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KiRa [710]
3 years ago
5

A random sample of 60 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact te

st, and on 16 of these helmets some damage was observed. (a) Find a 95% two-sided confidence interval on the true proportion of helmets of this type that would show damage from this test. Round your answers to 3 decimal places. ≤p≤ (b) Using the point estimate of p obtained from the preliminary sample of 60 helmets, how many helmets must be tested to be 95% confident that the error in estimating the true value of p is less than 0.02? n= (c) How large must the sample be if we wish to be at least 95% confident that the error in estimating p is less than 0.02, regardless of the true value of p?
Mathematics
1 answer:
Karolina [17]3 years ago
5 0

Answer:

a)0.154<p<0.48[

b)n≅ 6.57

c)n=2401

Step-by-step explanation:

n = 60

Sample proportion = p = 16/60 = 0.266

a) Find a 95% two-sided confidence interval on the true proportion of helmets

90 % confidence interval fro population is

p +- z0.05/2 * \sqrt{p(1-p)/n}

0.154

hence population proportion lies in the confidence interval

b) helmets must be tested to be 95% confident

point of estimates i s- 0.266

Margin of error = 0.02

ME=z0.05/2*\sqrt{p(1-p)n}

after putting values we get

n = 6.56

n≅ 6.57

c)large must the sample be

The sample is calculted as

n=(z0.025/E)^{2} 1/4

n=(1.96/0.02)^{2} 1/4

n=2401

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3 0
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An electrical rm manufactures light bulbs that have a life span that is approximately normally distributed. The population stand
stira [4]

Answer:

The 't' test statistic = 1.46 < 1.69

The test of hypothesis is H 0 :μ = 800 is accepted

A sample of 30 bulbs are found came from average µ= 800

Step-by-step explanation:

Step 1:-

Given population of mean μ = 800

given size of small sample n =30

sample standard deviation 'S' = 45

Mean value of the sample χ = 788

Null hypothesis H_{0} =µ  =800

alternative hypothesis  H_{1} = µ ≠ 800

<u>Step 2</u>:-

The 't' test statistic t =  \frac{x-μ}{\frac{sig}{\sqrt{n} } }

                             t = \frac{788-800}{\frac{45}{\sqrt{30} } }

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<u>Step 3</u>:-

The degrees of freedom γ = n-1 = 30-1 =29

From "t" value from table at 0.05 level of significance ( t = 1.69)

The calculated value t = 1.4607 < 1.69

Therefore The null hypothesis H_{0} ' is accepted.

<u>conclusion</u>:-

A sample of 30 bulbs are came found from average µ= 800

7 0
4 years ago
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