Question

A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 15 and the sample standard deviation is 2. Use a level of significance of 0.05 to conduct a two-tailed test of the claim that the population mean is 14.5.

(a) Is it appropriate to use a Student's t distribution? Explain.

Yes, because the x distribution is mound-shaped and symmetric and σ is unknown.No, the x distribution is skewed left.    No, the x distribution is skewed right.No, the x distribution is not symmetric.No, σ is known.

How many degrees of freedom do we use?


(b) What are the hypotheses?

H₀: μ > 14.5; H₁: μ = 14.5H₀: μ = 14.5; H₁: μ > 14.5    H₀: μ = 14.5; H₁: μ < 14.5H₀: μ < 14.5; H₁: μ = 14.5H₀: μ = 14.5; H₁: μ ≠ 14.5

(c) Compute the t value of the sample test statistic. (Round your answer to three decimal places.)

t =

(d) Estimate the P-value for the test.

P-value > 0.2500.100 < P-value < 0.250    0.050 < P-value < 0.1000.010 < P-value < 0.050P-value < 0.010

(e) Do we reject or fail to reject H₀?

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.    At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

(f) Interpret the results.

There is sufficient evidence at the 0.05 level to reject the null hypothesis.There is insufficient evidence at the 0.05 level to reject the null hypothesis.   

Answer 1

Solution:

Given:

Sample size = n = 16
Sample mean =
sample standard deviation = s = 2
level of significance = 0.05

Claim: The population mean is 14.5

Part a) Is it appropriate to use a Student's t distribution? Explain.

Yes, because the x distribution is mound-shaped and symmetric and σ is unknown.

How many degrees of freedom do we use?

df = n - 1= 16 - 1 = 15

Part b) What are the hypotheses?

H₀: μ = 14.5; H₁: μ ≠ 14.5

Part c) Compute the t value of the sample test statistic.

Part d) Estimate the P-value for the test.

df = 15

Look in t table for df = 15 row and find an interval in which t = 1.000 fall then find corresponding two tail area interval , which is p-value interval.

t = 1.000 fall between 0.866 and 1.074

corresponding two tail area interval is between 0.30 to 0.40

thus correct answer is: P-value > 0.250

Part e) Do we reject or fail to reject H₀?

Since  P-value > 0.250 > 0.05 level of significance, thus we failed to reject H0. Thus correct answer is:

.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

Part (f) Interpret the results.

There is insufficient evidence at the 0.05 level to reject the null hypothesis.