certain commodity is known to have a price that is stable through time and doesn't change according to any known trend. The price however, changes from day to day in a random fashion. If the price is at a certain level one day, it is likely to be at any level the next day within some probability bounds approximately given by a normal distribution. The mean price is believed to be Tshs. 1425. To test the null hypothesis that the average price is Tshs. 1425, a random sample of 40 daily prices is located. The results are; mean=Tshs. 1650 and S=Tshs. 58. Using 5% level of significance, can you reject the null hypothesis?
To test the null hypothesis, we can use a one-sample t-test. The null hypothesis states that the average price is Tshs. 1425, while the alternative hypothesis states that the average price is different from Tshs. 1425.
Given:
Sample mean (x̄) = Tshs. 1650
Sample standard deviation (S) = Tshs. 58
Sample size (n) = 40
Population mean (μ) = Tshs. 1425
Significance level (α) = 0.05
First, we need to calculate the t-test statistic using the formula:
t = (x̄ - μ) / (S / √n)
t = (1650 - 1425) / (58 / √40)
t = 6.034
The degrees of freedom for the t-test is given by n - 1 = 40 - 1 = 39.
Next, we need to find the critical value for the given significance level and degrees of freedom. Since this is a two-tailed test, we need to split the significance level in half. Using a t-table or a t-distribution calculator, we find the critical value to be approximately ± 2.023.
Since |t| > critical value, we reject the null hypothesis.
Therefore, we can conclude that based on the given sample data, there is sufficient evidence to reject the null hypothesis and conclude that the average price is different from Tshs. 1425.