Exercise 1: A random sample of 400 houscholds was drawn fromn a town and a survey generated
data on weekly earning. The mean in the sample was Birr 250 with a standard deviation Birr 80.
Test the hypothesis that the average weekly earnings is 280 birr at 5% level of significance and
also construct a 95% confidence interval for the population mean earning.
To test the hypothesis that the average weekly earnings is 280 Birr, we will use a t-test. The null hypothesis is that the average weekly earnings is 280 Birr, and the alternative hypothesis is that it is not 280 Birr.
Given:
Sample size (n) = 400
Sample mean (x̄) = 250 Birr
Sample standard deviation (s) = 80 Birr
Significance level (α) = 0.05
To test the hypothesis, we can calculate the t-statistic using the following formula:
t = (x̄ - μ) / (s / sqrt(n))
Where:
x̄ = sample mean
μ = population mean (280 Birr in this case)
s = sample standard deviation
n = sample size
t = (250 - 280) / (80 / sqrt(400))
t = -30 / (80 / 20)
t = -30 / 4
t = -7.5
Next, we need to find the critical value for a t-distribution with (n-1) degrees of freedom at a significance level of 0.05. Since the sample size is large (n > 30), we can approximate the critical value using a standard normal distribution.
The critical value for a 5% two-tailed test at α = 0.05 is approximately ± 1.96.
Since -7.5 is less than -1.96, we reject the null hypothesis. This means there is enough evidence to suggest that the average weekly earnings is not 280 Birr.
To construct a 95% confidence interval for the population mean earnings, we can use the following formula:
CI = x̄ ± (t * (s / sqrt(n)))
Where:
x̄ = sample mean
t = critical value for a t-distribution with (n-1) degrees of freedom at a 95% confidence level
s = sample standard deviation
n = sample size
CI = 250 ± (1.96 * (80 / sqrt(400)))
CI = 250 ± (1.96 * (80 / 20))
CI = 250 ± (1.96 * 4)
CI = 250 ± 7.84
The 95% confidence interval for the population mean earnings is (242.16, 257.84) Birr.