Binomial Distribution - Null Hypothesis - Statistics Assessment Answers

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• Subject Name : Marketing Proposal

Binomial Distribution Assignment Sample

Answer

1. 
a) Binomial Distribution would be applicable here. The conditions for the application of Binomial Distribution are

(i) All trials are independent of each other (Satisfied)

(ii) The number of trials is fixed. In this case, the number of trials is 10. (Satisfied)

(ii) Each outcome has only 2 outcomes, either success or failure. The probability of success is 40 which is fixed in this case.

Hence the random variable that drug is able to produce a satisfactory result follows Binomial Distribution with parameter n10 and p40

b) P(Xx) nCx px q(n-x)

P 40.00

n 10

P(Xlt3) P(X0) P(X1) P(X2) P(X3)

P(Xlt3) BINOM.DIST(3,10,0.4,TRUE)

P(Xlt3) 0.3823

2). 
a)
The sample size taken is a hundred drugs. The sample size is large enough for the application of the central limit theorem. The central limit theorem states that when the sample size is large, the distribution is approximately normally distributed. Hence a z test can be applied.

For the application, of the test, it is required that

1). The population is normally distributed
2). A method of simple random sampling is used to obtain the sample
3). The sample size is large or the population standard deviation is known

a) 95 confidence interval gives me a range of values in which the estimated population mean is expected to lie. In this case, a 95 confidence interval is it indicates that I am 95 confident that the estimated population means the time of effect will lie in this interval.

b). 
Mean 207
Sd sqrt(var) 65
n 100
Alpha5
Critical value, z(a/2) z(0.05/2) 1.960
CI mean - z(a/2,n-1)(sd/sqrt(n))
Lower 207 - 1.96(65/sqrt(100)) 194.26
Upper 207 1.96(65/sqrt(100)) 219.74
I am 95 confident that the estimated population means a time of effect will lie in (194.26, 219.74) seconds.

3).
a)
Using the critical value approach
The null hypothesis, Ho the painkiller drug needs to have a time of the effect of at most 200 seconds. u lt200
The alternative hypothesis, the painkiller drug needs to have a time of the effect of greater than 200 seconds. u gt200
Mean 207.00
sd 65.000
u 200.00
n 100.00
Alpha 5
Test statistic, z (mean-u)/(sd/sqrt(n))
z (207-200)/(65/sqrt(100))
z 1.0769
Critical value z(a) z(0.05) 1.645
Decision rule the decision rule is to reject the null hypothesis if z gt z(a).
Conclusion Since zlt z(a), I fail to reject the null hypothesis and conclude that The painkiller drug needs to have a time of the effect of at most 200 seconds. u lt200
Using the P-value approach
The null hypothesis, Ho the painkiller drug needs to have a time of the effect of at most 200 seconds. u lt200
The alternative hypothesis, the painkiller drug needs to have a time of the effect of greater than 200 seconds. u gt200
Mean 207.00
sd 65.000
u 200.00
n 100.00
Alpha 5
Test statistic, z (mean-u)/(sd/sqrt(n))
z (207-200)/(65/sqrt(100))
z 1.0769
P-value 1-P(Zltz)
P-value 1-P(zlt1.0769)
P-value 1-NORMSDIST(1.0769)
P-value 0.1408
Decision rule reject the null hypothesis if P-value is less than Alpha, 5
Conclusion since the P-value is greater than Alpha, 5 the null hypothesis is not rejected at 5 level of significance. I can conclude that the painkiller drugs need to have a time of effect at most 200 seconds.

b)
I prefer the value of a0.15 and b0.15 instead of a0.05 and b0.45 when Sample size is smaller (power decreases with the decrease in sample size)
The two-tailed hypothesis is preferred (One-tailed tests generally have more power)
Effect size is small (the small the effect size, the less is the power)
I prefer the value of a0.05 and b0.45 instead of a0.15 and b0.15 when Sample size is larger (power increases with the increase in sample size)
The one-tailed hypothesis is preferred (One-tailed tests generally have more power)
Effect size is large (the larger the effect size, the greater is the power)

4).

The process of AB testing includes
Data collection
Assign 100 individuals randomly to each of the two taglines.
Goal identification
I want to test if tagline 2 has a better conversion rate (from a visitor on the website to a paid customer).
 

Hypothesis Generation

The null hypothesis, ho there is no significant difference in the proportion of individuals to pay for the product between tagline 1 and tagline 2. P1 p2
The alternative hypothesis, h1 the proportion of individuals to pay for the product of tagline 1 is less than that using tagline 2. P1 lt p2
 

Variation creation

The tagline 1 is 50 off for 1st customer
The variation is created by changes in tagline 2.
The tagline 2 is 50 Credits for Sign Up

Experiment Run

p (p1 n1 p2 n2) / (n1 n2)
SE sqrt p ( 1 - p ) (1/n1) (1/n2)
z (p1- p2) / SE

Result Analysis

If p-value lt 5, I reject ho and conclude that the proportion of individuals to pay for the product of tagline 1 is less than that using tagline 2. P1 lt p2