If you answer this you'll get 50 points but it has to be correct.

The water diet requires you to drink two cups of water every half hour from the time you get up until you go to bed, but otherwi

Elan Coil [88]

Answer:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

Step-by-step explanation:

For this case we have the following info given:

Weight before diet 180 125 240 150

Weight after diet 170 130 215 152

We define the random variable $D = before-after$ and we can calculate the inidividual values:

D: 10, -5, 25, -2

And we can calculate the mean with this formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And the deviation with:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

And after replace we got:

$\bar D= 7, s_d = 13.638$

And the confidence interval for this case would be given by:

$\bar D \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}$

The degrees of freedom are given by:

$df = n-1= 4-1=3$

For the 95% of confidence the value for the significance is $\alpha=0.05$ and the critical value would be $t_{\alpha/2}= 3.182$ . And replacing we got:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

6 0

3 years ago

Ryan buys some jumpers to sell on a stall.

katovenus [111]

Answer:

£540

Step-by-step explanation:

I worked it out let me know if you want me to upload a picture of my workings

8 0

3 years ago

In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt

ankoles [38]

Answer:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

$df=n-1=44-1=43$

$p_v =P(t_{(43)}>1.436)=0.079$

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

$\bar X=13.88$ represent the sample mean

$s=3.14$ represent the sample standard deviation

$n=44$ sample size

$\mu_o =13.2$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:

Null hypothesis: $\mu \leq 13.2$

Alternative hypothesis: $\mu > 13.2$

The statistic for this case is:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculating the statistic

Replacing the info given we got:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

P-value

The degrees of freedom are:

$df=n-1=44-1=43$

Since is a right tailed test the p value is:

$p_v =P(t_{(43)}>1.436)=0.079$

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

8 0

3 years ago

Variable x is 7 more than variable y. Variable x is also 1 less than y. Which of the following pairs of equations best models th

Pepsi [2]

We know that
1) Variable x is 7 more than variable y
so
x=y+7

2) Variable x is also 1 less than y.
x=y-1

therefore

the answer is
the option
x = y + 7
x = y – 1

4 0

3 years ago

PLEASE HELP!!!!!! This season, the probability that the Yankees will win a game is 0.46 and the probability that the Yankees wil

tatuchka [14]

Answer:

Given:

Probability that the Yankees wins a game is

P(A) = 0.46

Probability that the Yankees loses a game is

P(A') = 1 - P(A') = 1 - 0.46 = 0.54

Probability that the Yankees scores 5 or more runs in a game is

P(B) = 0.59

Probability that the Yankees scores fewer than 5 runs in a game is

P(B') = 1 - P(B) = 1 - 0.59 = 0.41

Probability that the Yankees wins and scores 5 or more runs is

P(A⋂B) = 0.39

Applying the De Morgan's law, the probability that the Yankees scores fewer than 5 runs and they loses the game would satisfy:

1 - P(A'⋂B') = P(A) ⋃ P(B) = P(A) + P(B) - P(A⋂B)

or

1 - P(A'⋂B') = 0.46 + 0.59 - 0.39

or

1 - P(A'⋂B') = 0.66

=> P(A'⋂B') = 1 - 0.66 = 0.34

Applying the Bayes theorem, the probability that the Yankees would score fewer than 5 runs, given they lose the game:

P(B'|A') = P(A'⋂B')/P(A')

or

P(B'|A') = 0.34/0.54 = 0.630

Hope this helps!

:)

7 0

3 years ago