Mathematics, 28.02.2020 00:37 chonawilson4

Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, it uses different media to reach potential buyers. The mean annual family income for 12 people making inquiries at the first development is $153,000, with a standard deviation of $42,000. A corresponding sample of 24 people at the second development had a mean of $171,000, with a standard deviation of $30,000. Assume the population standard deviations are the same.

1.State the decision rule for .05 significance level: H0: ?1 = ?2; H1:?1 ? ?2. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)
Reject H0 if t is not between and .
2.Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
Value of the test statistic
3.At the .05 significance level, can Fairfield conclude that the population means are different?

Answers: 2

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Answers

Answer from: bellam302

(1) Null Hypothesis, H_0 : $\mu_1 = \mu_2$

Alternate Hypothesis, H_1 : $\mu_1\neq \mu_2$

(2) Test statistics = -1.48

(3) At the 0.05 significance level, Fair field conclude that the population means are same.

Step-by-step explanation:

Let $\mu_1$ = mean annual family income for 12 people making inquiries at the first development

$\mu_2$ = mean annual family income for 24 people making inquiries at the second development

s_1 = standard deviation of annual family income for 12 people making inquiries at the first development

s_2 = standard deviation of annual family income for 24 people making inquiries at the second development

n_1 = sample of people of first development i.e. 12

n_2 = sample of people of second development i.e. 24

(1) Null Hypothesis, H_0 : $\mu_1 = \mu_2$ {population means are same}

Alternate Hypothesis, H_1 : $\mu_1\neq \mu_2$ {population means are different}

DECISION RULE ;

If the test statistics is less than the critical value of t from table at 5% significance level, then we will accept null hypothesis, H_0

.If the test statistics is more than the critical value of t from table at 5% significance level, then we will reject null hypothesis, H_0

(2) The test statistics is given by;

$\frac{(X_1bar -X_2bar) - (\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ t_n__1+n_2-2

where, X_1bar = Sample mean income of people at first development

= $153,000

X_2bar = Sample mean income of people at second development = $171,000

s_1 = $42,000 and s_2 = $30,000

$s_p= \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2}}$ = $\sqrt{\frac{(12-1)42000^{2}+(24-1)30000^{2} }{12+24-2}}$ = 34344.30

Test statistics = $\frac{(153000 -171000) - 0}{34344.30\sqrt{\frac{1}{12}+\frac{1}{24} } }$ ~ t_3_4

= -1.48

(3) At 5% level of significance, t table gives critical value of 2.032 at 34 degree of freedom.Since our test statistics is less than the critical value of t so considering our decision rule, we will accept null hypothesis.

And conclude that population means are same.

Answer from: Quest

this is answer of 144 cubic meters

Ineed it’s for a test get back to be as soon as possible

Answer from: Quest

Answer

1 number

explanation

if we use a graphing calculator, there is only one such number $c\in[0.5,0.75]$ that satisfies the conclusion of the mean value theorem, that is, a number $c\in[0.5,0.75]$ such that

$f'(c) = \dfrac{f(0.75) - f(0.5)}{0.75-0.5},$

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.