A random sample has 61 values. The sample mean is 9.9 and the sample standard deviation is 2. Use a level of significance of a=0.01 to conduct a right-tailed test of the claim that the population mean is 9.2 ( i.e, H_0: u= 9.2, H_1: u>9.2).
A)Compute the sample test statistic t.
B) Do we reject or fail to reject H_0? Explain you reasoning.
To compute the sample test statistic t, we can use the formula:
t = (sample mean - population mean) / (sample standard deviation / √n)
Where:
- Sample mean is 9.9
- Population mean is 9.2
- Sample standard deviation is 2
- n is the sample size, which is 61 in this case
A) Compute the sample test statistic t:
t = (9.9 - 9.2) / (2 / √61)
= 0.7 / (2 / 7.8102)
= 0.7 / 0.2556
= 2.7354 (rounded to four decimal places)
Therefore, the sample test statistic t is approximately 2.7354.
B) To determine whether to reject or fail to reject the null hypothesis (H₀), we need to compare the calculated t-value with the critical value.
Since we're conducting a right-tailed test (H₁: µ > 9.2) at a significance level (α) of 0.01, we need to find the critical value from the t-distribution table or use t-distribution calculator.
The critical value for a right-tailed test with α = 0.01 and degrees of freedom (df) = n-1 (61-1 = 60) is approximately 2.660. (This value is obtained by finding the t-value that leaves a 0.01 area in the right tail)
Since the calculated t-value (2.7354) is greater than the critical value (2.660), it falls in the rejection region. This means we reject the null hypothesis (H₀) and conclude that there is enough evidence to support the claim that the population mean is greater than 9.2 at a significance level of 0.01.
In other words, we have sufficient evidence to suggest that the sample mean of 9.9 is significantly different from the hypothesized population mean of 9.2 and is likely to be greater.
To conduct a right-tailed test of the claim that the population mean is 9.2, we can follow these steps:
A) Compute the sample test statistic t:
The formula for calculating the sample test statistic t is given by:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Given:
Sample mean (x̄) = 9.9
Population mean (μ) = 9.2
Sample standard deviation (s) = 2
Sample size (n) = 61
Plugging in the values into the formula, we get:
t = (9.9 - 9.2) / (2 / √61)
Calculating this expression, we find:
t ≈ 1.644
B) To determine if we reject or fail to reject H₀, we need to compare the calculated t-test statistic to the critical value from the t-distribution at a significance level of α=0.01.
Since this is a right-tailed test, we are interested in the critical value at the 0.99th percentile.
Using a statistical table or calculator, we find the critical value to be approximately 2.602.
Since the calculated value of t (1.644) is less than the critical value of t (2.602), we fail to reject the null hypothesis H₀.
Therefore, there is not enough evidence to support the claim that the population mean is greater than 9.2 at the given significance level of α=0.01.