Question

A researcher would like to determine whether a new tax on cigarettes has had any effect on people’s behavior. During the year before the tax was imposed, stores located in rest areas on the state thruway reported selling an average of µ = 410 packs per day with σ = 60. The distribution of daily sales was approximately normal. For a sample of n = 9 days following the new tax, the researcher found an average of M = 386 packs per day for the same stores.

a. Is the sample mean sufficient to conclude that there was a significant change in cigarette purchases after the new tax. Use a two-tailed test with α = .05.
b. If the population standard deviation was σ = 30, is the result sufficient to conclude that there is a significant difference?
c. Explain why the two tests lead to different outcomes.

Answer 1

a)
`p_v =2*P(z<-1.2)=0.230`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the population mean is NOT significant different from 410 at 5% of significance.

b)
`p_v =2*P(z<-2.4)=0.016`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 410 at 5% of significance.

c) The expplanation why we have different outcomes is because for part a we use a higher standard error compared to part b. So we have enough evidence on part b to reject the null hypothesis that we no have significant difference from 410.

Explanation:

1) Part a

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=386` represent the sample mean

`\sigma=60` represent the population standard deviation

n=9 represent the sample selected

`\alpha=0.05` significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if we have significant difference on the mean, the system of hypothesis would be:

Null hypothesis:
`\mu = 410`

Alternative hypothesis:
`\mu \\eq 410`

If we analyze the size for the sample is < 30 but we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(386-410)/((60)/(√(9)))=-1.2`

P-value

Since is a two side test the p value would be:

`p_v =2*P(z<-1.2)=0.230`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the population mean is NOT significant different from 410 at 5% of significance.

2) Part b

State the null and alternative hypotheses.

The system of hypothesis not changes:

Null hypothesis:
`\mu = 410`

Alternative hypothesis:
`\mu \\eq 410`

Same statistic:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

Calculate the statistic

But now the population deviation changes
`\sigma=30`. We can replace in formula (1) the info given like this:

`z=(386-410)/((30)/(√(9)))=-2.4`

P-value

Since is a two side test the p value would be:

`p_v =2*P(z<-2.4)=0.016`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 410 at 5% of significance.