Question

A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 12 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 11.5.

(a) Is it appropriate to use a Student's t distribution? Explain. (Choose one of 5 listed below)

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.

(b) How many degrees of freedom do we use? ____
What are the hypotheses? (Choose one of five answers below)

H₀: μ = 11.5; H₁: μ > 11.5

H₀: μ < 11.5; H₁: μ = 11.5    

H₀: μ = 11.5; H₁: μ < 11.5

H₀: μ = 11.5; H₁: μ ≠ 11.5

H₀: μ > 11.5; H₁: μ = 11.5

(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. (choose one of five answers below)

P-value > 0.25

00.100 < P-value < 0.250    

0.050 < P-value < 0.100

0.010 < P-value < 0.050

P-value < 0.010

(e) Do we reject or fail to reject H₀? (Choose one of 4 answers below)

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. (Choose one of two answers below)

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 :

(a)

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

(b)

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :   =11.5

H_a :    11.5

= 12

= 11.5

s = 2

n = 16

df = n - 1 = 16 - 1 = 15

(c)

Test statistic = t

= ( - ) / s / n

= (12 - 11.5) / 2 / 16

= 1.000

Test statistic = 1.000

(d)

P-value > 0.25

= 0.05

P-value >

Fail to reject the null hypothesis .

(e)

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.