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2 years ago

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A scientist wishes to estimate the mean breaking strength of a certain type of steel rod within 3.0 psi with 95% confidence. Bas

ed on experience with a similar type of steel, she believes the standard deviation is approximately 15.0 psi in individual breaking strengths. How many steel rods should she test?

1 answer:

yaroslaw [1]2 years ago

6 0

Answer:97 rods

Step-by-step explanation:

The number of steel rods for 95% confidence level

$n=\left ( \frac{Z_\frac{\alpha }{2}\times \hat{\sigma }}{E} \right )^2$

$Z_\frac{\alpha }{2}=1.96$

E=3

$\hat{\sigma }=15$

Substituting values

$n=\left ( \frac{1.96\times 15}{3}\right )^2$

$n=96.04\approx 97$

Thus 97 rods is tested

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At a certain company, loan agents are paid based on the number of loans they close in a day. Based on company records, the numbe

Ilya [14]

Answer:

C. The mean daily salary is greater than $350 per day

Step-by-step explanation:

The computation is shown below:

Y = a + bX

where,

Y = money made by a random selected

a = $150

b = $50

X = number of loan

Now

E(x) = (1 × 0.05) + (2 × 0.10) + (3 × 0.22) + (4 × 0.30) + (5 × 0.18) + (6 × 0.12) + (7 × 0.03)

= 3.94

Now

E(y) = $150 + ($50 × 3.94)

= $347

hence, the option C is not correct

4 0

1 year ago

Read 2 more answers

A fisherman is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fis

maks197457 [2]

Answer:

p = 0.293

Conclusion:

"There is insufficient evidence to state that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake."

Step-by-step explanation:

------------R/trout--O/trout---bass--catfish---total

GVL - - - 105 - - - 27 - - - - - 35--- 24 - - - 191

EL - - - - 135------- 48-------- 67 - - 43 - - - 293

Total - - 240 - - - 75 - - - - - 102--- 67 - - - 484

Note : GVL = GREEN VALLEY LAKE ; EL = ECHO LAKE.

H0: The distribution of fish caught in green valley is the same as that caught in Echo lake.

H1: The distribution of fish caught in Green Valley lake is different from that caught in Echo valley lake.

Using the chi-squared test statistic calculator ;

χ2 = 3.7269

p-value = 0.2925.

p-value = 0.293

Conclusion:

"There is insufficient evidence to state that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake."

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1 year ago

What are the x- and y- coordinates of point E, which partitions the directed line segment from J to K into a ratio of 1:4? (–13,

Lana71 [14]

I added a screenshot of the complete question.

<u><em>Answer:</em></u>

(-7, -1)

<u><em>Explanation:</em></u>

<u>The formulas given to calculated the x and y coordinates are as follows:</u>

$x = (\frac{m}{m+n})(x_{2}-x_{1}) + x_{1}\\\\y = (\frac{m}{m+n})(y_{2}-y_{1}) + y_{1}$

<u>Let's define the variables used:</u>

(x₁ , y₁) are the coordinates of the first point while (x₂ , y₂) are the coordinates of the second point.

<u>We are given that the segment is directed from J to K, therefore:</u>

First point is J ..........> (x₁ , y₁) is (-15, -5)

Second point is K ....> (x₂ , y₂) is (25, 15)

m and n defined the portion of the partitioned segment (JE : EK). It is given that this ratio is 1:4. <u>Therefore:</u>

m = 1 and n = 4

<u>Finally, let's substitute with these variables in the equations as follows:</u>

$x = (\frac{1}{1+4})(25-(-15)) + (-15) = -7\\\\y = (\frac{1}{1+4})(15-(-5)) + (-5) = -1$

Based on the above, the coordinates of point E are (-7, -1)

Hope this helps :)

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2 years ago

Read 2 more answers

The Bay Area Online Institute (BAOI) has set a guideline of 60 hours for the time it should take to complete an independent stud

yarga [219]

Answer:

At the 5% level, BAOI can infer that the average time to complete does not exceeds 60 hours.

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 60 \ hr$

The sample size is $n = 16$

The sample mean is $\= x = 68 \ hr$

The standard deviation is $\sigma = 20 \ hr$

The null hypothesis is $H_o : \mu = 60$

The alternative $H_a : \mu > 60$

Here we would assume the level of significance of this test to be

$\alpha = 5\% = 0.05$

Next we will obtain the critical value of the level of significance from the normal distribution table, the value is $Z_{0.05} = 1.645$

Generally the test statistics is mathematically represented as

$t = \frac{ \= x - \mu}{ \frac{ \sigma }{\sqrt{n} } }$

substituting values

$t = \frac{ 68 - 60 }{ \frac{ 20 }{\sqrt{16} } }$

$t = 1.6$

Looking at the value of t and $Z_{\alpha }$ we see that $t< Z_{\alpha }$ hence we fail to reject the null hypothesis

This means that there no sufficient evidence to conclude that it takes more than 60 hours to complete the course

So

At the 5% level, BAOI can infer that the average time to complete does not exceeds 60 hours.

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1 year ago

Ephemeral services corporation (esco) knows that nine other companies besides esco are bidding for a $900,000 government contrac

EastWind [94]

Answer:

c. $100,000

Step-by-step explanation:

Calculation of the expected net profit of Ephemeral services corporation

Since we are been told that 9 other companies besides esco are as well bidding for the $900,000 government contract, it means we have to find the expected net profit by dividing 1 by 9×$900,000 .Thus ESCO can only expect to cover its sunk cost.

Hence ,

E(X) = (1/9) × $900,000

E(X)=0.111111111×$900,000

E(X)= $100,000

Therefore the expected net profit would be $100,000

4 0

1 year ago