Can someone please help me with this

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

3 years ago

Find the sum or product of 5×23×2 using properties

soldi70 [24.7K]

I think this is 230 ..

8 0

3 years ago

Y=2x-5 -3x - 4y=-2

Whitepunk [10]

Answer:

y=-x/5 =3/5. that is how it is.

8 0

3 years ago

Type the correct answer in each box. Use numerals instead of words. For this question, any answer that is not a whole number sho

egoroff_w [7]

Answer:

Step-by-step explanation:

Since we are not given the value of P, |Q|, R and S, we can as well assume values for them for the sake of this question.

Let P = 5, |Q| = 6, R =7 |S| = 2

Note that since Q and S are in modulus sign, they can return both positive and negative values.

P+Q = 5 + 6 (note that the positive value of Q is used since we need the greatest value of P+Q)

P+Q = 11

Hence the greatest value of P+Q is 11

For the least value of P+Q, we will use the negative value of Q as shown

P+Q = 5+(-6)

P+Q = 5-6

P+Q = -1

Hence the least value of P+Q is -1

Similarly:

R+S = 7 + 2 (note that the positive value of S is used since we need the greatest value of R+S)

R+S = 9

Hence the greatest value of R+S is 9

For the least value of R+S, we will use the negative value of S as shown

R+S = 5+(-2)

R+S = 5-3

R+S = 2

Hence the least value of P+Q is 2

NOTE THAT THIS ARE ASSUMED VALUES. ALL YOU NEED IS TO PLUG IN THE VALUES OF P, Q and R THAT YOU HAVE IN CASE THE VALUES DIFFERS.

5 0

3 years ago

5/2 c = 8 1/3 please help me

Diano4ka-milaya [45]

9514 1404 393

Answer:

3 1/3

Step-by-step explanation:

Multiply both sides by the reciprocal of the coefficient of c.

(5/2c)(2/5) = (8 1/3)(2/5)

c = (25/3)(2/5) = 50/(3·5) = 10/3

c = 10/3 = 3 1/3

3 0

3 years ago

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