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andreyandreev [35.5K]
3 years ago
13

A random sample of 23 items is drawn from a population whose standard deviation is unknown. The sample mean isx⎯ ⎯ x¯ = 840 and

the sample standard deviation is s = 15. Use Appendix D to find the values of Student's t. (a) Construct an interval estimate of μ with 98% confidence. (Round your answers to 3 decimal places.) The 98% confidence interval is from to
Mathematics
1 answer:
gayaneshka [121]3 years ago
4 0

Answer:

(832.156, \ 847.844)

Step-by-step explanation:

Given data :

Sample standard deviation, s = 15

Sample mean, \overline x = 840

n = 23

a). 98% confidence interval

$\overline x \pm t_{(n-1, \alpha /2)}. \frac{s}{\sqrt{n}}$

$E= t_{( n-1, \alpha/2 )} \frac{s}{\sqrt n}}

$t_{(n-1 , \alpha/2)} \frac{s}{\sqrt n}$

$t_{(n-1, a\pha/2)}=t_{(22,0.01)} = 2.508$

∴ $E = 2.508 \times \frac{15}{\sqrt{23}}$

  $E = 7.844$

So, 98% CI is

$(\overline x - E, \overline x + E)$

(840-7.844 ,  \ 840+7.844)

(832.156, \ 847.844)

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Answer:

If the warranty limits are at 41.12 months, only 10 percent of the HDTVs need repairs at the manufacturer's expense.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

For a new HDTV the mean number of months until repairs are needed is 36.84 with a standard deviation of 3.34 months. This means that \mu = 36.84, \sigma = 3.34.

Where should the warranty limits be set so that only 10 percent of the HDTVs need repairs at the manufacturer's expense?

This is the value of X when Z has a pvalue of 0.90.

Looking at the z-table, we get that this is between Z = 1.28 and Z = 1.29, so we use Z = 1.285.

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{X - 36.84}{3.34}

X - 36.84 = 1.28*3.34

X = 41.12

If the warranty limits are at 41.12 months, only 10 percent of the HDTVs need repairs at the manufacturer's expense.

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