In the year 2007, a company made $2.2 million in profit. For each consecutive year

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% 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 = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago

Read the question on the picture.

Marizza181 [45]

These two lines are congruent (the same) so we can set them equal to each other and solve.

5x-4=3x+6

The first thing we need to do is subtract 3x from both sides leaving us with 2x-4=6

Now we can add 4 to both sides leaving us with 2x=2

Now we need to divide both sides by 2 to get x alone.

Giving us our final answer of x=1

3 0

3 years ago

After every 4 days in abuja is holiday, if the last holiday was on friday which day will be holiday after 72 days

goldenfox [79]

Answer:

Sunday

Step-by-step explanation:

There is holiday after every 4 days

The last holiday was on Friday.

So,

Last holiday - Friday

After 4 days - Tuesday

After 8 days - Saturday

After 12 days - Wednesady

After 16 days - Sunday

After 20 days - Thursday

After 24 days - Monday

After 28 days - Friday

...... ............

After 56 days - Friday

After 60 days - Tuesday

After 64 days - Saturday

After 68 Days - Wednesday

After 72 days - Sunday

6 0

3 years ago

Consider the three functions below. f(x) = Negative StartFraction 6 Over 11 EndFraction (eleven-halves) Superscript x g(x) = Sta

Setler79 [48]

Answer:

The correct option is;

The ranges of f(x) and h(x) are similar and different from the ranges of g(x)

Step-by-step explanation:

Here we have the functions given as follows;

f(x) = -6/11 (11/2)ˣ

g(x) = 6/11 (11/2)⁻ˣ

h(x) = -6/11 (11/2)⁻ˣ

from the above equations, it can be seen that f(x) and h(x) are always negative while g(x) is always positive

f(x) and h(x) are symmetric about the y axis while g(x) and h(x) are symmetric about the x axis

Also h(x) is the inverse of f(x) hence the ranges of f(x) and g(x) are similar and are different from the ranges of g(x).

9 0

3 years ago

Read 2 more answers

An anthropologist finds bone that her instruments measure it as 0.061% of the amount of Carbon-14 the bones would have contained

julia-pushkina [17]

Let us assume that the rate of decay follows a 1st order reaction. Therefore the formula for rate of decay is:

A = Ao e^-kt ---> 1

Where,

A = final value after time t

Ao = initial value at t zero

k = rate constant

t = time elapsed

First let use find for the value of constant k. We know that the formula for half-life is:

t-half life = ln 2 / k

Finding for k:

5730 = ln 2 / k

k = 1.21 * 10^-4 yr^-1

Since it is stated that A = 0.00061 Ao, therefore substituting all the values into equation 1:

0.00061 Ao / Ao = e^(-1.21 * 10^-4 yr^-1 * t)

Finding for t:

t = 61,173.98 years

The person died about 61,174 years ago.

7 0

3 years ago