In mom's car, the number of miles driven is proportinal to the number of gallons of gas used.

A recent national survey found that high school students watched an average (mean) of 7.8 movies per month with a population sta

notka56 [123]

Answer:

Step-by-step explanation:

Given that :

Mean = 7.8

Standard deviation = 0.5

sample size = 30

Sample mean = 7.3 5.4772

The null and the alternative hypothesis is as follows;

$\mathbf{ H_o: \mu \geq 7.8}$

$\mathbf{ H_1: \mu < 7.8}$

The test statistics can be computed as :

$z = \dfrac{X- \mu}{\dfrac{\sigma}{\sqrt{n}}}$

$z = \dfrac{7.3- 7.8}{\dfrac{0.5}{\sqrt{30}}}$

$z = \dfrac{-0.5}{\dfrac{0.5}{5.4772}}$

$z = - 5.4772$

The p-value at 0.05 significance level is:

p-value = 1- P( Z < -5.4772)

p value = 0.00001

Decision Rule:

The decision rule is to reject the null hypothesis if p value is less than 0.05

Conclusion:

At the 0.05 significance level, there is sufficient information to reject the null hypothesis. Therefore ,we conclude that college students watch fewer movies a month than high school students.

8 0

3 years ago

The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

Alexandra [31]

Answer:

$\bar x = 260.1615$

$\sigma = 70.69$

The confidence interval of standard deviation is: $53.76$ to $103.25$

Step-by-step explanation:

Given

$n =20$

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

$\bar x = 260.1615$

$\bar x = 260.16$

Solving (b): The standard deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

So, the confidence interval of the standard deviation is:

$\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} }$ to $\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }$

$70.69 * \sqrt{\frac{20 - 1}{32.852}$ to $70.69 * \sqrt{\frac{20 - 1}{8.907}$

$70.69 * \sqrt{\frac{19}{32.852}$ to $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

8 0

2 years ago

I need answers to all please

valina [46]

Answer:

Step-by-step explanation:

2a.x=10+5

b. x=11-7

C. x=5*2

4 0

3 years ago

Read 2 more answers

Line segment RS is graphed with its end points at (1, 3) and (7, 11) If RS is dilated by a

Alchen [17]

For the answer to the question above, <span>what is the combined length of
the original line segment and its dilated image?
</span>RS = 10
image R'S' = 20
---> 10+20=30
I hope my answer helped you. Feel free to ask more questions. have a nice day!

7 0

2 years ago

Lilah completed the division problem below.

victus00 [196]

A - False

B - True

C - True

D - False

E - True

-----
7.84 / 2 = 3.92

6 0

3 years ago