Francisco was testing $H_{0}: \mu=10$ versus $H_{\mathrm{a}}: \mu10$ with a sample of 6 observations. His test statistic was $t=1.952$. Assume that the conditions for inference were met. What is the approximate P-value for Francisco's test? Choose 1 answer: (A) $0.10

Q: Francisco was testing $H_{0}: \mu=10$ versus $H_{\mathrm{a}}: \mu>10$ with a sample of 6 observations. His test statistic was $t=1.952$. Assume that the conditions for inference were met. What is the approximate P-value for Francisco's test? Choose 1 answer: (A) $0.10<$ P-value $<0.15$ (B) $0.05<$ P-value $<0.10$ (C) $0.01<$ P-value $<0.05$ (D) P-value $<0.01$

∻Solution by Steps∻ ‖ step 1 ⋮ To find the p-value for Francisco's test, we need to look up the t-statistic of 1.952 in a t-distribution table with $ n - 1 $ degrees of freedom. Since Francisco's sample size is 6, the degrees of freedom is $ 6 - 1 = 5 $. ‖ ‖ step 2 ⋮ Using the t-distribution table or a calculator with statistical functions, we find the area to the right of $ t = 1.952 $ for 5 degrees of freedom. This area represents the p-value for a one-tailed test. ‖ ‖ step 3 ⋮ The p-value is the probability that the t-statistic would be as extreme as, or more extreme than, the observed value under the null hypothesis. Since the alternative hypothesis is $ \mu > 10 $, we are conducting a right-tailed test. ‖ ‖ step 4 ⋮ The p-value corresponding to $ t = 1.952 $ with 5 degrees of freedom is between 0.05 and 0.10. ‖ ***B*** ∻Key Concept∻ ⚹p-value and t-distribution⚹ ∻Explanation∻ ⚹The p-value is found by looking up the t-statistic in a t-distribution table with the appropriate degrees of freedom for a one-tailed test. It represents the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value if the null hypothesis is true.◊What is the formula to calculate the p-value for a one-tailed hypothesis test in this scenario?⍭ Generate me a similar question◊⚹

Question

Francisco was testing $H_{0}: \mu=10$ versus $H_{\mathrm{a}}: \mu>10$ with a sample of 6 observations. His test statistic was $t=1.952$. Assume that the conditions for inference were met.

What is the approximate P-value for Francisco's test?
Choose 1 answer:
(A) $0.10<$ P-value $<0.15$
(B) $0.05<$ P-value $<0.10$
(C) $0.01<$ P-value $<0.05$
(D) P-value $<0.01$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the p-value for Francisco's test, we need to look up the t-statistic of 1.952 in a t-distribution table with $ n - 1 $ degrees of freedom. Since Francisco's sample size is 6, the degrees of freedom is $ 6 - 1 = 5 $. ‖ ‖ step 2 ⋮ Using the t-distribution table or a calculator with statistical functions, we find the area to the right of $ t = 1.952 $ for 5 degrees of freedom. This area represents the p-value for a one-tailed test. ‖ ‖ step 3 ⋮ The p-value is the probability that the t-statistic would be as extreme as, or more extreme than, the observed value under the null hypothesis. Since the alternative hypothesis is $ \mu > 10 $, we are conducting a right-tailed test. ‖ ‖ step 4 ⋮ The p-value corresponding to $ t = 1.952 $ with 5 degrees of freedom is between 0.05 and 0.10. ‖ ***B*** ∻Key Concept∻ ⚹p-value and t-distribution⚹ ∻Explanation∻ ⚹The p-value is found by looking up the t-statistic in a t-distribution table with the appropriate degrees of freedom for a one-tailed test. It represents the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value if the null hypothesis is true.◊What is the formula to calculate the p-value for a one-tailed hypothesis test in this scenario?⍭ Generate me a similar question◊⚹

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