Mathematics, 06.03.2020 08:08, jcarver2494

The nicotine content in cigarettes of a certain brand is known to be right-skewed with mean (in milligrams) μ and a known standard deviation σ = 0.17. The brand advertises that the mean nicotine content of its cigarettes is 1.5, but measurements on a random sample of 100 cigarettes of this brand give a sample mean of x ¯ = 1.53.
Is this evidence that the mean nicotine content is actually higher than advertised? To answer this, test the hypotheses H₀: μ = 1.5, Hₐ: μ > 1.5 at the 0.05 level of significance.
What do you decide?

Answers: 3

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answered: serenityarts123

There is enough evidence to support the claim the mean nicotine content is actually higher than advertised.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 1.5

Sample mean, $\bar{x}$ = 1.53

Sample size, n = 100

Alpha, α = 0.05

Population standard deviation, σ = 0.17

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 1.5\\H_A: \mu 1.5$

We use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{1.53 - 1.5}{\frac{0.17}{\sqrt{100}} } = 1.7647$

Now, we calculate the p-value from the standard normal table.

P-value = 0.038807

Since the p-value is less than the significance level we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim the mean nicotine content is actually higher than advertised.

answered: lsrgb

Yes, the mean nicotine content is actually higher than advertised.

Step-by-step explanation:

We are given that the nicotine content in cigarettes of a certain brand is known to be right-skewed with mean (in milligrams) μ and a known standard deviation σ = 0.17.

Also, the brand advertises that the mean nicotine content of its cigarettes, $\mu$ is 1.5, but measurements on a random sample of 100 cigarettes of this brand give a sample mean of x bar = 1.53.

Null Hypothesis, H_0 : $\mu$ = 1.5

Alternate Hypothesis, H_1 : $\mu$ > 1.5

The test statistics used here will be;

T.S. = $\frac{xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

Here, n = sample size = 100

So, test statistics = $\frac{1.53 - 1.5}{\frac{0.17}{\sqrt{100} } }$ = 1.765

Now, at 0.05 level of significance, the standard z table gives critical value of 1.6449. Since our test statistics is more than the critical value as 1.765 > 1.6449 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region.

Therefore, we conclude that the mean nicotine content is actually higher than advertised.

answered: Guest

What’s the question here? what have you been learning about?

answered: Guest

it is true

step-by-step explanation:

all the combinations add up i used calculator

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The nicotine content in cigarettes of a certain brand is known to be right-skewed with mean (in mill...