Question

1.
the unemployment and training administration reported u.s. mean unemployment insurance benefit was $238 per week. A researcher in the state of Virginia anticipated that sample data would show evidence that the mean weekly unemployment insurance benefit in virginia was significantly different. For sample of 100 individuals the sample mean weekly unemployment Insurance benefit was $231 with a sample standard deviation of $80. Is there sufficient evidence to support the researcher’s claim? Use a 0.10 level of significance.

a) Define the target parameter. Is it a large or small sample estimation exercise?
b) Set up the H0 and Ha. Is it one-tailed or two-tailed test? why?
c) Calculate the sample test statistic
d) Define the rejection region and use the graph make appropriate conclusions with response to H0 and Ha. provide the interpretation of result
e) Is it possible to reject the H0 with confidence 95%? why?

Answer 1

(a)

the target parameter = = the mean weekly unemployment insurance benefit in Virginia

It is a small sample estimation exercise because even though Sample Size = n = 100 > 30, Large Sample, the Population Standard Deviation = is not provided and the same is to be estimated from the given Sample Standard Deviation = s = 80.

(b)

H0: Null Hypothesis: = 238 ( The mean weekly unemployment insurance benefit in Virginia was significantly not different. from U.S. mean unemployment insurance benefit of $238 per week)

HA: Alternative Hypothesis: 238 ( The mean weekly unemployment insurance benefit in Virginia was significantly different. from U.S. mean unemployment insurance benefit of $238 per week) (Claim)

It is two-tailed test because the Alternative Hypothesis has "" sign and not < or > sign.

(c)

SE = s/

= 80/

= 8

Test Statistic is given by:

t = (231 - 238)/8

= - 0.125

So,

the sample test statistic = - 0.125

(d)

=0.10

ndf = 100 - 1 =99

From Table, critical values of t = 1.6604

The Rejection Region is defined as follows:

Reject H0 if t < - 1.6604 OR t > 1.6604

Since calculate value of t = - 0.125 is greater than critical value of t = - 1.6604, it is not in the Rejection Region. The difference is not significant. Fail to reject null hypothesis.

Interpretation:
The data do not support the claim that the mean weekly unemployment insurance benefit in Virginia was significantly different. from U.S. mean unemployment insurance benefit of $238 per week.

(e)

It is not possible to reject the H0 with confidence 95%

Reason is as follows:

=0.05

ndf = 100 - 1 =99

From Table, critical values of t = 1.9842

The Rejection Region is defined as follows:

Reject H0 if t < - 1.9842 OR t > 1.9842

Since calculate value of t = - 0.125 is greater than critical value of t = - 1.9842, it is not in the Rejection Region. The difference is not significant. Fail to reject null hypothesis.

Interpretation:
The data do not support the claim that the mean weekly unemployment insurance benefit in Virginia was significantly different. from U.S. mean unemployment insurance benefit of $238 per week.