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Aleksandr-060686 [28]
2 years ago
7

The mean income per person in the United States is $41,500, and the distribution of incomes follows a normal distribution. A ran

dom sample of 10 residents of Wilmington, Delaware, had a mean of $47,500 with a standard deviation of $10,600. At the .01 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?
(a) State the null hypothesis and the alternate hypothesis.
H0: ? ?
H1: ? >
(b) State the decision rule for .01 significance level. (Round your answer to 3 decimal places.)
Reject H0 if t >
(c) Compute the value of the test statistic. (Round your answer to 2 decimal places.)
Value of the test statistic
(d) Is there enough evidence to substantiate that residents of Wilmington, Delaware have more income than the national average at the .01 significance level?

Mathematics
1 answer:
Ymorist [56]2 years ago
5 0

Answer:

A) Null Hypothesis; H0: μ = $41,500

Alternative hypothesis; H1: μ > $41,500

B) Reject H0 is t > 2.821433

C) t = 1.79

D) there is no sufficient evidence to support the claim that residents of Wilmington, Delaware have more income than the national average

Step-by-step explanation:

A) The hypotheses is given as;

Null Hypothesis; H0: μ = $41,500

Alternative hypothesis; H1: μ > $41,500

B) From online t-score calculator attached using significance level of 0.01 and DF = n - 1 = 10 - 1 = 9, we have;

t = 2.821433

Normally, when the absolute value of the t-value is greater than the critical value, we reject the null hypothesis. However, when the absolute value of the t-value is less than the critical value, we fail to reject the null hypothesis.

Thus, if t > 2.821433, we will reject the null hypothesis H0.

C) Formula for the test statistic is;

t = (x' - μ)/(s/√n)

We have, μ = 41500, x' = 47500, s = 10600, n = 10

t = (47500 - 41500)/(10600/√10)

t = 1.79

D) So, 1.79 is less than the t-critical value of 2.821433. Thus, we will fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the claim that residents of Wilmington, Delaware have more income than the national average

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