Question

The mean income per person in the United States is $38,000, and the distribution of incomes follows a normal distribution. A random sample of 13 residents of Wilmington, Delaware, had a mean of $48,000 with a standard deviation of $9,200. At the 0.100 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?

State the null hypothesis and the alternate hypothesis.

State the decision rule for 0.100 significance level. (Round your answer to 3 decimal places.)

Compute the value of the test statistic. (Round your answer to 2 decimal places.)

Answer #1

Solution :

This is the right tailed test,

The null and alternative hypothesis is ,

H_{0} :
= 38000

H_{a} :
> 38000

= 0.100

degrees of freedom = n - 1 = 13 - 1 = 12

P( t > t ) = 0.100

= 1 - P( t < t ) = 0.100

= P( t < t ) = 1 - 0.100

= P( t < t ) = 0.900

= P( t < 1.356 ) = 0.900

t > 1.356

Test statistic = t =

= ( - ) / s / n

= (48000 - 38000) / 9200 / 13

Test statistic = t = 3.92

reject to the null hypothesis, because test statistic > critical value

There is sufficient evidence to conclude that residents of Wilmington, Delaware, have more income than the national average

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