How many yards in one foot

A researcher is interested in finding a 95% confidence interval for the mean number of times per day that college students text.

Oxana [17]

Complete Question

A researcher is interested in finding a 95% confidence interval for the mean number of times per day that college students text. The study included 210 students who averaged 28 texts per day. The standard deviation was 21 texts.

Round your answers to two decimal places.

A. The sampling distribution follows a [ Select ] ["F", "normal", "T", "Chi-square"] distribution.

B. With 95% confidence the population mean number of texts per day of is between [ Select ] ["24.92", "25.79", "27.37", "25.14"] and [ Select ] ["31.19", "31.20", "29.28", "30.86"] texts.

C. If many groups of 210 randomly selected students are studied, then a different confidence interval would be produced from each group. About [ Select ] ["5", "95", "1", "99"] percent of these confidence intervals will contain the true population mean number of texts per day and about [ Select ] ["95", "99", "5", "1"] percent will not contain the true population mean number of texts per day.

Answer:

a. With 95% confidence the population mean number of texts per day is between $I = 21.06$ and $I = 30.33$ texts.

b

If many groups of 132 randomly selected members are studied, then a different confidence interval would be produced from each group. About <u>95%</u> percent of these confidence intervals will contain the true population number of texts per day and about <u>5%</u> percent will not contain the true population mean number of texts per day.

Step-by-step explanation:

From the question we are told that

The sample size is $n = 102$

The population mean is $\mu = 25.7$

The standard deviation is $\sigma = 23.9$

Generally the standard error of mean is mathematically evaluated as

$\sigma_e = \frac{\sigma}{\sqrt{n } }$

substituting values

$\sigma_e = \frac{23.9}{\sqrt{102 } }$

$\sigma_e = 2.3665$

Given that the level of confidence is 95% then the level of significance will be

$\alpha = 100 -95$

$\alpha =$ 5%

=> $\alpha = 0.05$

Now the critical values of this level of significance is obtained from the critical value table as

$t_{n,\alpha } = t_{102,0.05} = 1.96$

Generally the 95% confidence interval is mathematically represented as

$I = \mu \pm t_{102, 0.05} * \sigma_e$

substituting values

$I = 25.7 \pm 1.96 *2.3665$

$I = 25.7 \pm 4.638$

$I = 30.33$ and $I = 21.06$

So the interval is

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Answer:

78

Step-by-step explanation:

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A motor boat and a raft start traveling downstream. After 1 hour, the boat turns around and travels back until it reaches the ra

Minchanka [31]

Answer:

Speed of the current is 3 miles per hour.

Step-by-step explanation:

Raft and a motor boat starts traveling downstream at the same time from point B towards point A downstream.

Let the speed of the motorboat = v miles per hour

and speed of the current = u miles per hour

Motorboat travels downstream with the speed = (v + u) miles per hour

Distance traveled by motorboat in 1 hour (Distance AC) = (v + u)×1 miles

Now motorboat turns around and reaches point B where it meats the raft against the river stream.

Speed of motorboat against the current = (v - u) miles per hour

Let it takes to reach at point B = t hours

Distance covered AB = (v - u)t miles

From the figure attached,

Distance BC = AC - AB

6 = (v + u) - (v - u)t ------(1)

Now raft covered the distance 6 miles in (1 + t) hours with the speed = u miles per hour.

Equation will be u(t + 1) = 6 --------(2)

Now by equating both the equations

(v + u) - (v - u)t = u(t + 1)

v + u - vt + ut = ut + u

v - vt = 0

vt = v

t = 1 hours

From equation (2)

u(1 + 1) = 6

2u = 6

u = 3 miles per hour.

Therefore, speed of the current will be 3 miles per hour.

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Answer:

-1

Step-by-step explanation:

1. 3x-1

2. -5+-3

3.7-8

4.=-1

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A radio station chose 1000 at random within it's broadcast area. it sent a survey to be filled out that asked questions about ra

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The answer is a because it is saying the corrext answer to the question

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