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Nadusha1986 [10]

3 years ago

6

An automobile manufacturer claims that its car has a 35.935.9 miles/gallon (MPG) rating. An independent testing firm has been co

ntracted to test the MPG for this car since it is believed that the car has an incorrect manufacturer's MPG rating. After testing 220220 cars, they found a mean MPG of 35.635.6. Assume the standard deviation is known to be 2.22.2. A level of significance of 0.020.02 will be used. Make a decision to reject or fail to reject the null hypothesis.

1 answer:

Serjik [45]3 years ago

5 0

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean.

For the given claim , we have

Null hypothesis : $H_0: \mu=35.9$

Alternative hypothesis : $H_a: \mu\neq35.9$

Since alternative hypothesis is two-tailed , so the test is a two-tailed test.

Given : Sample size : n=220 ;

Sample mean: $\overline{x}=35.6$ ;

Standard deviation: $s=2.2$

Test statistic for population mean:

$z=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

$z=\dfrac{35.6-35.9}{\dfrac{2.2}{\sqrt{220}}}\approx-2.02$

By using the standard normal distribution table of z , we have

P-value ( two tailed test ) : $2P(Z>|z|)=2(1-P(Z$

$=2(1-P(z$

Since , the P-value is greater than the significance level of $\alpha=0.02$ , it means we do not have sufficient evidence to reject the null hypothesis.

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Answer:

i) 0.1% probability that if the coin is actually fair, we reach a false conclusion.

ii) 0.05% probability that if the coin is actually unfair, we reach a false conclusion

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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:

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

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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

Fair coin:

Comes up heads 50% of the time, so $p = 0.5$

1000 trials, so $n = 1000$

So

$E(X) = np = 1000*0.5 = 500$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.5*0.5} = 15.81$

If the coin lands on heads 550 or more times, then we shall conclude that it is a biased coin.

(i) If the coin is actually fair, what is the probability that we shall reach a false conclusion?

This is the probability that the number of heads is 550 or more, so this is 1 subtracted by the pvalue of Z when X = 549.

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

$Z = \frac{549 - 500}{15.81}$

$Z = 3.1$

$Z = 3.1$ has a pvalue of 0.9990

1 - 0.9990 = 0.001

0.1% probability that if the coin is actually fair, we reach a false conclusion.

(ii) If the coin is actually unfair, what is the probability that we shall reach a false conclusion?

Comes up heads 60% of the time, so $p = 0.6$

1000 trials, so $n = 1000$

So

$E(X) = np = 1000*0.6 = 600$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.6*0.4} = 15.49$

If the coin lands on less than 550 times(that is, 549 or less), then we shall conclude that it is a biased coin.

So this is the pvalue of Z when X = 549.

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

$Z = \frac{549 - 600}{15.49}$

$Z = -3.29$

$Z = -3.29$ has a pvalue of 0.0005

0.05% probability that if the coin is actually unfair, we reach a false conclusion

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