Use technology and a t-test to test the claim about the population mean at the given level of significance using the given sample statistics. Assume the population is normally distributed. Claim: ;     Sample statistics: , s, n What are the null and alternative hypotheses? Choose the correct answer below. A. H0: HA: B. H0: HA: C. H0: HA: D. H0: HA:

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.

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

Step-by-step 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

the answer will be b x² +x-1

step-by-step explanation: