Answer to Question #293660 in Statistics and Probability for DSDASD

Question

A cigarette manufacturer wishes to test the claim that the variance of the nicotine content of

its cigarettes is 0.644. Nicotine content is measured in milligrams and assume that it is

normally distributed. A sample of 20 cigarettes has a standard deviation of 1.00 milligram.

At α = 0.05, is there enough evidence to reject the manufacturer’s claim?

Answer 1

We first define the null hypothesis and alternative hypothesis as below

H₀: variance of nicotine content is equal to 0.644 ( null hypothesis)

H₁: variance of nicotine content is not equal to 0.644 ( alternative hypothesis )

we then compute the test statistic where we use the chi-square test = ( ( n -1) / ( 0.644/1²) )

= ( ( 20 -1 ) / (0.644) )

= ( ( 19) / ( 0.644) ) = 29.50310559

Taking a 0.05 level of significance, the critical value at ( n - 1) degrees of freedom which is ( 20 - 1) = 19 degrees of freedom is 8.906525 to the left and 32.852327 to the right.

Thus we fail to reject the null hypothesis as the test value ( 29.50310559) falls within the non critical region.

Hence, we may conclude that there is no enough evidence to reject the manufacturers claim that the variance of the nicotine content is 0.644