The one in the hundred thousands place blake the vaule of the middle one

Historically, these bolts have an average thickness of 10.1 mm. A recent random sample of 10 bolts yielded these thicknesses:

Akimi4 [234]

Answer:

a) Sample mean = 9.99

Sample standard deviation = 0.3348

b) -1.0389

c) 0.1631

Step-by-step explanation:

We are given the following in the question:

9.7, 9.9, 10.3, 10.1, 10.5, 9.4, 9.9, 10.1, 9.7, 10.3

a) sample mean and standard deviation

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{99.9}{10} = 9.99$

Sum of squares of differences = 1.009

$S.D = \sqrt{\dfrac{1.009}{9}} = 0.3348$

b) observed value of the t-statistic

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{9.99 - 10.1}{\frac{0.3348}{\sqrt{10}} } = -1.0389$

c) probability of these statistics (or worse) if the true mean were 10.1 mm

Degree of freedom = n - 1 = 9

Calculating the value from the table

$P(x < 9.99) = 0.1631$

0.1631 is the the probability of these statistics (or worse) if the true mean were 10.1 mm

8 0

3 years ago

Find the percent of the number. 20% of 60 is

exis [7]

"of" means to multiply.
20% * 60
= 0.2 * 60 (since % means divide by 100)
= 12

3 0

3 years ago

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Faced with rising fax costs, a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail inste

Vladimir [108]

Answer:

The value of t test statistics is 5.9028.

We conclude that the true mean is greater than 10 at the .01 level of significance.

Step-by-step explanation:

We are given that a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail instead.

The firm examined 35 randomly chosen fax transmissions during the next year, yielding a sample mean of 14.44 with a standard deviation of 4.45 pages.

Let $\mu$ = true mean transmission of pages.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 10 pages {means that the true mean is smaller than or equal to 10}

Alternate Hypothesis, $H_A$ : $\mu$ > 10 pages {means that the true mean is greater than 10}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 14.44 pages

s = sample standard deviation = 4.45 pages

n = sample of fax transmissions = 35

So, test statistics = $\frac{14.44-10}{\frac{4.45}{\sqrt{35} } }$ ~ $t_3_4$

= 5.9028

(a) The value of t test statistics is 5.9028.

Now, at 0.01 significance level the z table gives critical values of 2.441 at 34 degree of freedom for right-tailed test.

Since our test statistics is more than the critical values of z as 5.9028 > 2.441, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true mean is greater than 10.

(b) Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_3_4$ > 5.9028) = Less than 0.05% {using t table)

7 0

3 years ago

ramone has 5 difficult questions left to answer on a multiple choice test. Each question has 3 choices. For the first 2 of these

olchik [2.2K]

1. a b c
2. a b c
3. a b c
4. a b c
5. a b c

then she eliminated 1 choice in 1 and 2, say as follows

1. b c
2. a b
3. a b c
4. a b c
5. a b c

Probability of answering correctly the first 2, and at least 2 or the remaining 3 is
P(answering 1,2 and exactly 2 of 3.4.or 5.)+P(answering 1,2 and also 3,4,5 )

P(answering 1,2 and exactly 2 of 3.4.or 5.)=
P(1,2,3,4 correct, 5 wrong)+P(1,2,3,5 correct, 4 wrong)+P(1,2,4,5 correct, 3 wrong)
also P(1,2,3,4 c, 5w)=P(1,2,3,5 c 4w)=P(1,2,4,5 c 3w )
so
P(answering 1,2 and exactly 2 of 3.4.or 5.)=3*P(1,2,3,4)=3*1/2*1/2*1/3*1/3*2/3=1/4*2/9=2/36=1/18

note: P(1 correct)=1/2
P(2 correct)=1/2
P(3 correct)=1/3
P(4 correct)=1/3
P(5 wrong) = 2/3

P(answering 1,2 and also 3,4,5 )=1/2*1/2*1/3*1/3*1/3=1/108

Ans: P= 1/18+1/108=(6+1)/108=7/108

5 0

4 years ago

Find 80 x 4 by rewriting 80 as 10 x 8 and using the associative property.

marusya05 [52]

Idk I think it is 32p

7 0

3 years ago

Read 2 more answers