay up all night