Question

(1 point) The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) μ and standard deviation σ=0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a random sample of 15 cigarettes of this brand. The sample yields an average of 1.4 mg of nicotine. Conduct a test using a significance level of α=0.01. (a) The test value (b) The critical value, z* = (c) The final conclusion is A. There is not sufficient evidence to show that the ad is misleading. B. The nicotine content is probably higher than advertised.

Answer 1

Solution :

= 1.4

= 1.5

= 0.1

n = 15

This is the right tailed test .

The null and alternative hypothesis is ,

H₀ :   = 1.4

H_a : > 1.4

a) Test statistic = z

= ( - ) / / n

= (1.5 - 1.4) /0.1 / 15

= 3.873

P(z > 3.873) = 1 - P(z < 3.873 ) = 0.0001

P-value = 0.0001

b) = 0.01

The critical value for a right-tailed test is z_c​=1.75.  

p = 0.0001< 0.04, it is concluded that the null hypothesis is rejected.

There is enough evidence to claim that the population mean μ is greater than 1.4, at the 0.04 significance level.

C) B. The nicotine content is probably higher than advertised.