A random sample has 49 values. The sample mean is 8.5 and the sample standard de

Question

A random sample has 49 values. The sample mean is 8.5 and the sample standard deviation is 1.5. use a level of significance of 0.01 to conduct a left-tailed test of the claim that the population mean is 9.2.

a) Is it appropriate to use a Student's t distribution? Explain. how many degrees of freedom do we use?

b) what are the hypotheses?

c) Compute the sample test statistic t.

d) Estimate the P-value for the test.

e) Do we reject or fail to reject H0?

f) The results

Please provide the following information for problems 11-22

a) What is the level of significance? State the null and alternate hypotheses.

b) What sampling distribution will you use? Explain that rationale for your choice of sampling distribution. Compute the value of the sample test statistic.

c) Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value.

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?

e) your conclusion in the context of the application. Note: for degrees of freedom d.f. not given in the student's t table, use the closest d.f. that is smaller. In some situation, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.

Explanation / Answer

Q1.

Given that,
population mean(u)=9.2
sample mean, x =8.5
standard deviation, s =1.5
number (n)=49
null, Ho: =9.2
alternate, H1: !=9.2
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.682
since our test is two-tailed
reject Ho, if to < -2.682 OR if to > 2.682
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =8.5-9.2/(1.5/sqrt(49))
to =-3.267
| to | =3.267
critical value
the value of |t | with n-1 = 48 d.f is 2.682
we got |to| =3.267 & | t | =2.682
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.2667 ) = 0.002
hence value of p0.01 > 0.002,here we reject Ho
ANSWERS
---------------
null, Ho: =9.2
alternate, H1: !=9.2
test statistic: -3.267
critical value: -2.682 , 2.682
decision: reject Ho
p-value: 0.002

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