A random sample of 46 adult coyotes in a region of northern Minnesota showed the average age to be x̄ = 2.07 years, with sample standard deviation s = 0.88 years. However, it is thought that the overall population mean age of coyotes is μ = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use α = 0.01.(a) What is the level of significance?State the null and alternate hypotheses.- O H₀: μ = 1.75 yr, H₁: μ > 1.75 yr- O H₀: μ < 1.75 yr, H₁: μ = 1.75 yr- O H₀: μ > 1.75 yr, H₁: μ = 1.75 yr- O H₀: μ = 1.75 yr, H₁: μ < 1.75 yr- O H₀: μ = 1.75 yr, H₁: μ ≠ 1.75 yr(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.- O The standard normal, since the sample size is large and σ is known.- O The standard normal, since the sample size is large and σ is unknown.- O The Student's t, since the sample size is large and σ is known.- O The Student's t, since the sample size is large and σ is unknown.What is the value of the sample test statistic? (Round your answer to three decimal places.)(c) Estimate the P-value.- O P-value > 0.250 - O 0.100 < P-value ≤ 0.250 - O 0.050 < P-value ≤ 0.100 - O 0.010 < P-value ≤ 0.050 - O P-value ≤ 0.010Sketch the sampling distribution and show the area corresponding to the P-value.The page includes four graphs featuring normal distribution curves. Each graph illustrates a normal distribution curve centered at 0 on the horizontal axis ranging from -4 to 4, with the shaded area indicating the P-value.(d) Based on your answers in parts (a) to (c), will

The sample size is n=46The sample mean is x¯=2.07 yearsThe sample standard deviation is s=0.88…

JL A random sample of 46 adult coyotes in a region of northern Minnesota showed the average age to be -2.07 years, with sample standard deviations-088 years. However, it is thought that the overal population mean age of coyotes is 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use -0.01. LAUSE SALT (a) What is the level of significance? - State the null and altermate hypotheses OM 1.75 yr ₁1.75 OM 1.75 yr; ₁-1.75 yr 1.75 yr; M: 1.75 yr 1.75 yr, M₁1.75 yr Og #1.75 yr, M₁:1.75 yr OM OM (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution O The standard nermal, since the sample size is large and is known O The standard narmal, since the sample size is large and is unknown. O The Student's f, since the sample size is large and a is known O The Student's t, since the sample size is large and d is unknown What is the value of the sample test statistic? (Round your answer to three decimal places) (c) Estimate the P-value OP-value 0.250 O 0.100P-value < 0.250 O 0.050 value 0.100 O 0.010 P-value < 0.050 O value 0.010 Sketch the sampling distribution and show the area corresponding to the value a O A (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a? At the 0.01 level, we reject the null hypothesis and conclude the data are statistically significant O At the 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant O At the O At the 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant (e) Interpret your conclusion in the context of the application O There is sufficient evidence at the 0.01 level to condude that cayotes in the specified region tend to live longer than 1.75 year O There is insuffident evidence at the 0.01 level to conclude that cayotes in the specified region tend to live langer than 1.75 year