HELP!!! ITS TIMED!!!! Please explain the answer!!

An article reported that for a sample of 52 kitchens with gas cooking appliances monitored during a one-week period, the sample

kolbaska11 [484]

Answer:

a) $654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

b) $n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189$

Step-by-step explanation:

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=52-1=51$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,51)".And we see that $t_{\alpha/2}=2.01$

Replacing we got:

$654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

Part b

The margin of error is given by :

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

The desired margin of error is ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , and we use an estimator of the population variance the value of 175 replacing into formula (b) we got:

$n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189$

6 0

3 years ago

Solve and check the following equation. p = −2p −15p =

masha68 [24]

Answer:

I think its -17p

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A village experienced 2% population growth, compounded continuously, each year for 10 years. At the end of the 10 years, the pop

pantera1 [17]

Answer:

The initial population at the beginning of the 10 years was 129.

Step-by-step explanation:

The population of the village may be modeled by the following function.

$P(t) = P_{0}e^{rt}$

In which P is the population after t hours, $P_{0}$ is the initial population and r is the growth rate, in decimal.

In this problem, we have that:

$P(10) = 158, r = 0.02$ .

So

$158 = P_{0}e^{0.02*10}$

$P_{0} = 158*e^{-0.2}$

$P_{0} = 129$

The initial population at the beginning of the 10 years was 129.

6 0

3 years ago

Read 2 more answers

Olivia made 14 outfits for her dolls. Each outfit used 3/5 of fabric. If she started with 10 3/4 yards of fabric, how much did s

Zepler [3.9K]

Answer:

0.1 yards of fabric

Step-by-step explanation:

Data provided in the question:

Number of outfits made for the doll = 14

Fabric used in each outfit = $\frac{3}{5}$ yards of fabric

Starting fabric available = 10 × $\frac{3}{4}$ or $\frac{30}{4}$ yards

Now,

The total fabric used in making the outfits of 14 dolls

= 14 × $\frac{3}{5}$ yards of fabric

= $\frac{42}{5}$ yards of fabric

Therefore,

The fabric left after making the outfits

= Fabric available in the start - Fabric used in making the outfits

= $\frac{30}{4}$ yards - $\frac{42}{5}$ yards

= $\frac{30\times5-42\times4}{4\times5}$ yards

= $\frac{150-168}{20}$ yards

= $\frac{2}{20}$ yards

= 0.1 yards of fabric

5 0

3 years ago

The manager of a bank recorded the amount of time each customer spent waiting in line during peak business hours one Monday. The

zavuch27 [327]

The standard deviation of the frequency distribution is 5.54

<h3>How to determine standard deviation?</h3>

The table of values is given as:

x f(x)

0-3 13

4-7 13

8-11 10

12-15 11

16-19 0

20-23 3

Rewrite the table by calculating the class midpoints:

x f(x)

1.5 13

5.5 13

9.5 10

13.5 11

17.5 0

21.5 3

Start by calculating the mean using:

$\bar x = \frac{\sum fx}{\sum f}$

This gives

$\bar x = \frac{1.5 * 13 + 5.5 * 13 + 9.5 * 10 + 13.5 * 11 + 17.5 * 0 + 21.5 * 3}{13 + 13 + 10 + 11 + 0 +3}$

Evaluate

$\bar x = 7.98$

The standard deviation is then calculated as:

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

So, we have:

$\sigma= \sqrt{\frac{13 * (1.5 - 7.98)^2 + 13 * (5.5 - 7.98)^2 + (9.5 - 7.98)^2 * 10 + (13.5 - 7.98)^2 * 11 + (17.5 - 7.98)^2 * 0 + (21.5 - 7.98)^2 * 3}{13 + 13 + 10 + 11 + 0 +3}}$

Evaluate

$\sigma= \sqrt{30.6496}$

Solve

$\sigma= 5.54$

Hence, the standard deviation is 5.54

Read more about standard deviation at:

brainly.com/question/15858152

#SPJ1

7 0

1 year ago