A random sample of 8 cigarettes of a certain brand has an average nicotine content of 4.2 milligrams and a standard deviation of 1.4 milligrams. Is this in line with the manufacturer’s claim that the average nicotine content does not exceed 3.5 mg? Use a 0.01 level of significance and assume the distribution of nicotine contents to be normal.

0.0793 > 0.01, which means that we have a result in line with the manufacturer's claim.

Step-by-step explanation:

Manufacturer’s claim that the average nicotine content does not exceed 3.5 mg

This means that the null hypothesis is given by:

And the alternate hypothesis is:

The test statistic is:

In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.

3.5 is tested at the null hypothesis

This means that

A random sample of 8 cigarettes of a certain brand has an average nicotine content of 4.2 milligrams and a standard deviation of 1.4 milligrams.

This means that

Value of the z-statistic:

Pvalue of the test:

We are testing if the mean is higher than 3.5.

The sample mean found is of 4.2, and we have to find the probability of finding a sample mean at least as large as this, which is 1 subtracted by the pvalue of z = 1.41.

z = 1.41 has a pvalue of 0.9207

1 - 0.9207 = 0.0793

0.0793 > 0.01, which means that we have a result in line with the manufacturer's claim.

answer:   14

step-by-step explanation:          

pu you stank bo u ugly bo lol

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.