Mathematics, 11.03.2020 23:07 erykp17

Mortgage rates: Following are interest rates (annual percentage rates) for a 30-year fixed rate mortgage from a sample of lenders in Macon, Georgia for one day. It is reasonable to assume that the population is approximately normal. 4.751 4.373 4.177 4.676 4.425 4.228 4.125 4.251 3.951 4.192 4.291 4.414 Send data to Excel Part: 0 / 2 Part 1 of 2 (a) Construct a 98% confidence interval for the mean rate. Round the answer to at least four decimal places. A 98% confidence interval for the mean rate is

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Answer from: alejachico2

98% Confidence interval: (4.144 ,4.498)

Step-by-step explanation:

We are given the following in the question:

4.751, 4.373, 4.177, 4.676, 4.425, 4.228, 4.125, 4.251, 3.951, 4.192, 4.291, 4.414

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where x_i are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{51.854}{12} = 4.321$

Sum of squares of differences = 0.5602

$s = \sqrt{\dfrac{0.5602}{11}} = 0.226$

98% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 11 and}~\alpha_{0.02} = \pm 2.718$

$4.321 \pm 2.718(\frac{0.226}{\sqrt{12}} ) = 4.321 \pm 0.1773 = (4.144 ,4.498)$

Answer from: haily13

98% confidence interval for the mean rate = [4.1437 , 4.4983]

Step-by-step explanation:

We are given the interest rates (annual percentage rates) for a 30-year fixed rate mortgage from a sample of lenders in Macon, Georgia for one day ;

4.751, 4.373, 4.177, 4.676, 4.425, 4.228, 4.125, 4.251, 3.951, 4.192, 4.291, 4.414

Now, Firstly we will find Mean of above data, Xbar ;

Mean, Xbar = $\frac{\sum X}{n}$ = $\frac{4.751 +4.373+ 4.177+ 4.676+ 4.425+ 4.228+ 4.125 +4.251 +3.951 +4.192+ 4.291 +4.414}{12}$ = 4.321

Standard deviation, s = $\sqrt{\frac{\sum (X-Xbar)^{2} }{n-1} }$ = 0.226

Now, the pivotal quantity for 98% confidence interval for the mean rate is;

P.Q. = $\frac{Xbar-\mu}{\frac{s}{\sqrt{n} }}$ ~ t_n_-_1

where, Xbar = sample mean

s = sample standard deviation

n = sample size = 12

So, 98% confidence interval for the mean rate, $\mu$ is ;

P(-2.718 < t_1_1 < 2.718) = 0.98

P(-2.718 < $\frac{Xbar-\mu}{\frac{s}{\sqrt{n} }}$ < 2.718) = 0.98

P(Xbar - 2.718 * ${\frac{s}{\sqrt{n} }}$ < $\mu$ < Xbar + 2.718 * ${\frac{s}{\sqrt{n} }}$ ) = 0.98

98% confidence interval for $\mu$ = [Xbar - 2.718 * ${\frac{s}{\sqrt{n} }}$ , Xbar + 2.718 * ${\frac{s}{\sqrt{n} }}$ ]

= [4.321 - 2.718 * ${\frac{0.226}{\sqrt{12} }}$ , 4.321 + 2.718 * ${\frac{0.226}{\sqrt{12} }}$ ]

= [4.1437 , 4.4983]

Therefore, 98% confidence interval for the mean rate = [4.1437 , 4.4983]

Answer from: Quest

the answer is b.)

< 1=< 2

Answer from: Quest

let x be the money she spent so that she can earn the free t-shirt.

since, she already spent $112.49

⇒ $x + 112.49$

also, as per the statement: jennifer must spend $250 or more in order to qualify for a free t-shirt at the store.

then,

we have an inequality i.,e

$x + 112.49 > 250$

subtract 112.49 from both sides we get;

$x > 137.51$

therefore, $137. 51 more money does jennifer need to spend so that she can earn the free t-shirt

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