The height of 10males of a given locality are found to be
70,67,62,68,61,68,70,64,64πππ66ππβππ . Is this reasonable to believe that average height is
greater than 64ππβππ at 5%level of significant?
To determine if it is reasonable to believe that the average height is greater than 64 inches at a 5% level of significance, we can perform a hypothesis test.
Null hypothesis (H0): The average height is equal to 64 inches.
Alternative hypothesis (Ha): The average height is greater than 64 inches.
To conduct the hypothesis test, we can use a one-sample t-test. The test statistic is calculated as:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
Using the given heights:
Mean height (xΜ) = (70 + 67 + 62 + 68 + 61 + 68 + 70 + 64 + 64 + 66) / 10 = 65 inches
Sample standard deviation (s) = sqrt(((70-65)^2 + (67-65)^2 + (62-65)^2 + (68-65)^2 + (61-65)^2 + (68-65)^2 + (70-65)^2 + (64-65)^2 + (64-65)^2 + (66-65)^2) / (10-1))
s = sqrt(200/9) = 4.71 inches
Sample size (n) = 10
The hypothesized mean is 64 inches.
t = (65 - 64) / (4.71 / sqrt(10))
t = 1 / (4.71 / sqrt(10))
t = 1 / 1.497
t β 0.668
To determine if this t-value is significant at a 5% level of significance, we compare it to the critical t-value for a one-sided test with 9 degrees of freedom (df = n - 1).
At a 5% level of significance, with 9 degrees of freedom, the critical t-value is approximately 1.833.
Since our calculated t-value (0.668) is not greater than the critical t-value (1.833), we fail to reject the null hypothesis.
Therefore, based on the data provided, there is not enough evidence to suggest that the average height is greater than 64 inches at a 5% level of significance.