Question

A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 8 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 7.5.

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

For a Student's t distribution with

d.f. = 11

and

t = 2.910,

consider the following.

(a) Find an interval containing the corresponding P-value for a two-tailed test.

0.0010 < P-value < 0.0100

.010 < P-value < 0.020    

0.020 < P-value < 0.0500

.050 < P-value < 0.1000

.100 < P-value < 0.1500

.150 < P-value < 0.2000

.200 < P-value < 0.2500

.250 < P-value < 0.500

(b) Find an interval containing the corresponding P-value for a right-tailed test.

0.0005 < P-value < 0.0050.

005 P-value < 0.010     

0.010 < P-value < 0.0250

.025 < P-value < 0.0500

.050 < P-value < 0.0750

.075 < P-value < 0.1000

.100 < P-value < 0

.1250.125 < P-value < 0.250

Answer 1

(a) It is appropriate to use a Student's t distribution because the x distribution is mound-shaped and symmetric and σ is unknown.

The hypotheses are

Rejection region:

Reject Ho if |t|>t_0.025,15=2.131

(b) Test statistic t is calculated as

Conclusion:

Since the test statistic calculated t=1 is less than t_0.025,15=2.131 then we fail to reject the null hypothesis and we do not have enough evidence to support the claim

For a Student's t distribution with

d.f. = 11

and

t = 2.910,

a) Then for the two-tailed test, the p-value computed using excel tool or by t distribution table

0.010 < P-value < 0.020

b) the corresponding P-value for a right-tailed test.

0.050 < P-value < 0.075

T table as