Probability and Statistics Lab 10

Hypothesis Small Sample

Question 1

The annual rainfall at a certain place is normally distributed with mean 30. If the rainfall during the past 8 years are 31.1, 30.7, 24.3, 28.1, 27.9, 32.2, 25.4 and 29.1, can we conclude that average rainfall during the past 8 years is less than the normal rainfall? Write R program for above problem.

$\mu =30$ $\alpha =0.05$

# Sample data
rainfall <- c(31.1, 30.7, 24.3, 28.1, 27.9, 32.2, 25.4, 29.1)
mu <- 30           # Population mean
alpha <- 0.05       # Level of significance
 
# t-test
t_test <- t.test(rainfall, mu = mu, alternative = "less")
 
# Display results
print(t_test)
 
# Decision
if (t_test$p.value < alpha) {
  print("Reject Null Hypothesis: The average rainfall is less than the normal rainfall.")
} else {
  print("Fail to Reject Null Hypothesis: The average rainfall is NOT significantly less than the normal rainfall.")
}
 

Question 2

Two random samples gave the following data:

Sample	Size $n$	Mean $\bar{X}$	Variance $s^2$
1	16	440	40
2	25	460	42
Can we conclude that the means of the two samples differ Significantly?

Solution

No significance given $\alpha =0.05$

Hypothesis:

• $H_0:{\mu }_1={\mu }_2$ → The means of the two samples are equal.
• $H_1:{\mu }_1\ne {\mu }_2$ → The means of the two samples differ significantly (two-tailed test).

# Given data
n1 <- 16
n2 <- 25
mean1 <- 440
mean2 <- 460
var1 <- 40
var2 <- 42
alpha <- 0.05
 
# Calculate the pooled variance
pooled_var <- ((n1-1)*var1 + (n2-1)*var2) / (n1 + n2 - 2)
 
# Calculate the t-statistic
t_stat <- (mean1 - mean2) / sqrt(pooled_var * (1/n1 + 1/n2))
 
# Calculate degrees of freedom
df <- n1 + n2 - 2
 
# Calculate critical t-value for two-tailed test
t_critical <- qt(1 - alpha/2, df)
 
# Calculate p-value
p_value <- 2 * pt(-abs(t_stat), df)
 
# Print results
cat("t-statistic:", t_stat, "\n")
cat("Degrees of freedom:", df, "\n")
cat("Critical t-value (two-tailed):", t_critical, "\n")
cat("p-value:", p_value, "\n")
 
# Decision
if(abs(t_stat) > t_critical) {
  cat("Reject H0: The means differ significantly.\n")
} else {
  cat("Fail to reject H0: No significant difference between means.\n")
}
 
# Alternative using t.test function
result <- t.test(x = NULL, y = NULL, 
                 alternative = "two.sided", 
                 mu = 0, paired = FALSE, 
                 var.equal = TRUE,
                 conf.level = 0.95,
                 data.frame(
                   x = rep(mean1, n1),
                   y = rep(mean2, n2),
                   var.x = rep(var1, n1),
                   var.y = rep(var2, n2)
                 ))
 
# Print t.test results
print(result)
 

Output:

[1] "Accept Null Hypothesis."
Reject H0: The means differ significantly.
t-statistic: -9.728757 
> cat("Degrees of freedom:", df, "\n")
Degrees of freedom: 39 
> cat("Critical t-value (two-tailed):", t_critical, "\n")
Critical t-value (two-tailed): 2.022691 
> cat("p-value:", p_value, "\n")
p-value: 5.542818e-12

Question 3

The following data relate to the marks obtained by 11 students in 2 tests, one held at the beginning of a year and the other at the end of the year after intensive coaching. Do the data indicate that the students have benefited by coaching? Write R program for above problem.

Given Data

Student	Test 1 (Before Coaching)	Test 2 (After Coaching)
1	55	63
2	60	70
3	65	70
4	75	81
5	49	54
6	25	29
7	18	21
8	30	38
9	35	32
10	54	50
11	61	70
12	72	80

Solution

Hypothesis:

• $H_0:{\mu }_d=0$ → No significant difference in scores before and after coaching.
• $H_1:{\mu }_d>0$→ Significant improvement in scores after coaching (right-tailed test).

Data

# Given data
test1 <- c(55, 60, 65, 75, 49, 25, 18, 30, 35, 54, 61, 72)  # Before coaching
test2 <- c(63, 70, 70, 81, 54, 29, 21, 38, 32, 50, 70, 80)  # After coaching
 
# Paired t-test (right-tailed)
result <- t.test(test2, test1, paired = TRUE, alternative = "greater", conf.level = 0.95)
 
print(result)

Output

Paired t-test
 
data:  test2 and test1
t = 3.8178, df = 11, p-value = 0.001427
alternative hypothesis: true mean difference is greater than 0
95 percent confidence interval:
 2.603905      Inf
sample estimates:
mean difference 
       4.916667

---

Question 4

In a test given to two groups of students, the marks obtained were as follows:

First Group:

$18,20,36,50,49,36,34,49,41$

Second Group:

$29,28,26,35,30,44,46$ Examine whether there is a significant difference between the means of marks secured by students of the above two groups.

Solution

Hypothesis:

• $H_0:{\mu }_1={\mu }_2$ → The means of the two groups are equal.
• $H_1:{\mu }_1\ne {\mu }_2$ → The means of the two groups differ significantly (two-tailed test).

R Program

# Data for the two groups
group1 <- c(18, 20, 36, 50, 49, 36, 34, 49, 41)
group2 <- c(29, 28, 26, 35, 30, 44, 46)
 
# Two-sample t-test (two-tailed)
result <- t.test(group1, group2, var.equal = TRUE, alternative = "two.sided")
 
# Display result
print(result)

Output:

Two Sample t-test
 
data:  group1 and group2
t = 0.57131, df = 14, p-value = 0.5768
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -8.262417 14.262417
sample estimates:
mean of x mean of y 
       37        34

References

Information

• date: 2025.03.25
• time: 13:09