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LenaWriter [7]
3 years ago
15

A company manufactures 2,000 units of its flagship product in a day. The quality control department takes a random sample of 40

units to test for quality. The product is put through a wear-and-tear test to determine the number of days it could last. If the product has a lifespan of less than 26 days, it is considered defective. The table gives the sample data that a quality control manager collected.
39 31 38 40 29
32 33 39 35 32
32 27 30 31 27
30 29 34 36 25
30 32 38 35 40
29 32 31 26 26
32 26 30 40 32
39 37 25 29 34

Which sample size would you use to get the best point estimate?
A.
90
B.
70
C.
150
D.
500
E.
280
Mathematics
2 answers:
shepuryov [24]3 years ago
5 0

Answer:

B

Step-by-step explanation:

The recommended sample size n for point estimates is:

n = NX / (N + X - 1)

where N is the population size, and X is defined as:

X = Z² p (1 - p) / E²

where Z is the critical value, p is the sample proportion, and E is the margin of error.

Assume a confidence level of level of 95% and a margin of error of 5%.

α = 0.05, Z(α/2) = 1.96

E = 0.05

Of the 40 units tested, 2 had lifespans less than 26 days.  So the proportion is:

p = 2/40 = 0.05

Therefore:

X = (1.96)² (0.05) (1 - 0.05) / (0.05)²

X = 73

Given N = 2000:

n = (2000) (73) / (2000 + 73 - 1)

n = 70.45

Rounding, the recommended sample size is 70.

Nezavi [6.7K]3 years ago
5 0

Answer:

B

Step-by-step explanation:

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Find the slope of the line.
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I'll show you two methods.

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6 0
3 years ago
Find the value of r so the line that passes through (-5,2) and (3,r) has a slope of -1/2
soldier1979 [14.2K]

The value of r so the line that passes through (-5,2) and (3,r) has a slope of -1/2 is -2

<u>Solution:</u>

Given that line is passing through point (-5, 2) and (3, r)

Slope of the line is \frac{-1}{2}

Need to determine value of r.

Slope of a line passing through point \left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right)  is given by following formula:

\text { Slope } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}  --- eqn 1

\text { In our case } x_{1}=-5, y_{1}=2, x_{2}=3, y_{2}=\mathrm{r} \text { and } m=-\frac{1}{2}

On substituting the given value in (1) we get

\begin{array}{l}{-\frac{1}{2}=\frac{r-2}{3-(-5)}} \\\\ {\text { Solving the above expression to get value of } r} \\\\ {=>-\frac{1}{2}=\frac{r-2}{3+5}} \\\\ {=>-8=\frac{r-2}{3+5}} \\\\ {=>-8=2(r-2)} \\\\ {=>-8=2 r-4} \\\\ {=>2 r=-8+4} \\\\ {=>2 r=-4} \\\\ {=>r=\frac{-4}{2}=-2}\end{array}

Hence the value of "r" is -2

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