Complete question is;
Use a t-test to test the claim about the population mean μ at the given level of significance α using the given sample statistics. Assume the population is normally distributed.
Claim: μ ≥ 8300,α= 0.10
Sample statistics: ¯x = 8000, s = 440, n = 24
a. What are the null and alternative hypotheses?
b. What is the value of the standardized test statistic? (Round to 2 decimal places as needed.)
c. What is the p-value? (Round to three decimal places as needed.)
d. Decide whether to reject or fail to reject the null hypothesis.
Answer:
A) Null hypothesis:H0: μ ≥ 8300
Alternative Hypothesis:Ha:μ < 8300
B) t = -3.34
C) p-value = 0.001
D) we will reject the null hypothesis
Step-by-step explanation:
A) We are told that the claim is: μ ≥ 8300. Thus, the null hypothesis would be the claim. So;
Null hypothesis:H0: μ ≥ 8300
Also, alternative hypothesis would be;
Alternative Hypothesis:Ha:μ < 8300
B)Formula for standardized test statistic with a t-test is;
t = (¯x - μ)/√(s/n)
Plugging in the relevant values, we have;
t = (8000 - 8300)/√(440/24)
t = -3.34
C) From online p-value from t-score calculator attached using t = -3.34, n = 24, significance level = 0.01, DF = 24 - 1 = 23 and a one - tailed test, we have;
p-value = 0.001421 ≈ 0.001
D) The p-value of 0.001 is less than the significance value of 0.01,thus we will reject the null hypothesis
