The Employment and Training Administration reported that the U. S. mean unemployment insurance benefit was $238 per week (The World Almanac, 2003). Aresearcher in the state
of Virginia anticipated that sample data would show evidence that the mean weekly unemployment
insurance benefit in Virginia was below the national average.
a. Develop appropriate hypotheses such that rejection of H0 will support the researcher’s
contention.
b. For a sample of 100 individuals, the sample mean weekly unemployment insurance
benefit was $231 with a sample standard deviation of $80. What is the p-value?
c. At αα = .05, what is your conclusion?
d. Repeat the preceding hypothesis test using the critical value approach.

Step-by-step explanation:

a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 238

For the alternative hypothesis,

H1: µ < 238

This is a left tailed test

b) Since the population standard deviation is not given, the distribution is a student's t.

Since n = 100

Degrees of freedom, df = n - 1 = 100 - 1 = 99

t = (x - µ)/(s/√n)

Where

x = sample mean = 231

µ = population mean = 238

s = samples standard deviation = 80

t = (231 - 238)/(80/√100) = - 0.88

We would determine the p value using the t test calculator. It becomes

p = 0.19

c) Since alpha, 0.05 < than the p value, 0.19, then we would fail to reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed insignificant evidence that the mean weekly unemployment insurance benefit in Virginia was below the national average.

d) Since α = 0.05, the critical value is determined from the t distribution table. Recall that this is a left tailed test. Therefore, we would find the critical value corresponding to 1 - α and reject the null hypothesis if the test statistic is less than the negative of the table value.

1 - α = 1 - 0.05 = 0.95

The negative critical value is - 1.66

Since - 0.88 is greater than - 1.66, then we would fail to reject the null hypothesis.

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