Question

# A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample...

A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 13 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 12.5.

(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 = 12.5; H1: μ > 12.5H0: μ < 12.5; H1: μ = 12.5    H0: μ = 12.5; H1: μ < 12.5H0: μ > 12.5; H1: μ = 12.5H0: μ = 12.5; H1: μ ≠ 12.5

(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.

(a) Yes, because the x distribution is mound-shaped and symmetric and σ is unknown

Degree of freedom = n-1 = 16-1 = 15

(b) H0: μ = 12.5; H1: μ ≠ 12.5

Since it is a two tailed test, we are saying that mean is not equal; it can be less than or greater than 12.5

(c) t value

(d) P value

P value = TDIST (t statistics, df, 2) = TDIST (1, 15, 2) = 0.333

P-value > 0.250

(e) level of significance = 0.05 and P value os 0.333

Since P value is greater than level of significance, we fail to reject the Null hypothesis.

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

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

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