Lab-3

1.For a binomial distribution with n = 7 and p = 0.2 find (a) P(X = 5). (b) P(X > 2). (c) P(X < 8). (d) P(X ≥ 4). (e) Plot pmf and CDF.

n <- 7
p <- 0.2
# (a) P(X = 5)
dbinom(5, size = n, prob = p)
## [1] 0.0043008
# (b) P(X > 2)
sum(dbinom(3:n, size = n, prob = p))
## [1] 0.148032
# (c) P(X < 8)
sum(dbinom(0:7, size = n, prob = p))
## [1] 1
# (d) P(X ≥ 4)
sum(dbinom(4:n, size = n, prob = p))
## [1] 0.033344
# (e) Plot pmf and CDF
x <- 0:n
plot(x, dbinom(x, size = n, prob = p), type = "h", main = "Binomial pmf", xlab = "x", ylab = "P(X=x)")

plot(x, pbinom(x, size = n, prob = p), type = "s", main = "Binomial CDF", xlab = "x", ylab = "P(X≤x)")

  1. Given λ = 3, for a Poisson distribution, find (a) P(x ≤ 2). (b) P(x ≥ 5). (c) P(x = 8).(d) Plot pmf and CDF.
lambda <- 3
# (a) P(x ≤ 2)
ppois(2, lambda)
## [1] 0.4231901
# (b) P(x ≥ 5)
1 - ppois(4, lambda)
## [1] 0.1847368
# (c) P(x = 8)
dpois(8, lambda)
## [1] 0.008101512
# (d) Plot pmf and CDF
x <- 0:10
plot(x, dpois(x, lambda), type = "h", main = "Poisson pmf", xlab = "x", ylab = "P(X=x)")

plot(x, ppois(x, lambda), type = "s", main = "Poisson CDF", xlab = "x", ylab = "P(X≤x)")

  1. The probability that an evening college student will graduate is 0.4. Determine the probability that out of 5 students (a) none, (b) one, and (c) at least one will graduate.
p <- 0.4
n <- 5
# (a) none will graduate
dbinom(0, size=n , prob=p)
## [1] 0.07776
# (b) one will graduate
dbinom(1,size=n ,prob=p)
## [1] 0.2592
# (c) at least one will graduate
1-dbinom(0,size=n ,prob=p)
## [1] 0.92224
  1. One-fifth per cent of the blades produced by a blade manufacturing factory turn out to be defective. The blades are supplied in packets of 10. Use Poisson distribution to calculate the approximate number of packets containing no defective, one defective, and two defective blades respectively in a consignment of 1,00,000 packets.
p <- 0.001/5 # probability of defective blade
n <- 10 # blades per packet
lambda <- n * p # parameter for Poisson distribution
N <- 100000 # number of packets in consignment

# number of packets containing no defective blades
dpois(0, lambda) * N
## [1] 99800.2
# number of packets containing one defective blade
dpois(1, lambda) * N
## [1] 199.6004
# number of packets containing two defective blades
dpois(2, lambda) * N
## [1] 0.1996004
  1. In an intelligence test administered to 1000 students, the average score was 42 and standard deviation 24. Find (a) the number of students exceeding a score of 50, (b) the number of students lying between 30 and 54, (c) the value of the score exceeded by the top 100 students.
n <- 1000 # number of students
mu <- 42 # mean score
sigma <- 24 # standard deviation

# (a) the number of students exceeding a score of 50,
pnorm(50,mu,sigma ,lower.tail=FALSE)*n
## [1] 369.4413
# (b) the number of students lying between 30 and 54,
(pnorm(54,mu,sigma)-pnorm(30,mu,sigma))*n
## [1] 382.9249
# (c) the value of the score exceeded by the top 100 students.
qnorm((n-100)/n,mu,sigma)
## [1] 72.75724

Lab-4

1.A random sample of 10 boys had the following IQ.’ s : 70, 120, 110, 101, 88, 83, 95, 98, 107, 100. Do these data support the assumption of a population mean I.Q. of 100 ? Find a reasonable range in which most of the mean I.Q. values of samples of 10 boys lie.

iq <- c(70, 120, 110, 101, 88, 83, 95, 98, 107, 100)
t.test(iq, mu = 100)
## 
##  One Sample t-test
## 
## data:  iq
## t = -0.62034, df = 9, p-value = 0.5504
## alternative hypothesis: true mean is not equal to 100
## 95 percent confidence interval:
##   86.98934 107.41066
## sample estimates:
## mean of x 
##      97.2
# To find a reasonable range for the mean IQ values of samples of 10 boys
mean(iq) + c(-1,1) * qt(0.975, df = length(iq)-1) * sd(iq)/sqrt(length(iq))
## [1]  86.98934 107.41066
  1. In a survey of buying habits. 400 women shoppers are chosen at random in super market ‘A’ located in a certain section of the city. Their average weekly food expenditure is Rs. 250 with a standard deviation of Rs. 40. For 400 women shoppers chosen at random in super market ‘B’ in another section of the city. the average weekly food expenditure is Rs. 220 with a standard deviation of Rs. 55. Test at 1% level of significance whether the average weekly food expenditure of the two populations of shoppers are equal.
# create variables for sample 1 (market A)
n1 <- 400
xbar1 <- 250
s1 <- 40

# create variables for sample 2 (market B)
n2 <- 400
xbar2 <- 220
s2 <- 55

# perform the two-sample t-test
t_test <- t.test(x = c(xbar1, xbar2), n = c(n1, n2), sd = c(s1, s2), alternative = "two.sided", conf.level = 0.99)
t_test
## 
##  One Sample t-test
## 
## data:  c(xbar1, xbar2)
## t = 15.667, df = 1, p-value = 0.04058
## alternative hypothesis: true mean is not equal to 0
## 99 percent confidence interval:
##  -719.8511 1189.8511
## sample estimates:
## mean of x 
##       235
# extract the p-value
p_value <- t_test$p.value

# check whether the null hypothesis is rejected
if (p_value < 0.01) {
  cat("The average weekly food expenditure of the two populations of shoppers are significantly different at the 1% level of significance.")
} else {
  cat("There is no significant difference in the average weekly food expenditure of the two populations of shoppers at the 1% level of significance.")
}
## There is no significant difference in the average weekly food expenditure of the two populations of shoppers at the 1% level of significance.
  1. A sample of 900 members has a mean 3·4 cms, and s.d. 2·61 cms. Is the sample from a large population of mean 3·25cms. and s.d. 2·61cms ? If the populations is normal and its mean is unknown, Find the 95% and 98’% confidence limits of true mean.
sample_mean <- 3.4
pop_mean <- 3.25
sd <- 2.61
n <- 900

t_value <- (sample_mean - pop_mean) / (sd / sqrt(n))
p_value <- 2 * pt(-abs(t_value), df = n-1)

xbar <- 3.4
s <- 2.61
n <- 900

# 95% confidence interval
alpha <- 0.05
t_critical <- qt(alpha/2, df = n-1, lower.tail = FALSE)
margin_of_error <- t_critical * (s / sqrt(n))
lower_bound <- xbar - margin_of_error
upper_bound <- xbar + margin_of_error
cat("95% confidence interval: (", lower_bound, ", ", upper_bound, ")\n", sep="")
## 95% confidence interval: (3.229253, 3.570747)
# 98% confidence interval
alpha <- 0.02
t_critical <- qt(alpha/2, df = n-1, lower.tail = FALSE)
margin_of_error <- t_critical * (s / sqrt(n))
lower_bound <- xbar - margin_of_error
upper_bound <- xbar + margin_of_error
cat("98% confidence interval: (", lower_bound, ", ", upper_bound, ")\n", sep="")
## 98% confidence interval: (3.197246, 3.602754)
  1. Experience has shown’ that 20% of a manufactured product is of the top quality. In one day’s production of 400 articles only 50 are of top quality. Show that either the production of the day taken was not a representative sample or the hypothesis of 20% was wrong.
p0 <- 0.2
n <- 400
x <- 50

binom.test(x,n,p0)
## 
##  Exact binomial test
## 
## data:  x and n
## number of successes = 50, number of trials = 400, p-value = 9.994e-05
## alternative hypothesis: true probability of success is not equal to 0.2
## 95 percent confidence interval:
##  0.09421851 0.16145866
## sample estimates:
## probability of success 
##                  0.125
  1. In a random sample of 100 men taken from village A, 60 were found to be consuming alcohol. In another sample of 200 men taken form village B, 100 were found to be consuming alcohol. Do the two villages differ significantly in respect of the proportion of men who consume alcohol?
n1 <-100
x1 <-60

n2<-200
x2<-100

prop.test(x=c(x1,x2),n=c(n1,n2))
## 
##  2-sample test for equality of proportions with continuity correction
## 
## data:  c(x1, x2) out of c(n1, n2)
## X-squared = 2.2919, df = 1, p-value = 0.1301
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  -0.02591167  0.22591167
## sample estimates:
## prop 1 prop 2 
##    0.6    0.5

Lab-5

  1. In an experiment to study the dependence of hypertension on smoking habits, the following data were taken on 180 individuals: Non- smokers Moderate Smokers Heavy Smokers Hypertension 21 36 30 No hypertension 48 26 19 Test the hypothesis that the presence or absence of hypertension is independent of smoking habits. 10 5 Use a 0.05 level of significance
# Input data
smoking <- matrix(c(21, 36, 30, 48, 26, 19), nrow = 2, byrow = TRUE)
colnames(smoking) <- c("Non-smokers", "Moderate Smokers", "Heavy Smokers")
rownames(smoking) <- c("Hypertension", "No Hypertension")

# Perform chi-squared test
chisq.test(smoking)
## 
##  Pearson's Chi-squared test
## 
## data:  smoking
## X-squared = 14.464, df = 2, p-value = 0.0007232
  1. Two independent groups of IO children were tested to find how many digits they could repeat from memory after hearing them. The results are as follows: . Group A 8 6 5 7 6 8 7 4 5 6 Group B 1 0 6 7 8 6 9 7 6. 7 7 Is the difference between the mean scores of the two groups significant ?
# Input data
groupA <- c(8, 6, 5, 7, 6, 8, 7, 4, 5, 6)
groupB <- c(10, 6, 7, 8, 6, 9, 7, 6.7)

# Perform two-sample t-test
t.test(groupA, groupB)
## 
##  Welch Two Sample t-test
## 
## data:  groupA and groupB
## t = -1.9251, df = 14.501, p-value = 0.07406
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -2.6645423  0.1395423
## sample estimates:
## mean of x mean of y 
##    6.2000    7.4625
  1. Two random samples drawn from two normal populations are : Sample 1 : 20, 16, 26, 27, 23. 22, 18, 24, 25, 19 Sample 2 : 27, 33, 42, 35, 32, 34, 38, 28, 41, 43, 30, 37 Obtain estimates of the variances of the populations and test whether the populations have same variances.
# Input data
sample1 <- c(20, 16, 26, 27, 23.22, 18, 24, 25,19)
sample2 <- c(27,33 ,42 ,35 ,32 ,34 ,38 ,28 ,41 ,43 ,30 ,37)

# Perform F-test
var.test(sample1,sample2)
## 
##  F test to compare two variances
## 
## data:  sample1 and sample2
## F = 0.52759, num df = 8, denom df = 11, p-value = 0.3735
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.1440008 2.2387943
## sample estimates:
## ratio of variances 
##          0.5275929

Lab-6

  1. A research company has designed three different systems to clean up oil spills. The following table contains the results, measured by how much surface area (in square meters) is cleared in 1 hour. The data were found by testing each method in several trials. Are the three systems equally effective at 5% LOS? System A 55 60 63 56 59 55 System B 57 53 64 49 62 System C 66 52 61 57
# Input data
system_a <- c(55, 60, 63, 56, 59, 55)
system_b <- c(57, 53, 64, 49, 62)
system_c <- c(66, 52, 61, 57)

# Combine data into a data frame
data <- data.frame(
    value = c(system_a, system_b, system_c),
    system = factor(rep(c("A", "B", "C"), times = c(length(system_a), length(system_b), length(system_c))))
)

# Perform ANOVA test
result <- aov(value ~ system, data = data)

# Print summary of ANOVA test
summary(result)
##             Df Sum Sq Mean Sq F value Pr(>F)
## system       2   8.93   4.467   0.172  0.844
## Residuals   12 312.00  26.000
# Check if p-value is less than 0.05
if (summary(result)[[1]][["Pr(>F)"]][1] < 0.05) {
    cat("The three systems are not equally effective at a 5% level of significance.\n")
} else {
    cat("The three systems are equally effective at a 5% level of significance.\n")
}
## The three systems are equally effective at a 5% level of significance.
  1. The following table gives the gains in weights of 4 different types of pigs fed on three different rations over a period. Test whether (i) the difference in the rations significant (ii) the four types of pig differ significantly in gaining weight.

Ration I II III IV A 13.8 15.7 16.0 20.2 B 8.7 11.8 9.0 12.9 C 12.0 16.5 13.3 12.5

# create the data frame
pigs <- data.frame(
  ration = rep(c("A", "B", "C"), each = 4),
  type = rep(c("I", "II", "III", "IV"), times = 3),
  weight_gain = c(13.8, 15.7, 16.0, 20.2, 8.7, 11.8, 9.0, 12.9, 12.0, 16.5, 13.3, 12.5)
)

# check the data
pigs
# test whether the difference in the rations is significant
anova_lm <- aov(weight_gain ~ ration, data = pigs)
summary(anova_lm)
##             Df Sum Sq Mean Sq F value Pr(>F)  
## ration       2  67.87   33.94   6.496 0.0179 *
## Residuals    9  47.01    5.22                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# test whether the four types of pigs differ significantly in gaining weight
anova_lm2 <- aov(weight_gain ~ type, data = pigs)
summary(anova_lm2)
##             Df Sum Sq Mean Sq F value Pr(>F)
## type         3  26.35   8.784   0.794  0.531
## Residuals    8  88.53  11.067

3.Analyse the variance in the following Latin square:

20 B 17 C 25 D 34 A 23 A 21 D 15 C 24 B 24 D 26 A 21 B 19 C 26 C 23 B 27 A 22 D

freq=c(20,17,25,34,23,21,15,24,24,26,21,19,26,23,27,22)
col=c(rep("col1",1),rep("col2",1),rep("col3",1),rep("col4",1))
row=c(rep("rowA",4),rep("rowB",4),rep("rowC",4),rep("rowD",4))
seed=c("B","C","D","A","A","D","C","B","D","A","B","C","C","B","A","D")
mydata=data.frame(row,col,seed,freq)
mydata
myfit=anova(lm(freq~row+col+seed,mydata))
print(myfit)
## Analysis of Variance Table
## 
## Response: freq
##           Df  Sum Sq Mean Sq F value Pr(>F)
## row        3  34.187  11.396  0.7058 0.5826
## col        3  22.688   7.563  0.4684 0.7151
## seed       3 141.188  47.063  2.9148 0.1227
## Residuals  6  96.875  16.146