Complete question is;
Use technology and 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 > 71; α = 0.05
Sample statistics:
¯x = 73.9, s = 3.7, n = 25
A) What are the null and alternative hypotheses?
Choose the correct answer below.
A. H0:μ = 71 ; HA:μ ≠ 71
B. H0:μ ≤ 71; HA:μ > 71
C. H0: μ ≥ 71; HA: μ < 71
D. H0: μ ≠ 71; HA: μ = 71
B) What is the value of the standardized test statistic? The standardized test statistic is (Round to two decimal places as needed.)
C) What is the P-value of the test statistic? P-value = Round to three decimal places as needed.)
D) Decide whether to reject or fail to reject the null hypothesis.
Answer:
A) Option B - Alternative hypothesis: HA: μ > 71
Null hypothesis: H0: μ ≤ 71
B) t = -7.54
C) p-value = 0.000
D) we reject the null hypothesis
Explanation:
A) We are told that the claim is: μ > 71. Thus, due to the sign, the alternative hypothesis would be the claim. So;
Alternative hypothesis: HA: μ > 71
Null hypothesis: H0: μ ≤ 71
B)Formula for standardized test statistic with a t-test is;
t = (¯x - μ)/√(s/n)
Plugging in the relevant values, we have;
t = (71 - 73.9)/√(3.7/25)
t = -7.54
C) From online p-value from t-score calculator attached using t = -7.54, n = 25, significance level = 0.05, DF = 25 - 1 = 24 and a one - tailed test, we have;
p-value = 0.00001 ≈ 0.000
D) The p-value of 0.000 is less than the significance value of 0.05,thus we will reject the null hypothesis