What is the value of x?<br> Enter your answer in the box.<br><br> x =

Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. Th

Verizon [17]

Answer:

$(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422$

$(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778$

And we are 95% confident that the true difference means are between $1.422 \leq \mu_1 -\mu_2 \leq 2.778$

Step-by-step explanation:

We know the following info:

$\bar X_1 = 21.9$ sample mean for group 1

$\bar X_2 = 19.8$ sample mean for group 2

$s_1 = 3.4$ sample standard deviation for group 1

$s_2 = 3.5$ sample standard deviation for group 2

$n_1 = 200$ sample size group 1

$n_2 = 200$ sample size group 2

We want to find a confidence interval for the difference of means and the correct formula to do this is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

Now we just need to find the critical value. The confidence level is 0.95 then the significance is $1-0.95 =0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df= n_1 +n_2 -2= 200+200-2= 398$

The critical value for this case would be : $t_{\alpha/2}=1.966$

And replacing into the confidence interval formula we got:

$(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422$

$(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778$

And we are 95% confident that the true difference means are between $1.422 \leq \mu_1 -\mu_2 \leq 2.778$

5 0

3 years ago

Tim Worker expects Social Security benefits of $937 per month. His wife, Mary, is eligible for a wife's benefit of 50 percent of

mihalych1998 [28]

Answer:

468.5

Step-by-step explanation:

so that would be $468 and 50 cents

8 0

3 years ago

You are planning a May camping trip to Denali National Park in Alaska and want to make sure your sleeping bag is warm enough. Th

Goshia [24]

The probability that this bag will be warm enough on a randomly

selected May night at the park is 0.8106 ⇒ answer C

Step-by-step explanation:

You are planning a May camping trip to Denali National Park in Alaska

and want to make sure your sleeping bag is warm enough.

The average low temperature in the park for May follows a normal

distribution

The given is:

1. The mean is 32°F

2. The standard deviation is 8°F

3. One sleeping bag you are considering advertises that it is good for

temperatures down to 25°F (x ≥ 25)

At first let us find the z-score

∵ z = (x - μ)/σ, where x is the score, μ is the mean, σ is standard deviation

∵ μ = 32°F , σ = 8°F , x = 25°F

- Substitute these values in the rule

∴ z = $\frac{25-32}{8}=-0.875$

Now let us find the corresponding area of z-score in the normal

distribution table

∵ The corresponding area of z = -0.875 is 0.18943

∵ For P(x ≥ 25) the area to the right is needed

∵ P(x ≥ 25) = 1 - 0.18943 = 0.8106

∴ P(x ≥ 25) = 0.8106

The probability that this bag will be warm enough on a randomly

selected May night at the park is 0.8106

Learn more:

You can learn more about mean and standard deviation in brainly.com/question/6073431

#LearnwithBrainly

3 0

3 years ago

Need this very badly

Fofino [41]

Can u show us how the location looks please?

7 0

3 years ago

What is the distributive property of 75x6

ohaa [14]

That would be 60 and 15

5 0

3 years ago

Read 2 more answers

What is the value of x? Enter your answer in the box. x =