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denis-greek [22]
3 years ago
8

3. The average annual income of 100 randomly chosen residents of Santa Cruz is $45,221 with a standard deviation of $30,450. a.

What is the standard deviation of the mean annual income? b. Test the 1-sided hypothesis that the average annual income is $40,000 against the alternative that it is less than $40,000 at the 10% significance level. c. Test the 2-sided hypothesis that the average annual income is equal to $42,000 against the alternative that it is not at the 5% significance level. d. What is the confidence interval for part (c)? e. What is the p-value for part (b)? f. Do you reject or fail to reject part (c)? Please write your answer and a complete, accurate, sentence.
Mathematics
1 answer:
Sergeu [11.5K]3 years ago
5 0

Answer:

Step-by-step explanation:

Given that the average annual income of 100 randomly chosen residents of Santa Cruz is $45,221 with a standard deviation of $30,450.

a) Std deviation of mean = \frac{\sigma}{\sqrt{n} } \\=3045

b) H_0: x bar =40000\\H_a: x bar

Test statistic = Mean diff/std dev of mean = \frac{-5221}{3045} \\=-1.715

p value = 0.043

Since p >0.01 we accept null hypothesis.

c) For two sided p value = 0.086

Here we have

H_0: x bar =42000\\H_a: x bar \neq 42000\\\alpha = 0.05

Test statistic = \frac{-3000}{3045} \\\\=-0.99

p value = 0.3221

Since p >alpha, we accept null hypothesis.

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