Suppose that when a signal having value mu is transmitted from location A the va

Question

Suppose that when a signal having value mu is transmitted from location A the value received at location B is normally distributed with mean mu and variance 4. That is, if mu is sent, then the value received is mu + N where N, representing noise, is normal with mean 0 and variance 4. To reduce error, suppose the same value is sent 9 times. The successive values received are 5, 8.5, 12, 15, 7, 9, 7.5, 6.5, and 10.5. a. What is the 95% confidence interval for mu? b. What is the 99% confidence interval for mu? c. Using your point estimate for mu, calculate the probability of the signal received exceeding the value of 10.

Explanation / Answer

Solution

Back-up Theory

Let X = value received at location B. Then, X ~ N(µ, 2).

100(1 – ) % confidence interval for µ when 2 is known is: {Xbar ± (/n)(Z/2)},…(1)

where

Xbar = sample mean,

= population standard deviation,

n = sample size and

Z/2 = upper (/2) % point of Standard Normal Distribution.

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(2)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(3)

Now, to work out solution,

We are given

n = 9, 2 = 4 or = 2,

From the given data, Xbar = 81/9 = 9

Part (a)

95% CI => = 5% or 0.05. Upper 2.5% point of N(0, 1) = 1.96 [using Excel Function]

[vide (1) under Back-up Theory],

95% CI for µ = {9 ± (2/9)(1.96)} = 9 ± 1.307 ANSWER

Part (b)

99% CI => = 1% or 0.01. Upper 0.5% point of N(0, 1) = 2.575 [using Excel Function]

[vide (1) under Back-up Theory],

99% CI for µ = {9 ± (2/9)(2.575)} = 9 ± 1.717 ANSWER

Part (c)

Estimate of µ = Xbar = 9.

We want probability signal received exceeds the value 10, i.e., P(X > 10)

[vide (3) under Back-up Theory], P(X > 10) = P[Z > {(10 - 9)/2}] = P(Z > ½)

= 0.3085 ANSWER [using Excel Function]

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