23. A certain type of gasoline is supposed to have a mean octane rating of at le

Question

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 randomly selected samples, giving ratings of 89.7 89.4 87.4 89.888.5 ssume that the distribution of ratings is in fact known to be normal, and that the true standard deviation is known to be 0,8 a) For the alternative hypothesis that the mean octane rating is less than 90, compute the p- b) For significance level 0.01, what do we conclude. If we made an error, is it type I or II? c) For a significance level 0.01 test and for a sampling of size 5, what is the probability of concluding that the mean is really 90 or more, when in fact the true mean is 89? d) If we are allowed to increase the sample size, what minimum size do we need so that the probability calculated in part c is 0.10 or less?

Explanation / Answer

(a) Here sample mean x = 88.96

True standard deviation = 0.8

standard error of sample mean se0 = / sqrt(n) = 0.8/ sqrt(5) = 0.3578

Test statistic

Z = (x - 90)/se0 = (88.96 - 90)/ 0.3578 = - 2.9067

P - value = Pr(Z < -2.9067) = 0.00183

(b) Here significance level 0.01, we conclude that p - value is less than the significance level so we reject the null hpothesis and can say that mean octane rating is less than 90. We made an error, it is a type I error.

(c) Here true mean = 89

Standard error of mean = 0.3578

Here the sample mean above which we will say that mean is reallly 90 or more.

x > 90 - Z0.01 se0

x > 90 - 2.326 * 0.3578

x > 89.1678

so probability of concluding that the mean is really 90 or more is

Pr(x >= 89.1678 ; 89; 0.3578)

Z = (89.1678 - 89)/ 0.3578 = 0.4690

Pr(x >= 89.1678 ; 89; 0.3578) = 1 - Pr(Z > 0.4690) = 1 - 0.6805 = 0.3195

(d) Here if the probability calculated in part c is 0.10 or less.

so if sample size is n

the sample mean x above which we will say that probability of soncluding that the mean is really 90 or more.

x > 90 - 2.326 * / n

x > 90 - 2.326 * 0.8/n

x > 90 - 1.8608/n

so as given,

Pr(x > 90 - 1.8608/n ; 89 ;  0.8/n ) = 0.1

Z = 1.645

(90 - 1.8608/n - 89)/ ( 0.8/n ) = 1.645

1 = 3.1768/ n

n = 10.0

or n = 10

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